Here is the AMC 12A 2007.

EDIT: Removed for copyright purposes until someone confirms it is OK to post the test.  If not, I'll send it to KSU and let them post it on their website.

[quote="paladin8"]I don't think we're allowed to reproduce the test in a PDF format (remember 4everwise's AIME PDF's?).[/quote]

[quote="MAA"]
The publication, reproduction or communication of the problems of the problem or solutions of the AMC 12 during the period when students are eligible to participate seriously jeopardizes the integrity of the results.  Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during the period is a violation of the competition rules.  After the contest period, permission to make copies of individual problems in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and copies bear the copyright notice.

Copyright © 2007, The Mathematical Association of America
[/quote]
(yes i typed all of that)  So... I think he's okay if he adds the copyright line.

I wonder if the policy for AIME's is the same (it makes sense that it should be, but I'm not sure). Then 4everwise could just add a little copyright note to this AIME's and we could use them :!:.

[quote="ccy"][quote="paladin8"]I don't think we're allowed to reproduce the test in a PDF format (remember 4everwise's AIME PDF's?).[/quote]

[quote="MAA"]
The publication, reproduction or communication of the problems of the problem or solutions of the AMC 12 during the period when students are eligible to participate seriously jeopardizes the integrity of the results.  Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during the period is a violation of the competition rules.  After the contest period, permission to make copies of [b]individual problems[/b] in paper or electronic form including posting on web-pages for educational use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and copies bear the copyright notice.

Copyright © 2007, The Mathematical Association of America
[/quote]
(yes i typed all of that)  So... I think he's okay if he adds the copyright line.[/quote]
I don't think he's okay regardless. These aren't individual problems.

OK, so I'm taking it down.  Perhaps if I add the copyright notice, I can put it back up, but I will wait for confirmation.  The reason I was pretty sure this was OK was the KSU [url=http://www.math.ksu.edu/main/handbook/ProblemSets/AMC12Problems]website[/url] .

The KSU website is violating the AMC's copyright.

Please do not post PDFs of the entire contest.

[quote="DPatrick"]The KSU website is violating the AMC's copyright.

[/quote]

Are you sure about this?

Prof Tom Muenzenberger is the State Director of the American Mathematics Contests for Kansas.

I'm certain that the tests are copyrighted.

Whether KSU has permission from the AMC to post them, I have no way of knowing.  They might be claiming "fair use" (as they are a nonprofit educational institution), but I don't personally believe that it is fair use (then again, I'm not a lawyer).  And there's no indication of the MAA's copyright notice on the KSU site.


geometry
rectangle
similar triangles
AMC

You are given the rectangle ABCD such that the lengths of sides AB and CD are 5, and the lengths of sides BC and DA are 2.4=12/5. We construct a right triangle with hypotenuse AB such that the last vertex of the right triangle lies on CD. Find the sum of the legs of this right triangle.

Ah I'll do this.

The legs in the right triangle I will label a and b.  We're looking for the sum of the two, a + b.

a:^2: + b:^2: = 25
2ab = [ABCD] * 2 = 24
a:^2: + 2ab + b:^2: = 49
(a + b) = 7      *negative roots are unnecessary*

By the way, 2ab is twice the area of rectangle ABCD because if you see, the right triangle can have its area expressed as ab/2.  It will also be half of the area of ABCD because the area can also be displayed as (AB)(BC)/2.

The sum of the legs is 7 units.

yupyup. most of the peeps get caught up in similar triangles and end up doing a lot of algebra instead of just going whoa we can square this blurgle


function
trigonometry
calculus
real analysis
algebra
functional equation

Suppose $ f$ and $ g$ are continuous nonconstant functions such that $ f(x \minus{} y) \equal{} f(x)f(y) \plus{} g(x)g(y)$. Find all possible ordered pairs $ (f, g)$.

I've found a solution, but I... abused the continuity fact. I ended up using a bit of real analysis to finish this problem off completely, so I'm wondering if there's a completely elementary solution.

I got this from Problem-Solving Strategies; the author started off with some progress and asked the reader to finish the rest, though in my opinion, he left it at the hardest part.

[hide="What Engel mentioned"]$ f(x)^2 \plus{} g(x)^2 \equal{} 1$, $ f(0) \equal{} 1$, and $ f( \minus{} x) \equal{} x$. He then said something about $ f(2x) \equal{} f(x)^2 \plus{} g(x)g( \minus{} x)$, but I didn't see how much that would help considering we don't know that $ g$ is odd. [/hide]

[hide="What I did"]I observed as well that $ f(x)^2 \plus{} g(x)^2 \equal{} 1$, $ f(0) \equal{} 1$, and $ f( \minus{} x) \equal{} x$. I realized that if we show that $ g$ is odd, by substituting $ \minus{} y$ for $ y$ in our original equation and adding that to our original, we get D'alembert's functional equation, with possible solutions $ \cos ax$ and $ \cosh ax$ (since $ f$ is continuous.) $ f(x)^2 \plus{} g(x)^2 \equal{} 1$ quickly rules out $ \cosh ax$, which is unbounded, and expanding the trig formula and fixing $ y$ very quickly gets you $ g(x) \equal{} \sin ax$.

Now, for showing that $ g$ is odd. From $ f(x)^2 \plus{} g(x)^2 \equal{} 1 \equal{} f( \minus{} x)^2 \plus{} g( \minus{} x)^2$, we get that $ g( \minus{} x) \equal{} \pm g(x)$ for all $ x$. Also, since $ f(0) \equal{} 1$, $ g(0) \equal{} 0$.

Let $ S \equal{} \{x : g(x) \equal{} 0\}$. We observe that $ S$ is a group (just putting all 4 properties together, I'm not going to use any group theory...), since $ g(x) \equal{} 0 \to g( \minus{} x) \equal{} \pm 0 \equal{} 0$, and if $ g(x) \equal{} g(y) \equal{} 0$, then $ f(x \minus{} y)^2 \equal{} f(x)^2 f(y)^2 \equal{} 1 \to g(x \minus{} y)^2 \equal{} 0 \to g(x \minus{} y) \equal{} 0$; setting $ y' \equal{} \minus{} y$, we get $ g(x \plus{} y') \equal{} 0$, as desired.

We now have two cases: for any positive real number $ \epsilon$, $ g$ has a zero in the interval $ ( \minus{} \epsilon, 0)$ or $ g$ does not. In the first case, observe that it follows that $ g$ must have infinitely many zeroes in the given interval. For simplicity, we may choose $ \epsilon$ s.t. $ g(\epsilon) \equal{} 0$. Let $ T \equal{} S \cap [ \minus{} \epsilon, \epsilon]$. If we can show $ S$ is dense, then since $ g$ is continuous over a dense set that agrees with a constant function, $ g$ itself must be a constant.

For any $ \delta > 0$, we have some $ r$ and $ s$ in $ T$ such that $ |r \minus{} s| < \delta$. It is easy to see that otherwise, $ T$ must be finite. Therefore, for any $ \delta > 0$, there exists some $ t$ in $ T$ such that $ |t| < \delta$. Since $ T$ is closed under addition and subtraction, we may add/subtract as many arbitrarily small real numbers as we like, so it follows easily that $ T$ is dense, as desired. (I am aware this isn't rigorous, but I'm slightly too lazy to get that done now.)

Therefore, it is impossible for there to exist any interval $ ( \minus{} \epsilon, 0)$ where $ g$ has infinitely many roots. We must have that there exists some interval $ ( \minus{} a, \minus{} b)$ where $ g$ has no roots, $ a > b > 0$. Assume now for the sake of contradiction that there is some $ x \in ( \minus{} a, \minus{} b)$ with $ g(x) \neq 0$ such that $ g( \minus{} x) \equal{} x$. It follows by that that $ g( \minus{} y) \equal{} g(y)$ for all $ y$ in $ ( \minus{} a, \minus{} b)$, since $ g( \minus{} y) \equal{} \pm g(y)$, and $ \minus{} g(y)$ and $ g(x)$ have different signs, so by the intermediate value theorem, there must be some $ c$ in $ ( \minus{} a, \minus{} b)$ such that $ g(c) \equal{} 0$, a contradiction.

But then, for any $ x$ in $ ( \minus{} a, \minus{} b)$, set $ y \equal{} \minus{} x$; we get that $ f(2x) \equal{} f(x)^2 \plus{} g(x)^2 \equal{} 1$! Thus, $ f(x) \equal{} 1$ for all $ x$ in $ ( \minus{} a/2, \minus{} b/2)$. Furthermore, $ g$ must be constant in this interval as well. We can easily use the original functional equation to show that since $ f$ and $ g$ are continuous (I don't even think that's necessary), $ f$ and $ g$ must be constant everywhere, a contradiction.

Thus, ina ny interval $ ( \minus{} a, \minus{} b)$ with $ a > b > 0$, and $ g( \minus{} a) \equal{} g( \minus{} b) \equal{} 0$, $ g( \minus{} x) \equal{} \minus{} g(x)$ in that interval. This covers all the negative intervals, so we are done.

As you can see, it was a bit... long-winded, and it used some analysis. There's probably an easier way, and I'm really curious as to what it is. Any response would be appreciated. [/hide]

[hide]If all you wanted to was show $ g$ was odd, consider the given equation for $ (x,y)$ and $ (\minus{}x,\minus{}y)$. This implies $ g(x)g(y) \equal{} g(\minus{}x)g(\minus{}y)$ (*). $ x \equal{} y$ shows $ g(x) \pm g(\minus{}x)$. (*) also shows if we find $ g(x) \equal{} \minus{}g(\minus{}x)$ (or $ g(x) \equal{} g(\minus{}x)$) for one $ x$ with $ g(x) \neq 0$, then it holds for every $ x$.

So suppose $ g(x) \equal{} g(\minus{}x)$. Then $ f(2x) \equal{} f(x)^2 \plus{} g(x)^2$, so $ f(2x) \equal{} 1$, contradicting $ f$ nonconstant. And so we get $ g$ odd.[/hide]

Darn. I now vow to myself never to use calculus on functional equations ever so I don't miss little jewels like this.

Thank you Palmer.

Where are the solutions $f(x)=cosax, g(x)=sinax$?  :o

OP said that if he could prove (which he could) that $g$ is odd, then $P(x,y)+P(x,-y)$ gives the D'alembert's functional equation, which, when combining with non-constant and ontinuous constraint, give solutions $f(x)\equiv cos(ax)$ and $f(x)\equiv cosh(ax)$, the latter ruled out as it's unbounded.

Yes, I want to see a solution where is proved this for continuous fonction only, because I know a proof only if $f$ is differentiable.

It's easy to prove by induction that any non-constant continuous function $f$ that satisfied the D'Alembert functional equation will also satisfied the equation $f(0)=1,$ $f(-x)=f(x),$ and $f(ny)=T_n(f(y))$ for all $n\in\mathbb{N},$ where $T_n$ is the Chebyshev polynomial of degree $n.$ In particular, $f(2x)=2f(x)^2-1.$

Since $f$ is continuous, there is an $\epsilon>0$ such that $f(x)>0$ for all $x$ such that $|x|<\epsilon.$ Choose a random $k$ in the interval $(0,\epsilon)$ and let $a=\arccos(f(k))/k$ or $a=arccosh(f(k))/k$, whichever is defined. (From now on, we will assume that $ak=\arccos(f(k))$ as the another case can be handled similary.)

Since $f(x)>0$ for all $0<x<k<\epsilon,$ we can see that $f\left(\frac{k}{2^n}\right)=\cos\left(\frac{ak}{2^n}\right)$ for all $n\in\mathbb{N}\cup\{0\},$ this implies that $f(kq)=\cos(akq)$ for all rational numbers $q$ of the form $\frac{m}{2^b}$ for $b\in\mathbb{N}\cup\{0\}$ and $m\in\mathbb{Z}.$ Continuity thus gives $f(x)=\cos(ax)$ for all $x\in\mathbb{R}.$

[quote=TuZo]Thank you! So this equation it is equivalent with the D'Alembert functional equation. It is true?[/quote]

Yes, it is. And it's quite easy to prove. The main steps are proving f is even and g is odd

Yes, exatly what I used to pove! Thank you.


Show this multiplication in the simplest form:

\[ \left( {1 \plus{} \frac{1}{a}} \right)\left( {1 \plus{} \frac{1}{{a^2 }}} \right)\left( {1 \plus{} \frac{1}{{a^4 }}} \right) \cdots \left( {1 \plus{} \frac{1}{{a^{2^{100} } }}} \right)\]

Thanks "tonypr"  :)  I was missing note (2)  :oops:

Could I do it ?

$ 1\plus{}(\frac{1}{a})^{2e}$

e = the number that you like to see.

$ 1\plus{}(\frac{1}{a})^{2e}$

$ 1\plus{}(\frac{1}{a}^{2\times2})$

$ 1\plus{}\frac{1}{a^{4}}$


prove that if  (p-1)! +1 is divisible by p then p is a prime.

hello,here you will find the proof
http://www.maths.nott.ac.uk/personal/pmznd/g12nth/n13.pdf
Sonnhard.


I'm prepping for a proficiency exam in the fall and I came across these two problems in the set of notes I borrowed:

[18(7x-3)]/(2x+1)=[200sqrt(2x+1)]/[3sqrt(7x-3)]

and

(a+x)^2/3 +4(a-x)^2/3=5(a^2-x^2)^1/3

I'm not sure how to solve them.

Any help/input is greatly appreciated.

1)
$ \frac{18(7x\minus{}3)}{2x\plus{}1}\equal{}\frac{200\sqrt{2x\plus{}1}}{3\sqrt{7x\minus{}3}}\Rightarrow 54(7x\minus{}3)^{\frac{3}{2}}\equal{}200(2x\plus{}1)^{\frac{3}{2}}\Rightarrow\left(\frac{7x\minus{}3}{2x\plus{}1}\right)^{\frac{2}{3}}\equal{}\frac{100}{27}$
$ \Rightarrow\frac{7x\minus{}3}{2x\plus{}1}\equal{}\frac{10\sqrt[3]{10}}{9}\Rightarrow x\equal{}\frac{27\plus{}10\sqrt[3]{10}}{63\minus{}20\sqrt[3]{10}}$

2) Let $ (a\plus{}x)^{\frac{1}{3}}\equal{}u, \ (a\minus{}x)^{\frac{1}{3}}\equal{}v$

Then $ u^2\plus{}4v^2\minus{}5uv\equal{}0$

Divide by $ v^2$:

$ \left(\frac{u}{v}\right)^2\minus{}5\frac{u}{v}\plus{}4\equal{}0\Rightarrow\frac{u}{v}\equal{}1$ or $ \frac{u}{v}\equal{}4$.

If $ \frac{u}{v}\equal{}1\Rightarrow x\equal{}0$

If $ \frac{u}{v}\equal{}4\Rightarrow\left(\frac{a\plus{}x}{a\minus{}x}\right)^{\frac{1}{3}}\equal{}4\Rightarrow\frac{a\plus{}x}{a\minus{}x}\equal{}64\Rightarrow x\equal{}\frac{63a}{65}$


geometry
3D geometry
combinatorics unsolved
combinatorics

Show that a cube can be dissected into $ n$ cubes for any positive integer $ n>54$.

Check this link for minimum $ n\ge 48$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=307546[/url]


[quote="keta"]1. a. How many photons (wavelength = 620 nm) must be absorbed to melt a 2.0-kg block of ice at 0 degrees C into water at 0 degrees C?

b. On the average, how many H20 molecules does one photon convert from the ice phase to the water phase?

2. An incident X-ray photon of wavelength 0.2800 nm is scattered from an electron that is initially at rest. The photon is scattered at an angle of  :theta:  = 180.0 degrees and has a wavelength of 0.2849 nm. Use the conservation of linear momentum to find the momentum gained by the electron.

Can anyone provide some help on these 2 problems? Especially for the first one, do I have to set E = hf = mc (delta T)?[/quote]

keta,

I had thought I answered this yesterday, but apparently not. I shall try again.

1. When water, or any other chemical compound, undergoes a phase change, e.g. solid-to-liquid or liquid-to-solid, its temperature does not change during the transition. One has to wait for the entire sample to undergo the transition before its temperature is free to change again.  As such, the regions of the Energy vs. Temperature graph where the slope is zero, are called latent heats; the solid/liquid transition, the latent heat of fusion ($L_f$), and the liquid/gas transition, the latent heat of vaporization ($L_v$).  [b]Latent heats are dependent only upon the mass of a given sample, not its temperature.[/b]

The latent heat of fusion for water is: $L_f = 6.013 kJ/mol$.

The trick here is to equate the photon energy with the latent heat of fusion.

2. Problem #2 is an example of  [url=http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/comptint.html]Compton scattering[/url].

I hope this helps.


Let $m\geq n\geq1$ be real number. Prove or disprove that: for any $a,b,c>0$ then we must have
\[ \frac{(a^m+b^m)(b^m+c^m)(c^m+a^m)}{(abc)^m}\geq \frac{(a^n+b^n)(b^n+c^n)(c^n+a^n)}{(abc)^n} \]

How about $1\geq m\geq n\geq0$ ?

By multiplying it out you get:
\[ \sum_{sym}a^{2m+n}b^{m+n}c^n\geq\sum_{sym}a^{2n+m}b^{n+m}c^m \]
Which is true by Muirhead (sorry, whats the correct spelling of this name?), cause $2m+n,m+n,n$ majorize $2n+m,n+m,m$ :)


According to the Wiki, the words of the week for March 1-8 are "Gravity" and "AMC 8." However, the forum says the words are still "Momentum, and CTY."


[quote="nust2007"]3 books on maths 4 books on physics , 6 books on chemistry, ,, how many sets of books can be choosen so that each set has 2 books of different subjects[/quote]

[hide]he has to choose 1 from two categories so $ 3\cdot 4+4\cdot 6+3\cdot 6=54$.[/hide]

[quote="nust2007"]3 books on maths 4 books on physics , 6 books on chemistry, ,, how many sets of books can be choosen so that each set has 2 books of different subjects[/quote]
quick question are the physics (and the other subject) books dictinct from eachother?

I got a different answer for this..
I assumed they were different because it said the books were "on math" so they can be different from eachother.


[quote="stvs_f"]$ f: Z \to R$  is a function such that :
\[ f(n) = \{\begin{array}{cc}n - 1, & \mbox{if}n > 100 \\
f(f(n + 1)) , & \mbox{ if } n \leq 100 \end{array}
\]
prove : $ \forall 100 \geq n$ f(n)=91[/quote]

i think you forget somthing,
if we take $ f(n)=n-1: \ \forall n\in Z$ we have $ f(n) = \{\begin{array}{cc}n - 1, & \mbox{if}n > 100 \\
f(f(n + 1)) , & \mbox{ if } n \leq 100 \end{array}$

$ f(n)\neq 91$ for $ n<80$

[quote="stvs_f"]why $ f(n) \neq 91$[/quote]
$ g(n)\equal{}n\minus{}1$ for all $ n\in N$ is solution,because $ g(n)\equal{}gog(n\plus{}1)$
but for exemple $ g(79)\equal{}78\neq 91$


Does string theory touted as TOE offer a possible explanation of what happens beyond the event horizon?

If not what may?

hep-th archive (arxiv.org)   has papers/proposals by Samir Mathur on this.

Thank's  for the location of the article.


what do people think of geometry revisited?

It seems more like an olympiad preperation book. Also, the material covered doesn't seem to come up too often in competitions. Although, if you just want to learn more geometry, not necessarily for competitions, it's a great book.

Challenging problems in geometry is a good book if you're looking for problems. It doesn't cover much material by itself however. I haven't looked at in a while; it's probably somewhere in my room, mixed in with all the books and paper on the floor :P

Oh, and my opinion doesn't matter much, since I haven't actually worked through most, or indeed, much, of these books, since I thought that geometry revisited was too challenging for me at the time (a few months back) and I guess I was too lazy to do CPIG. Maybe I should actually get back to work on them....

[quote="Jongao"]Are there any other books besides Volume 1/2 that can help me with AIME-level geometry?[/quote]

Introduction to Geometry?  AMC-easy AIME level, I think.  Some MC level too.

For me, I'm just going to do the online geometry course that AoPS offers on Feb 4.
This is a good preparation for AIME / USAMO

( I hope I make USAMO... )

Hmm... if you want a geometry book that focuses on OLYMPIADS, there is a stickied link to Kiran Kedlaya's Geometry Unbound in the Olympiad Geometry Thereoms forum.


Ruth picks pairs of numbers from the set $ S$ which is : $ S \equal{} [1,2,3,4,5,6,7,8,9,10]$ and writes in her notebook the biggest element of the pair that she chose. After picking every possible pairs (without repeating any), Ruth added all the numbers she wrote. What is the answer that she got?

I need a help on where to start with this problem...

Try to construct pairs in an orderly manner. You'll see a pattern.

It's weird. I keep trying to do it, yet, my answer is not in the options...

I keep getting $ 325$

and the answer options are:

$ (a)$ $ 250$      $ (b)$ $ 330$    $ (c)$ $ 350$     $ (d)$ $ 430$    $ (e)$ $ 450$

Anyone can do this please?

Good solution FantasyLover. xcenario, this is what I meant by constructing pairs in an orderly manner and counting them.

-rts2007

This is almost the exact same problem as AIME 2003, Problem 3.


Compute the number of positive integers that have the same number of digits expressed in base 3 and base 5.

For a given number of digits, the lower bound is dictated by powers of 5 and the upper bound is dictated by powers of 3.

One digit: $[5^{0},3^{1}) \to \{ 1,2 \}$
Two digits: $[5^{1},3^{2}) \to \{ 5,6,7,8 \}$
Three digits: $[5^{2},3^{3}) \to \{ 25,26 \}$
Four digits: $5^{3}>3^{4}$, so nothing with 4 digits or above.

The intersection set, $\{ 1,2,5,6,7,8,25,26 \}$, has $\boxed{8}$ elements.

This was number 1 on the ARML Individual Round this year.


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

I'm just playing with ideas here, I don't have an answer yet...

It almost looks like a trig substitution problem...
$ \ln(x) \equal{} \frac{\pi}{2}\tan(\theta)$
$ x \equal{} e^{\frac{\pi}{2}\tan(\theta)}$
$ dx \equal{} e^{\frac{\pi}{2}\tan(\theta)}\frac{\pi}{2}\sec^{2}(\theta)d\theta$
Substituting and simplifying:
$ \int\frac{\frac{\pi}{2}\tan(\theta) \cos(e^{\frac{\pi}{2}\tan(\theta)}) \plus{}\frac{\pi}{2}\sin(e^{\frac{\pi}{2}\tan(\theta)})}{(\frac{\pi}{2}\tan(\theta))^{2}\plus{}\frac{\pi^{2}}{4}}e^{\frac{\pi}{2}\tan(\theta)}\frac{\pi}{2}\sec^{2}(\theta)d\theta$
$ \equal{} \int (\tan(\theta) \cos(e^{\frac{\pi}{2}\tan(\theta)}) \plus{}\sin(e^{\frac{\pi}{2}\tan(\theta)})e^{\frac{\pi}{2}\tan(\theta)})d\theta$
Because:
$ tan^{2}(\theta)\plus{}1 \equal{} sec^{2}(\theta)$

That will get rid of the denominator, but you can see things turn ugly in the cos(x) and sin(x).

I don't think that there exist solution without complex analysis. :(


modular arithmetic
function
arithmetic sequence
number theory
prime numbers

$ \Delta a_{n}: \equal{} d\implies a_{n}\equal{} a_{0}\plus{}(n\minus{}1)d\implies a_{(a_{0}\plus{}1)}\equal{} (1\plus{}d) a_{0}\equiv 0\pmod{1\plus{}d}$

Thus, if $ d\in\mathbb{N}$, $ a$ cannot be a sequence of primes only.

[quote="mdk"]Prove that, if the terms of an infinite arithmetic progression of natural numbers are not all equal, they cannot all be primes.[/quote]

[hide="eh."]
Let $ p$ be a prime and let it be the first term of our progression, with $ d$ being the common difference. Then the $ p\plus{}1$st terms of the sequence is $ p\plus{}pd\equal{}p(d\plus{}1)$ which is composite since $ d$ is not 0 (which would make all terms equal)
[/hide]

Interestingly enough, there do exist arbitrarily long arithmetic progressions of primes, so here is an interesting question:

What constraints on $ d$ are necessary for an arithmetic progression of primes of length at least $ L$?

[quote="Lazarus"]$ d > A(69,L)$ should be satisfactory.[/quote]

He said necessary, not sufficient.

[quote="Lazarus"]$ d > A(69,L)$ should be satisfactory.[/quote]

What does the $ A(69,L)$ mean?

He's probably talking about the [url=http://mathworld.wolfram.com/AckermannFunction.html]Ackermann Function[/url].  In any case, the problem I proposed is not terribly difficult; try working it out for small examples, say $ L \equal{} 3, 4, 5$.

[hide="solution"]Say that the first term is a, and the common difference is d.

If a≠1, then the a+1st term is a(1+d), which is composite.
If a=1, it works for d=1. If d=k+1, where k≥1, then the k-1st term is k^2, which works. So if the terms of an infinite arithmetic progression of natural numbers are not all equal, they cannot all be primes.[/hide]

[quote="mdk"]Prove that, if the terms of an infinite arithmetic progression of natural numbers are not all equal, they cannot all be primes.[/quote]
[hide="Solution"]If $ a$ and $ d$ are respectively the first term and the common difference, then the $ (a\plus{}1)$-th term is $ a\plus{}(a\plus{}1\minus{}1)d \equal{} a(d\plus{}1)$, which cannot be prime for $ a > 1$. On the other hand, if $ a \equal{} 1$ then the sequence is $ 1,d\plus{}1,2d\plus{}1,3d\plus{}1,\dots$, of which the $ (d\plus{}2)$-th term is $ (d\plus{}2)d\plus{}1 \equal{} (d\plus{}1)^{2}$, which is not prime either. Hence the proof is complete.[/hide]

[quote="t0rajir0u"]What constraints on $ d$ are necessary for an arithmetic progression of primes of length at least $ L$?[/quote]

$ \gcd(d,L) > 1$. Is this correct? :maybe:

[hide="Proof"] It is well known (and can be proven easily) that if $ (d,L) \equal{} 1$ then the numbers $ d,2d,3d,\dots, (L\minus{}1)d$ are all non-congruent modulo $ L$. Thus $ a,a\plus{}d,a\plus{}2d,\dots, a\plus{}(L\minus{}1)d$ are also non-congruent modulo $ L$, which implies one of them must be divisible by $ L$. So in order to get $ L$ consecutive prime numbers in the A.P. we must have $ (d,L)\neq 1$.[/hide]

Yes, that is a necessary condition, but you can do better.  Think about it.  An arithmetic progression of primes of length $ L$ also contains a progression of length $ L\minus{}1$, $ L\minus{}2$... ;)

I had to think about this for some time and finally came to this. I'm not sure if this is what you wanted though.

[hide]So from my work and your hint we must have $ (k,d) > 1$, for $ k \equal{} 1,2,\dots,L$. This clearly implies that $ p_{i}\mid L$ for all primes $ p_{1},p_{2},\dots p_{n}\le d$. And thus $ p \equal{} p_{1}p_{2}\cdots p_{n}\mid L$.[/hide]

Actually I found this long ago, and I did some problems with it too. For example about one year or so ago I found the six primes $ 7,37,67,97,127,157$ in A.P. using this fact. But it seems that I forgot it so easily. :blush:

EDIT: Yes t0rajir0u is right. Sorry about that. :oops:

[quote="nayel"]I had to think about this for some time and finally came to this. I'm not sure if this is what you wanted though.

[hide]So from my work and your hint we must have $ (k,L) > 1$, for $ k \equal{} 1,2,\dots,d$. This clearly implies that $ p_{i}\mid L$ for all primes $ p_{1},p_{2},\dots p_{n}\le d$. And thus $ p \equal{} p_{1}p_{2}\cdots p_{n}\mid L$.[/hide]

Actually I found this long ago, and I did some problems with it too. For example about one year or so ago I found the six primes $ 7,37,67,97,127,157$ in A.P. using this fact. But it seems that I forgot it so easily. :blush:[/quote]

You seem to have your variables switched... I think you mean $ (d, k) > 1, k \equal{} 1, 2, ... L$.  :)


Using induction, prove a_1 <= a_n. We also know that a_n<=a_(n+1).  I am not so familiar with induction so anything would help. Thanks

Base of induction: $ a_1\le a_2$.
The inductive step: assuming $ a_1\le a_{n\minus{}1}$ (the induction hypothesis) and using the inequality $ a_{n\minus{}1}\le a_{n}$, we obtain $ a_1\le a_n$.

[url=http://en.wikipedia.org/wiki/Mathematical_induction]Mathematical induction[/url]


LaTeX
linear algebra
matrix
conics
ellipse
algorithm
function

Hi
at first i want to say that i am very happy that i have found this forum! now to my question: i am 14 and im very interested in programming and math. two weeks ago i read "the da vinci code" and so i found the fibonacci numbers. since then im fascinated in them. so i made a small php programm to generate them, but php isnt very good for math and so i want to know which language you would recommend me.
flo

Matlab is a very nice language and environment for math.

Among applied mathematicians, Matlab and C++ are both very popular.

What about LaTeX? (I think it counts as a language...)

Fortran is meant for math, too, but it's *ahem* [i]slightly[/i] outdated. *ahem*

of course! It has a very pure mathematical structure.

I think $\LaTeX$ is just a way to type stuff (more specifically, math) up, so it'd be more like a font than a language.
:Read above statement
:If
:you disagree
:Then
:write a program in $\LaTeX$ that computes the $n^{th}$ term of the Fibonacc sequence using Binet's formula (or a similar/explicit one)
:Else
:My point is made and I win!
:End
 :D

I also recommend using Matlab. It depends on what you want to do. I am thinking from the statistician and machine learning perspective where you need a lot of matrix computations to do all kind of fancy transformations of fancy datasets. But Matlab is also strange because you have to get rid of thinking in loops but have to try to implement iterations by means of matrix operations which takes some time to get used to if coming from a different world as me, more or less having applied languages as Ruby, C, C# and Java. Loops are very inefficient in Matlab.
As said before in Matlab you basically get a lof of matrix stuff for free as computation of eigenvalues/eigenvectors, Cholesky Factorization, Singular Value Decomposition and many more. But on the other hand it is like programming in the 70s. Matlab does not have any convincing object-oriented approach. C++ and those others languages are mentioned above are rather suited for traditional application development. My advice is to start learning Java. The best book to learn Java, in my opinion, can be found [url=http://www.amazon.co.uk/exec/obidos/ASIN/0131249339/202-7873790-0543068]here.[/url] It treats the beginner gently. It does not teach you hacking but to write good programs in the way of how to structure your programs, when to re-factor classes, when it is reasonable to apply inheritance etc. Or in other words don't forget about software engineering. And the most important point is that it is easy to get started. You don't need programming environments (integrated development environment) as Eclipse or anything like that which is confusing for beginners due to its broad range of settings. So you are starting with the BlueJ environment.
And you should not forget about the fact that Matlab is not a free software, e.g. you have to buy it first. But, of course, you could use its free alternative Octave.
Basically you should know different kind of languages, e.g. knowing Java it is easy to learn C. If you want to take a look at a functional programming language I recommend OCaml.
Hope this helps.

[quote]I think $\LaTeX$ is just a way to type stuff (more specifically, math) up, so it'd be more like a font than a language [/quote]

FYI: Languages which are merely used to display certain texts/fonts (like LaTeX and HTML) are called markup languages whereas languages that are used to do computations (like C++ and Matlab) are called programming langauges.  :)

[quote="Aunt Sally"][quote]It does not teach you hacking but to write good programs in the way of how to structure your programs[/quote]

What do you mean by "hacking" here?

A lot of people talk about "hacking" at AoPS but I'm never quite sure what that is.[/quote]

Hacking = writing really quick and poorly structured and commented and not that re-usable code. Also known as the sort of programs that everyone writes for Topcoder (if this sentence confuses you, ignore it).

To answer original question, C++ = best all-purpose language to use most likely.

I like C, it is a great balance between speed and ease of use. But I guess C++ kind of makes it obselete.  And Matlab?  That's just very slow BASIC like programming, even though its computational speeds are quick, its structures like loops are quite slow and inefficient.

[quote="schaf88"]thank you very much for the answers. i posted this question in an other forum and they recommended haskell. what do you say about that language?[/quote]

Yeah Haskell is a good choice as well. It is a functional programming language that can be learned quite easily. Many universities start with a mixture of Java and Haskell. At first they teach Haskell and then they swap over to Java during the second part of the semester when algorithms come into play. If you want to learn some Haskell look [url=http://www.doc.ic.ac.uk/~ajf/Teaching/Haskell2005.html]here.[/url] I don't recommend C, C++ or C# when starting to learn a programming language. You can learn it later.

[url=http://mitpress.mit.edu/sicp/full-text/book/book.html]"Structure and Interpretation of Computer Programs"[/url] is a must reading when starting to program.

@pieterminate: I guess you never really needed to work out fancy matrix computations. Thus you just see the disadvantages which I already mentioned. I promise that nobody ever wanted to do that in C or whatever language you like.

[quote="schaf88"]thx. do you think oo-languages (like ruby) are better for maths or functional languages? I think ruby is qite easier to understand.[/quote]

Well Ruby or Haskell are not quite mathematical programming languages. Well a program in a functional programming language is a (nested) function but of course you can do imperative stuff by means of side-effects in those languages.
Of course, you can program recursion easily as you intended to do. But for the sake of math I am experimenting around a lot of time and you quickly want to have something plotted (displayed) and that is really easy in Matlab, e.g. you want to have a look at the growth of the Fibonacci numbers etc. Ruby is a pure object oriented languages and basically it is advancement compared to Java. You can program in Ruby like Java. But that is not quite the purpose you intend to follow. Ruby offers so many things that are very tedious in Java. And with the event of the framework "Ruby on Rails" Ruby gets more and more widely spread. But if I were you in the course of some self-study I would try learning Java.

Maybe anyone wants to comment on his experiences with Mathematica or Maple. If you are doing statistics I recommend R.


Hi guys.

Unfortunately it seems that phpBB does not (yet) support daylight savings time settings, so for now, all users in GMT 0,+1,+2,+3 areas, are urged to change their time zones with +1, that is in GTM +1,+2,+3,+4 respectively.

to be more exact, all users from Romania, who want the correct time displayed on stuff, should go to Profile -> Date Zone -> and select +3 GMT.

this will be in effect untill the next daylight saving time change occurs.

[quote="RC-7th"]Why is it that people in North America don't have that problem?[/quote]

We don't have that problem here in the U.S. because we're the ones who live the anachronistic timezone nation.


In how many ways can you divide 6 people into two teams, with each team having at least one member?

If the number of teams were 7, C(7,1)+C(7,2)+c(7,3)=63 provides the correct answer. I'm not sure why that works and can someone help me figure how to do the problem with six people?

[quote="rofler"]For each player, you can pick him for the first team, or for the second team. That gives 2^6=64 possible combinations of teams, eliminating the two possiblilties of all the players on one team or the other, gves 62 possibilites.[/quote]

you overcounted that two teams are indistinguishable..
it is $6c1+6c2+6c3=41$

because if theres 1 people on a team, then there will be 5 people on the other team, so it is 6 choose 1 to ensure that happen

if there are 2 people onm a team, there will be 4 people on the other team. 6 choose 2...

etc..

The series goes individuals-ways
2-1
3-3
4-7
5-15
6-31
7-63
I think (2^n-2)/2 gets you the right answer each time.
Like (2^6-2)/2=31.
The C(7,1)+C(7,2)+C(7,3) works for all odd number of individuals such as C(5,1)+C(5,2)=15. I'm not sure why C(6,1)+C(6,2)+C(6,3)=41 doesn't work. That's what I'm wondering about. Thanks!


how would an 8th grader do at the camp, like which track would i be on?

It depends, there were people going into 9th grade in pretty much all tracks, don't worry about it too much, once you're there, if your track is too easy/hard then they'll let you change.

[quote="CatalystOfNostalgia"]Tracks don't necessarily depend on ability, mostly interests.[/quote]
They depend on ability somewhat too though, I mean, track 1 Number Theory (Combinatorial Nullstellensatz, root flipping, Dirichlet series, Hensel's Lemma, Quadratic Reciprocity, primitive roots etc.) was a lot lot harder than track 2 Number Theory.

NOTE: The track system has been abolished, people keep making references to it, but it won't be the system for this year.
http://www.awesomemath.org/AMSP_dailysch.pdf

[quote="CatalystOfNostalgia"]Hence the word necessary. What I'm hinting at is that if you are interested in geometry, you shouldn't take track 1 just because it is harder than track 2, even if you can handle track 1.[/quote]
Except that since they don't have tracks, it doesn't matter anymore karel.


Find the number of 10-letter permutations comprising 4 $ a$'s, 3 $ b$'s and 3 $ c$'s such that no two adjacent letters are identical.

I'm not sure if this is right, but...

[hide]
Let's start with the sequence abc. There are 6 initial ways to arrange this. We will build adding an a, b, c, a, b, c, a in that order. There are two places to put the first a, 3 places to put the first b, 4 places to put the first c, three places to put the second a, etc... In the end, it will be

$ 6 \cdot 2 \cdot 3 \cdot 4 \cdot 3 \cdot 4 \cdot 5 \cdot 4 \equal{} 34560$

But we overcounted for the 4 a's, 3 b's, and 3 c's. Thus we divide by $ 4! \cdot 3! \cdot 3!$ to arrive at an answer of $ \boxed {40}$.

I'm not sure if that's right.[/hide]

My approach was a bit different.

[hide]Consider how we can arrange 3 c's and 3 b's such that no two identical letters are adjacent.  There are two, either bcbcbc or cbcbcb.  Now we have 4 A's to distribute so that they are not paired with each other.  There are seven possible places to place the a's either to the left or right of the letters b and c, and we want to find how we can choose four of them.  This becomes 7C4, or 35, and since there are two original sequences of b's and c's, we arrive at an answer of 70.[/hide]

I think I'm missing some solutions when bc and cb can be interchanged, but I am not sure.

Yongyi, the sequence is not necessarily abc; it could be aba.

azn, there are a lot more ways than u found; u could have cbbcbc, for example, and then add an a between the 2 adjacent b's.

[hide]
I used pretty complicated casework to get $ \boxed{248}$...
Probably wrong.[/hide]

That could be remedied by putting the a between the b and c like abc -> abac.

I DID say that you could arrange the abc 6 ways, right?

I'll try to add on to my answer of 70, which occurred when b and c were already separated.

Case 1:  The string has a pair of b's but no other pairs.

There are two strings of letters where this occurs, and since an a must divide the pair, there are only 6C3 strings now, or 2X20 more words.  We multiply this by two again because the same holds true for a pair of c's, resulting in 80 more words.

Case 2:  There is a pair of c's and a pair of b's in the string.

If we look at the sequence bcbc, there are two places each for the c and for the b to form two pairs of letters, and the same holds true for cbcb, so we have 8 new possibilities.  2 a's will be used breaking these pairs, so we have 5C2 places left for the others, resulting in 8x(5C2), or another 80 words.

Case 3: There is a triple of one letter and a pair of the other letter.

There are two ways to arrange this for one triple (the pair comes before or after it), so we have 2X2X4, or 16 more words.

Case 4:  There are two triples.

This occurs twice and every a must break a pair, resulting in 2 new words.

So the answer is 248.  Yay!

[hide="this has got to be wrong, probably missing something obvious"]Define $ p_{a,b,c}$ as the number of permutations with $ a$ as, $ b$ bs, and $ c$ cs. Adding a new letter to a string, there are $ a\plus{}b\plus{}c\minus{}2$ ways to place it, so $ p_{a,b,c}\equal{}(a\plus{}b\plus{}c\minus{}2)p_{a\minus{}1,b,c}$. Using this recursion, we have that the answer is $ 9!\cdot6$, since $ p_{1,1,1}\equal{}6$.[/hide]

A counterexample is $ p_{3, 1, 1}$. Clearly, there are only 2 ways to arrange aaabc, which are abaca and acaba. And $ p_{4, 1, 1} \equal{} 0$.


A trigonometric polynomial is nonnegative on R. Then it is of the form $ abs( P^2(cosx+isinx))$ for a polynomial P. Here abs is the modulus function.

It was also question 1a/ in part IV of written problem
Ecoles Normales Suprieures sujet commun Paris/Lyon 1999.


okay, so just a few days ago i began doing AoPS vol 1. yesterday i was doing the end of chapter problems for the proportions section. one of the problems (an AHSME from some year) brought up a question for me.
if some City A has 1000 dollars in its budget and has 100 citizens and each citizen gets the same amount of money, then to find the wealth of each citizen you would divide the budget by the # of citizens. this is obvious.
however, if you had what percentage City A has of the entire world's population and what percentage City A has of the entire world's wealth, could you do the same thing and divide the percentage of population by the percentage of wealth to get the wealth of each citizen. (this is assuming the problem doesn't give you any numerical values for either). I did this and got the right answer to the problem, but the solutions manual gave a different method. i just want to be sure if i got the answer correct by luck or you can use my method in such problems.

Sorry if the questions are a little confusing.
Thanks!

you're right. they probably did it more algebraically.

wait, shouldn't it be wealth/population? instead of population/wealth
i'm guessing that's a typo

you're right indianamath, it was a typo.

thanks!


all into base b:

we have $ \frac{log_bx}{2}+\frac{1}{2log_bx}=1$

           let $y=log_bx$

so   $  \frac{y}{2}+\frac{1}{2y}=1$

       $y^2-2y+1=0$
       $(y-1)^2=0$

hence   $log_bx=1$   $x=b$

ans is $D$

You can apply change of base formula, and have them all in log base b:

So:

$\frac{\log_{b}{x}}{\log_{b}{b^2}}+\frac{\log_{b}{b}}{\log_{b}{x^2}}=1$
$\frac{\log_{b}{x}}{2}+\frac{1}{2(\log_{b}{x})}=1$
From here, it's fairly easy to deduce that both of those fractions being added together must equal $1$ and since they are both $\frac{1}{2}$, then $x$ must equal $b$ to cancel the logs out. So $D$.

[quote="riddler"]any more log problems??? :?  :([/quote]

calm down and relax :sleep2:


vector
function
induction
floor function
algebra
functional equation
algebra unsolved

Find all $ f: R\rightarrow R$ such that
i)$ f(xy)=f(x)f(y)$ for all real x,y
ii)$ f(x+1)=f(x)+1$ for all real x

Obviosly $ f(1)=1\to f(n)=n \ \forall n\in Z \to f(r)=r \ \forall r\in Q$.
But there are infinetely many solutions in R.

Algebra R(Q) had infinetely many automorfisms. For example $ Q[\sqrt 2 ]$ had 2 automorfisms.

[quote="snowyowl"]If $ f$ is continuous, are there finite solution(s)?[/quote]

If $ f(x)$ is continuous in all $ \mathbb{R}$ (including $ 0$), it is rather well known that the functional equation $ f(xy)=f(x)f(y)$ has only solutions $ f(x)=0$ and $ f(x)=x^{a}$ for any real $ a\geq 0$.

If we add then the requirement $ f(x+1)=f(x)+1$, the only available continuous solution is $ f(x)=x$.

But I don't understand Rust's demo saying that it exist infinitely many non continuous solutions (at least we need axiom of Choice, and even with this axiom, I don't see).

Rust, I would be very interested in some examples.

[quote="Rust"][quote="pco"]
But I don't understand Rust's demo saying that it exist infinitely many non continuous solutions (at least we need axiom of Choice, and even with this axiom, I don't see).

Rust, I would be very interested in some examples.[/quote]
All automorfism $ f: R/Q\to R/Q$ ($ f|_{Q}=id$) is solution. By axiom of Choice we get infinetely many automorfisms $ R/Q\to R/Q$. construct[/quote]

I'm sorry but I still don't understand. May be a language problem
Assume (with axiom of Choice) that it exist a $ \mathbb{Q}$-vector space $ A$ such that $ \mathbb{R}=A+\mathbb{Q}$ and  $ A\cap\mathbb{Q}=\{0\}$.
Then every real $ x$ can be written in a unique way as $ x=q(x)+a(x)$ with $ q(x)\in\mathbb{Q}$  and $ a(x)\in A$

Let then $ f(x)=q(x)+3a(x)$

$ f(x)$ is an automorphism from the $ \mathbb{Q}$-vector space $ \mathbb{R}$ into itself with restriction of f in $ \mathbb{Q}$ being the identity function (as you seemed to say) but $ f(x)$ is absolutely not a solution of the problem, isn't it?

[quote="pco"]I'm sorry but I still don't understand. May be a language problem
Assume (with axiom of Choice) that it exist a $ \mathbb{Q}$-vector space $ A$ such that $ \mathbb{R}=A+\mathbb{Q}$ and  $ A\cap\mathbb{Q}=\{0\}$.
Then every real $ x$ can be written in a unique way as $ x=q(x)+a(x)$ with $ q(x)\in\mathbb{Q}$  and $ a(x)\in A$

Let then $ f(x)=q(x)+3a(x)$

$ f(x)$ is an automorphism from the $ \mathbb{Q}$-vector space $ \mathbb{R}$ into itself with restriction of f in $ \mathbb{Q}$ being the identity function (as you seemed to say) but $ f(x)$ is absolutely not a solution of the problem, isn't it?[/quote]
By transfinite induction we had $ Q=A_{0}\subset A_{1}\subset A_{2}\subset ...=R.$
Were $ A_{i+1}=A_{i}[x_{i+1}], \ x_{i+1}\not \in A_{i}$ and for infinite ordinal i:  $ A_{i}=\bigcup_{j<i}A_{j}$.
We construct automorfism $ f_{0}=id: Q\to Q$, $ f_{i}: A_{i}\to A_{i}$ if $ i$ is infinite ordinal, then $ f_{i}=\bigcup_{j<i}f_{j}$, else
automorfism $ f_{i}: A_{i}\to A_{i}$, suth that $ f_{i}|_{A_{i-1}}=f_{i-1}$. If $ x_{i}$ is trancendental, then work $ x_{i}\to x+a_{i-1}, \ a_{i-1}\in A_{i-1}$ if $ x_{i}$ is algebraic, then work all automorfism Galua.
See  book: Zariski C, Samuel P. "Commutative algebra".

[quote="harazi"]Unless I made a terrible mistake, in the next thread, your construction is definitely wrong.
 http://www.mathlinks.ro/viewtopic.php?t=19193[/quote]

very nice ! and RUST'assertion is completely wrong !

Here is a simpler demo adapted from yours using our stronger conditiopn $ f(xy)=f(x)f(y)$ :

We have clearly $ f(x)=x$ for $ x\in\mathbb{Q}$
We have $ f(x)\geq 0$ $ \forall x\geq 0$ (from $ f(x)=(f(\sqrt{x}))^{2}$)
So we have $ f(x)=\lfloor x\rfloor+f(\{x\})\geq \lfloor x\rfloor$ $ \forall x\in\mathbb{R}$

Suppose now we have an $ x$ such that $ f(x)<x$. Then $ nx-f(nx)=n(x-f(x))$ ($ n$ positive integer) could be as great as we want and we could have $ nx-f(nx)>1$ which is in contradiction with $ f(nx)\geq \lfloor nx\rfloor$. So $ f(x)\geq x$ $ \forall x\in\mathbb{R}$.

Then $ f(-1)=-1$ implies $ f(-x)=-f(x)$ and so $ f(-x)\geq-x$ implies $ -f(-x)\leq x$ and so $ f(x)\leq x$

So we have $ f(x)\geq x$ and $ f(x)\leq x$ and the only solution is $ f(x)=x$

Thanks Harazi for your demo.

Yes, I am wrong. There are unique automorfism $ R/Q\to R/Q$.
My method work only for C.


How do you find the motivation for such factorizations?


calculus
derivative
real analysis
real analysis unsolved

Find the sum, using derivative $ {C_n}^1\plus{}2{C_n}^2\plus{}3{C_n}^3\plus{}...\plus{}n{C_n}^n$

Let $ S(x)\equal{}\sum_{k\equal{}1}^n\binom{n}{k}x^k\equal{}(1\plus{}x)^n\minus{}1$.  Then, $ S'(x)\equal{}n(1\plus{}x)^{n\minus{}1}$.  We want $ S'(1)\equal{}n2^{n\minus{}1}$.  And the conclusion follows.


set theory
real analysis
real analysis theorems

PROVE :

If.$ x\in A\Longrightarrow y\in A$,THEN x=y ,where $ A\neq\emptyset$

where all the sets live.

it is not so hard to find it :)

This problem makes no sense.  As written, the set $ A \equal{} \{x, y \}$ satisfies the hypotheses of the question without satisfying the conclusion.

Maybe in math my favorite subject is logic, but not strickt one (justifying facts due to axioms etc.). What this problem asks to prove is very "logical". I mean that if y was not equal to x then how could we know that if x belongs to A then y belongs as well? :wink:

[quote="JBL"]I finally figured out what the question is supposed to mean:

Fix $ x$.  Suppose that $ \forall A$, $ x \in A$ implies $ y \in A$.  Then $ x \equal{} y$.

(Quantifiers are important!!)

Anyhow, since $ x \in A$ implies $ y \in A$ for all $ A$, we have $ y \in \{x\}$ and so $ y \equal{} x$.  (It shouldn't be hard to justify each of those steps directly in terms of ZF, but I don't think I really want to go to the effort.)[/quote]

 :)  :) [b]But toraiir0u thinks otherwise[/b]

In proving the forward implication of : (SxT)xU = Sx(TxU)$ \Longleftrightarrow S\equal{}\emptyset\vee T\equal{}\emptyset U\equal{}\emptyset$ i was forced to accept the above problem. you can check it up maybe i am wrong.

Anyway ,next step to check  your proof i suppose will be a formal proof justifying each step like you said

[quote="JBL"]t0r doesn't think otherwise.  He interpreted your problem literally (with no quantifier) and demonstrated (correctly) that it's false, while I gave a guess about what non-false statement you might have meant (which involved adding a quantifier) and gave a proof of it.[/quote]

Yes ,the right formula for the above problem is:

$ \forall A\forall x\forall y[(x\in A\Longrightarrow y\in A)\Longrightarrow x\equal{}y]$

But whenever i write this kind of formula in a proof people get annoyed.

Do you thing formalization will solve many mistakes in mathematics??

You don't have to write it that formally. There is a balance between over formalization and insufficient formalization. Your first post had insufficient formalization, your last post had too much. The way that JBL formulated the problem was clear, easy to understand, and sufficiently precise to comunicate the problem.

As a rule of thumb, just try to keep your mathematical writing as simple as possible without leaving it open to misinterpretation.

WHY ?

If i write the  problem that formally ,will mathematics loose the power of its axioms??

I REALLY do not understand why complete formalization is the red cloth for mathematicians ,while at the same time they try to formalize a problem as possible??

Is not that a schizophrenic behavior??

Besides   [b]only[/b] complete formalization can give a formal proof .

Taking out of a formula the logical symbols will prevent the formal proof show the rules of logic involved  in the proof

You [u]could[/u] write your mathematics that formally, and in some instances you might want to -- for instance, if you wanted your proof to be checked by a computerized proof-checking system (yes, such things do actually exist). However, over formalization and added symbols often make math harder to understand and slower to read. So if your goal is to communicate your ideas (which is often the case in mathematical writing), then you have to strike a balance with your intended audience in mind. If it is communicated well, then the audience should be able to implicitly fill in the needed formalities without being led to any possible misinterpretations of what you are trying to say.

[quote="τρικλινο"]Yes ,the right formula for the above problem is:
$ \forall A\forall x\forall y[(x\in A\Longrightarrow y\in A)\Longrightarrow x \equal{} y]$[/quote]

No, it's not.  This is the incorrect statement that t0r disproved.  (In English, it says, "For any set and any pair $ (x, y)$, we have that "$ x \in A$ implies $ y \in A$" if and only if $ x \equal{} y$," and this is demonstrated false by the triple $ A \equal{} \{1, 2\}$, $ x \equal{} 1$, $ y \equal{} 2$.)  The correct formalization of the statement is
\[ \forall x \forall y ((\forall A(x \in A \Longrightarrow x \in y)) \Longrightarrow x \equal{} y).
\]This adds nothing at all to my phrasing of the statement, and in fact my earlier statement could be "deformalized" by replacing the symbolic quantifier $ \forall$ with its word-equivalent with no loss.  (Indeed, it might have been better had I written it that way to begin with.)

[quote="τρικλινο"]But whenever i write this kind of formula in a proof people get annoyed.
Do you thing formalization will solve many mistakes in mathematics??[/quote]  Perhaps this is because you are sloppy and/or not very good at correctly formalizing what you mean.  Formalization is useful in theory because it shows that our intuitive notions of proof really can be mechanically checked.  In practice, it serves almost no purpose: it would take a huge amount of time to properly formalize even fairly simple statements in many branches of mathematics, and doing so largely obscures the actual meaning of what you're trying to say.  This is particularly try if you formalize improperly, as in this case.  Your original problem was wrong because you were imprecise, not because you were informal.

To the extent that I haven't actually rewritten what he wrote, I agree with gauss.

I'd say that i also agree that over-formalization helps noone and serves no purpose. Mathematics is all about using your mind. When you make a mathematical statement your only goal is to commute your thought. As far as this happens there is no need of further formalizations. Further formalizations are a matter of literature not math(at least thats my opinion).

[u]However[/u], i do not understand why the initial statement of this problem statement is wrong. My proof(scroll up) that uses just common sense seems to prove the statement(or do i miss something).

P.S. I'd like to give an example of the logic i like.

[u]Statement:[/u] All criminals are bad people and all bad people are on jail.

Then it follows using common sense that all criminals are in jail. One could say that i haven't referred to ZF etc. but what would by the point of doing so? Would this require thought or sharpen my mind :?:

[quote="filosofimenos"] i do not understand why the initial statement of this problem statement is wrong. My proof(scroll up) that uses just common sense seems to prove the statement(or do i miss something).[/quote]
This is a perfect demonstration of why it's problematic to rely on common sense.  The original problem doesn't have a quantifier on $ A$ and you interpreted it as having a quantifier on $ A$ in the correct location (the statement JBL gave).

Your "proof" was not a proof at all. It was a question. Your explaination was vacuous because it did not address the original issue. It simply restated the problem in contrapositive form.

This shows where logical formalization can have some advantages. It can be helpful as a teaching tool to train you to see logical fallacies and to teach you how to avoid bad and/or vacuous logical arguments.

Chill guys, don't shoot. :surrender: Yea, OK i restated the problem but i did it in order to reveal the obvious of the claim. That, i call proof.

We are all brothers :)  :cool:

[quote="gauss202"]You [u]could[/u] write your mathematics that formally, and in some instances you might want to -- for instance, if you wanted your proof to be checked by a computerized proof-checking system (yes, such things do actually exist).

[b]Why let the computer check my proof ,when i can write my own formal proof??[/b]

"Check" = "Verify that it is correct after it is written down."


MATHCOUNTS Putnam geometry AMC USA(J)MO USAMO AIME

Pretty much every community has an introduction topic, so I thought, why not start one here?
Just say your name, where you live, your school, or anything you want to say about yourself.

I'm Kriti, and I live in Cupertino.  I want to be on the school MathCounts team in 7th grade.

Hi guys! Name is Ricardo, and I am an undergraduate student at Stanford.

Yep -> getting ready for PUTNAM 09 and IMC 09 =)

It's kindof obvious where I live from my username...  :D  I'm going to ninth grade. I'm female and would rather not say my name here. Made USAMO but failed it.

Um...I'm going to 10th grade...from San Jose, CA...kind of failed USAMO, but that's okay...

I'm going to be a junior, and I go to a school in Menlo Park.

Hi, my name is (name censored), nice to meet you!

I'm going into 9th grade, near San Jose.

if you went to AMP 08, you probably heard of me, one way or another.

[quote="Nerd_of_the_Ages"][quote="Ubemaya"]I'm Fat.

I live in California (Bay Area, to be more specific... any more specificness would be stalkerish).[/quote]
Ditto.[/quote]

You are definitely not Fat.

ubey you should really get some more self-esteem

for example:
a) you are NOT THAT FAT
b) you are NOT BOTTOM FIVE IN RED MOP
c) you do NOT SUCK AT PROGRAMMING AS MUCH AS I DO

Hehe I've seen a good deal of people say he's not that fat, but sorry Albert I still think you're really fat :P

Hmmm so why did someone suddenly revive this 3 year old thread?
Circlesquared I'll put something into the name censored for you shall I?

I'm Andrew and I failed AIME so I didn't get to take USAMO (I think I would've gotten like around 14).

We'll have to see a picture of that.
Ubey is pretty fat, but I've seen worse so it's possible

After MOP and this summer, I'm fatter than anyone here can even imagine. The only person fatter in the whole Bay Area at this moment would be my sister. (At this moment, she is sitting beside me patting her belly).

[quote="hunter34"]Ubey is everyone fat in your family? It kinda be ironic if you parents were anorexic.[/quote]

No they'd just stuff the excess food in ubey's room or something.

Hello I have a story about Ubey and food.

Ubey and Toan were roommates. On one of the last few nights of MOP, he, Toan, Jeffrey S, and Charles X were in our room talking about some problem on our door given by Hamster. I hadn't finished my inequalities write-ups, and the teacher (Ali) was trying to get me to do them. Toan and Ubey kept trying to steal my granola bars, and succeeded. Then I made them open their door so I could search their room. I failed miserably to find it, and then said that they could each have one if they returned the box (with all the rest of them). It turns out that the granola bar was in one of the drawers I had searched 3 times :ninja:

[quote="serialk11r"][quote="Nerd_of_the_Ages"]Actually, my name is albert, too (different last name, though). I'm pretty much Ubemya except I suck at math and typing, and am even fatter.

The guy revived it, since he's a new member and doesn't know about this forum yet.[/quote]
You're Albert Wu right?
Albert Gu is pretty hard to beat in fatness, at least that's what it seems to me.[/quote]
Who's Albert Wu?

Nope he's not Albert Wu. Albert Wu is most definitely not that fat.

[quote="not_trig"]Toan[/quote]
OH MAN TOAN WAS SO AWESOME

"NO NO WAIT, FAT" (talking to David Zeng) "LEMME CONFISTICATE YOUR CRACK"

Name- Matthew
Live- East Side San Jose. Don't attempt to visit.
School- Chaboya Middle School (Besides, only school in East San Jose to have a math program!) Eighth grade

The fattest mopper eh? How many pounds do you weigh? Less than me I bet. I weigh over 9000

David Shi wasn't on the MOP list of 2008...

where does it say that I ever went to mop? You misread.

Fattest mopper doesn't necessarily mean really fat, so it would make more sense to compare if you did make mop.

I know. The whole post was a joke, just like most of the bay area threads you know?

I know, I was trying to add to the joke (in a failing manner).

Name: Eugene
School harvest park middle school
grade 7
wise-ness STUPID
anyone go to san jose math circle here?

[quote="dragon96"]I never visit this forum much, but anyways...

name: dragon96
location: refer to location (antarctica)
phone number: (911)-911-0000
emailaddress: PM->dragon96@artofproblemsolving.com[/quote]

Notice this is the bay area forum?
No peopel from antarctica should be here.

[quote="Ubemaya"][quote="xscapezaer"]Basically, what Ubemaya said in his first post. Except I'm way, way worse at math[/quote]

No, you're way better.[/quote]

Heh heh... Ubey, you enjoy your little joke...

I lost.

hi i am rt_08
i live in india
and i will be giving my rmo for my region

[quote="AIME15"]Rofl xscape...didn't u make usamo?

anyway, do you go to san jose math circle?

or berkely math circl[/quote]

He did it seems
[quote]
Succeeded - AIME: 9(8); USAMO - Qualified
Failed - AMC10:120(135); AMC12:114(132)[/quote]


linear algebra
matrix
function
superior algebra
superior algebra solved

If $A,B \in M_n(C)$ such that $e^A = e^B$

Prove or disprove that

for any $x \in R \;  e^{xA}=e^{xB}$

That isn't going to work. Let

$A=\left[\begin{array}{cc} 0&-2\pi \\ 2\pi&0\end{array}\right]$ and $B=\left[\begin{array}{cc} 0&-4\pi \\ 4\pi&0\end{array}\right].$

Then $e^A=e^B=I$ but for real $x$,

$e^{xA}=\left[\begin{array}{cc} \cos(2\pi x)&-\sin(2\pi x) \\ \sin(2\pi x)&\cos(2\pi x)\end{array}\right]$ and

$e^{xB}=\left[\begin{array}{cc} \cos(4\pi x)&-\sin(4\pi x) \\ \sin(4\pi x)&\cos(4\pi x)\end{array}\right].$

Those aren't the same functions of $x$.


calculus
derivative
geometry
trigonometry
function
calculus computations

I'm not sure how to go about doing related rates problems where you have to take the derivative of 3 variables being multipled (i.e. $ D_x(xyz)$). If anyone can help me walk through how to do these types of problems, I would be very thankful:

1. If $ a$ and $ b$ are lengths of two sides of a triangle, and $ \theta$ the measure of the included angle, the area $ A$ of the triangle is $ A\equal{}\frac{1}{2}ab \sin \theta$. How is $ \frac{dA}{dt}$ related to $ \frac{da}{dt}$, $ \frac{db}{dt}$, and $ \frac{d\theta}{dt}$?

2. Suppose that the edge lengths $ x$, $ y$, and $ z$ of a closed rectangular box are changing at the following rates:

$ \frac{dx}{dt} \equal{} 1$ m/s

$ \frac{dy}{dt} \equal{} \minus{}2$ m/s

$ \frac{dz}{dt} \equal{} 1$ m/s

Find the rates at which the box's volume, surface area, and diagonal length $ s\equal{}\sqrt{x^2\plus{}y^2\plus{}z^2}$ are changing at the instant when $ x\equal{}4$, $ y\equal{}3$, and $ z\equal{}2$.

For the first question: $ a$, $ b$ and $ \theta$ are functions of $ t$.  You just need to use the product and chain rules.  Likewise for the others.  If you're not comfortable with the product rule for a product of three terms, just remember associativity: $ xyz \equal{} (xy)z$, so $ D_t(xyz) \equal{} D_t((xy)z) \equal{} D_t(xy) \cdot z \plus{} xy \cdot D_t(z)$, and then just use the (two-term) product rule again.  This idea should make it easy to state the rule in general (for a product of any number of terms).

With that hint, why don't you show how you would proceed?


algebra
polynomial
geometry
superior algebra
superior algebra unsolved

We discussed this problem in class and the solution was really anything but easy. Does anyone have a natural solution (ok, the one I have is also natural, but not very...)?
    If you have $n<m$ homogeneous nonconstant polynomials in $m$ variables over an algebraically closed field, then they have a nontrivial common zero.

This is a consequence of the following theorem (Theorem I.6.4 in Shafarevich's [i]Basic Algebraic Geometry[/i]):

If a form $F$ is not $0$ on an irreducible projective variety $X$ then $\dim X_{F}=\dim X-1$, where $X_{F}\subsetneq X$ is the closed subvariety defined by $F=0$.

To prove this, one first shows that if $X\subsetneq Y$ is a closed subvariety of $Y$ irreducible, then $\dim X < \dim Y$.  One then forms a chain of varieties

$X=X^{(0)}\supsetneq X_{F}=X^{(1)}\supsetneq X^{(2)}\supsetneq \ldots,$

where $X^{(i+1)}=X_{F_{i}}^{(i)}$, where $F_{i}$ are forms of the same degree not vanishing on any component of $X^{(i)}$.  Suppose $\dim X=n$.  Then $X^{(n+1)}=\emptyset$.  This implies that $\varphi: X\to\mathbb{P}^{n}$, $\varphi(x)=(F_{0}(x): \ldots: F_{n}(x))$, is a finite map, and hence that $\dim\varphi(X)=\dim X = n$.  Since $\varphi(X)\subset\mathbb{P}^{n}$ is closed, $\varphi$ is surjective.  In particular, $(0: \ldots: 0: 1)\in\varphi(X)$, i.e., $F_{0},\ldots,F_{n-1}$ have a common zero on $X$, i.e., $X^{(n)}\neq\emptyset$.  It follows that $X^{(1)}=X_{F}$ has dimension $\geq \dim X-1$, proving the theorem.


if there are 2 cards drawn from a standard deck of cards, what is the probability that at least one card will not be a king or at least one will be red?

EDIT: Fixed a lot of things.

[hide="I like fancy symbols"]If $ \rm{A} \equal{} \text{at least one card is not a king}$ and $ \rm{B} \equal{} \text{at least one is red}$, then what we want is $ P(\rm{A}\wedge \rm{B}).$

By De Morgan's law,
\[ \neg(\text{A}\vee \text{B}) \equal{} \neg\text{A}\wedge \neg\text{B}\iff P(\text{A}\vee\text{B}) \equal{} 1 \minus{} P(\neg\text{A}\wedge\neg\text{B}).
\]
In English, we want the complement of the probability that both cards are kings and that neither is red.

That probability is $ \frac{2}{52}\cdot\frac{1}{51}\equal{}\frac{1}{1326}$
Answer: $ \boxed{\frac {1325}{1326}}$.[/hide]

I think the question would be more fun if we were drawing 3 cards instead of 2.

[quote="grn_trtle"]That probability can be further broken down as $ P(\neg \text{A})P(\neg \text{B})$.[/quote]  The probability that I brush my teeth tomorrow morning and I brush my teeth tomorrow morning is [b]not[/b] the same as the probability that I brush my teeth tomorrow morning multiplied by the probability that I brush my teeth tomorrow morning.

[hide="Hint"]Looking at the complement was a good idea.  Looking at it symbolically instead of writing down in words what it means was a bad idea.[/hide]

Hmm. Isn't the complement what I wrote down? (Both cards are kings and neither is red)

Wait, your example is confusing me. Are you brushing your teeth twice tomorrow morning?

[quote="grn_trtle"]Hmm. Isn't the complement what I wrote down? (Both cards are kings and neither is red)[/quote]  Yes.  Now, in how many ways may you draw two cards from the deck such that both are kings and neither is red?

[quote="grn_trtle"]Wait, your example is confusing me. Are you brushing your teeth twice tomorrow morning?[/quote]  You (appeared to) claim that for arbitrary events $ X, Y$ we have $ \Pr(X \wedge Y) \equal{} \Pr(X) \cdot \Pr(Y)$.  This is not necessarily true if $ X$ and $ Y$ are not independent -- for example, if $ X \equal{} Y \equal{} \textrm{I brush my teeth tomorrow morning}$ has probability strictly between 0 and 1 of occurring, then the two probabilities are not equal.  (Note that if $ \Pr(X) \equal{} 1$ or $ \Pr(X) \equal{} 0$ then actually $ X$ would be independent of itself, and of everything else (at least in a finite probability space).)

Ahh, so the kings can't be red as well, of course.

So then we have $ \frac{2}{52}\cdot\frac{1}{51}\cdot\frac{24}{50}\cdot\frac{23}{49}\equal{}\frac{46}{270725}$?

Huh?  What are those extra two factors there for?

Oh, wait, for some reason I thought we were picking 4 cards :blush:

Ok, ok, so $ 1\minus{}\frac{2}{52}\cdot\frac{1}{51}\equal{}\boxed{\frac{1325}{1326}}$

[quote="JBL"]And if we chose 3 cards like I suggested?[/quote]

i think thats right because thats what i got and so did some onelse.

another way you could have solved it is by using alot of combination crunching but i would not prefer that way.

[quote="1=2"]The desired probability is the inverse of the probability that both cards are kings and both of them are not red.[/quote]

Why is this so?  I was wondering where did you guys learn all these probabilities/

Look at my first post; it comes from De Morgan's law.
Do Intermediate Counting and Probability to learn more.

The word "inverse" is completely wrong, though -- "complement" would be better, and that fact was well-covered by earlier posts in the thread.

iYOA: what are you saying?  Yes it's correct, and I suggested a further exercise to reinforce the idea.


Putnam
induction
modular arithmetic
greatest common divisor
quadratics
calculus
integration

Let $ S$ be a set of rational numbers such that

(a) $ 0\in S;$

(b) If $ x\in S$ then $ x\plus{}1\in S$ and $ x\minus{}1\in S;$ and

(c) If $ x\in S$ and $ x\notin\{0,1\},$ then $ \frac{1}{x(x\minus{}1)}\in S.$

Must $ S$ contain all rational numbers?

I answered no, because while a) and b) get you the set of all integers, there are some rational numbers which could not be found from the three conditions.  For example, (1/13).

1/13 is definitely achievable. In fact, 1/k is achievable for all k/inN.

My argument for this question was to basically characterize any term that could be in $ S$. Here's a sketch:
- The integers are all in $ S$ (proof by induction)
- All rationals of the form $ \dfrac{1}{x(x\minus{}1)} \plus{} p$ are in $ S$, where $ x$ and $ p$ are integer
- The remaining elements of $ S$ are $ \dfrac{n\minus{}1}{n} \plus{} p$, where $ p,n$ are integers. This is because if you substitute $ y\equal{}\dfrac{1}{x(x\minus{}1)}$ into $ \dfrac{1}{y(y\minus{}1)}$, you get something of the form $ \dfrac{n\minus{}1}{n} \plus{} r$, where $ n,r$ are integers... then you can show that if we substitute in $ y\equal{}\dfrac{n\minus{}1}{n}$, we get something else of the form $ \dfrac{z\minus{}1}{z} \plus{} s$, again where $ z,s$ are integers...so we are done... Note that we only have to consider the rationals in (0,1), because we can just add any integer to these to "translate" the interval to any other one of unit length.

So, for example, $ \dfrac{2}{5}$ will not be contained here.

I couldn't come up with a full solution, but I showed that you can get $ \frac{1}{n}$ for any positive integer n.

Start with induction, we can clearly get 1. We get $ \frac{1}{2}$ by putting the integer 2 into identity C).

Suppose we can get $ \frac{1}{k}$. Then we can get:

$ \frac{k\plus{}1}{k}$

from this we can get:

$ \frac{1}{\frac{k\plus{}1}{k} \cdot \frac{1}{k}}$

$ \equal{} \frac{k^2}{k\plus{}1}$

Also, we have:

$ k\plus{}1 \equiv 0 \pmod{k\plus{}1}$
$ (k\plus{}1)(k\minus{}1) \equiv k^2 \minus{} 1 \equiv 0 \pmod {k\plus{}1}$
$ k^2 \equiv 1 \pmod{k\plus{}1}$

Thus there exists an integer m such that $ k^2 \minus{} m(k\plus{}1) \equal{} 1$.

This means that:

$ \frac{k^2}{k\plus{}1} \minus{} m \equal{} \frac{1}{k\plus{}1}$

$ 2/5$ does not have to be in $ S$. It suffices to show we can reach no number of the form $ \frac{5k\plus{}2}{5}$ from the operation in (c). Writing $ x \equal{} p/q$ for $ p,q$ nonzero relatively prime integers with $ p \neq q$, we have $ \frac{1}{x(x\minus{}1)} \equal{} \frac{q^2}{p(p\minus{}q)}$. Notice that since $ p,q$ are relatively prime we also get $ q^2, p(p\minus{}q)$ relatively prime, so this fraction is in simplest form. For it to be equal to $ \frac{5k\plus{}2}{5}$ one of the following holds.
$ q^2 \equal{} 5k\plus{}2, p(p\minus{}q) \equal{} 5$
$ q^2 \equal{} \minus{}5k\minus{}2, p(p\minus{}q) \equal{} \minus{}5$
In both cases we have a square equivalent to $ 2$ or $ 3$ mod $ 5$, which is impossible.

[quote="Kent Merryfield"]You need to prove that, of course. What's your argument for why $ \frac1{13}$ is inaccessible?[/quote]

Sounds like I may have been incorrect on this one.  Maybe there's something else I'm missing here, but I could not find a way to construct that fraction.

for $ \frac1{n(n\minus{}1)}$ to equal $ \frac1{13}$, n would be irrational, which is not in our set.

[quote="Kierhammer"][quote="Kent Merryfield"]You need to prove that, of course. What's your argument for why $ \frac1{13}$ is inaccessible?[/quote]

Sounds like I may have been incorrect on this one.  Maybe there's something else I'm missing here, but I could not find a way to construct that fraction.

for $ \frac1{n(n \minus{} 1)}$ to equal $ \frac1{13}$, n would be irrational, which is not in our set.[/quote]

You only need operation (c) to yield 1/13 + n for some integer n, not 1/13 specifically. x = 13/12 yields such a number.

I had the same as MellowMelon, S could be the set of all rationals but 2/5+N.

[quote="Nawahd Werdna"]Note that we only have to consider the rationals in (0,1), because we can just add any integer to these to "translate" the interval to any other one of unit length.[/quote]
What about 1/(7/6 * 13/6) = 36/91 ?

[quote="MellowMelon"]$ 11 \plus{} \frac {1}{13}$ is constructible by choosing $ x \equal{} \frac {13}{12}$, and then you can use (b) to shift it to $ \frac {1}{13}$.[/quote]

I was about to ask how you obtained $ x \equal{} \frac {13}{12}$ s in the set, but I can see that from x=4 is in the set, I can get 1/12 is in the set, which gives $ \frac {13}{12}$ .  And now I can see that 144-k(13) can get me to 1, so 1/13 is in the set.

Yes, if you plug in a rational number you're going to get a rational number.  I still feel that there's some numbers you cannot obtain but I'm not sure how to show which ones they are.  If they exist.

Consider $ S$ to be the set of all $ \frac{a}{b}$ with $ |a|$ is a square $ \pmod{|b|}$ with 0. It satisfies the conditions but $ \frac25\not \in S$

[quote="Kent Merryfield"]Let $ S$ be a set of rational numbers such that

(a) $ 0\in S;$

(b) If $ x\in S$ then $ x \plus{} 1\in S$ and $ x \minus{} 1\in S;$ and

(c) If $ x\in S$ and $ x\notin\{0,1\},$ then $ \frac {1}{x(x \minus{} 1)}\in S.$

Must $ S$ contain all rational numbers?[/quote]
No.

specifically, $ S$ will contain ony the rational numbers $ \frac {p}{q}$ [where GCF($ p$,$ q$)=0] for which $ \sqrt {p^2 \plus{} 4pq}$ is an integer.

Proof:
Let $ \frac {p}{q} \equal{} \frac {1}{x(x \minus{} 1)}$, where x is in S, and we wish to show what x in S proves $ \frac {p}{q}$ in $ S$
Then, since $ x\notin{0,1}$, we can multiply both sides by $ x(x \minus{} 1)$ and solve for 0:
$ px^2 \minus{} px \minus{} q \equal{} 0$
The solutions for x are:
$ x \equal{} \frac { \minus{} p\pm\sqrt {p^2 \plus{} 4pq}}{q^2}$
Since x is in $ S$, it can only be a rational number (shown below), which means that the discriminant must be a perfect square.

S contains only rational numbers:
We begin with only rational numbers, specifically $ \mathbb{Z}$.  Let $ y \equal{} \frac {1}{\frac {p}{q}(\frac {p}{q} \minus{} 1)} \equal{} \frac {q^2}{p^2 \minus{} q^2}$
Because $ \frac {p}{q}$ can be any rational number, the solution y must also be, meaning S is limited to rational numbers.

After jmerry sent me a counterexample for my argument I've given it some thought.

The problem with my mod 3 argument is that you can arrive at 2/0 mod 3 by multiplying 1/0 by -1/-1.  So essentially my proof relied on you not being able to get a square to be 2 mod 3.  The nice thing about 2/5 is that neither 2 or 3 are squares mod 5, so you wont get to 2 or -2, 3 or -3.  This is the quirk that destroys my mod 3 argument.

[quote="Albanian Eagle"]Consider $ S$ to be the set of all $ \frac {a}{b}$ with $ |a|$ is a square $ \pmod{|b|}$ with 0. It satisfies the conditions but $ \frac25\not \in S$[/quote]
I wrote this after solving the way MellowMelon did.  It's sort of amazing how short this solution is because verifying that this particular choice of $ S$ satisfies the problem conditions takes hardly any time.

[quote="jmerry"]No, those aren't the only values. We can shift by integers as well.

Pretty much anything that works is going to get there through the impossibility of $ \frac25$.[/quote]
Actually they ARE the only values, although I was a little imprecise in my math.  Let me fix the proof:
Answer: No.

specifically, $ S$ will contain ony the rational numbers $ \frac {p}{q} \plus{} n \equal{} \frac {p \plus{} nq}{q}$ [where GCF($ p$,$ q$)=0 and $ 0 < p < q$ and n is an integer] for which $ \sqrt {p^2 \plus{} 4pq}$ is an integer.

Proof:
Let $ \frac {p \plus{} nq}{q} \equal{} \frac {1}{x(x \minus{} 1)}$, where x is in S, and we wish to show what x in S proves $ \frac {p \plus{} nq}{q}$ in $ S$
Then, since $ x\notin{0,1}$, we can multiply both sides by $ x(x \minus{} 1)$ and solve for 0:
$ (p \plus{} nq)x^2 \minus{} (p \plus{} nq)x \minus{} q \equal{} 0$
The solutions for x are:
$ x \equal{} \frac { \minus{} p \minus{} nq \pm\sqrt {(p \minus{} nq)^2 \plus{} 4q(p \minus{} nq)}}{q^2} \equal{} \frac { \minus{} p \minus{} nq \pm\sqrt {q^2n^2 \plus{} 2nq(p \plus{} 2q) \plus{} (p^2 \plus{} 4pq))}}{q^2}$
Since x is in $ S$, it can only be a rational number (shown below), which means that the discriminant must be a perfect square.  Note that in the second way written, the discriminant can be clearly seen as a quadratic itself, which is an the square of an integral degree-one polynomial if the $ n^1$ coefficient is equal to half the product of the square roots of the other two terms' coefficients OR if both $ n \equal{} 0$ and $ p^2 \plus{} 4pq \equal{} 0$; that is, the discriminant is a perfect square if $ q(p \plus{} 2q) \equal{} q\sqrt {p^2 \plus{} 4pq}$ OR if both $ n \equal{} 0$ and $ p^2 \plus{} 4pq \equal{} 0$.

For the former, when both sides are squared and the equation is reduced, it is found that $ q \equal{} 0$, an impossibility for a rational number's denominator.  Thus, $ \sqrt {p^2 \plus{} 4pq}$ is an integer for EVERY member of $ S$.

Proof that S contains only rational numbers:
We begin with only rational numbers, specifically $ \mathbb{Z}$.  Let $ y \equal{} \frac {1}{\frac {p}{q}(\frac {p}{q} \minus{} 1)} \equal{} \frac {q^2}{p^2 \minus{} q^2}$
Because $ \frac {p}{q}$ can be any rational number, the solution y must also be, meaning S is limited to rational numbers.

... and then you add an integer to [i]that[/i]. As Luize posted, any such set contains $ \frac1n$ for all $ n$, and $ 4n\plus{}1$ is not always a perfect square.

[quote="jmerry"]... and then you add an integer to [i]that[/i]. As Luize posted, any such set contains $ \frac1n$ for all $ n$, and $ 4n \plus{} 1$ is not always a perfect square.[/quote]
I can't see how that extends to all integers.

I can see how that works for 2 and 6, but what about 3? 5? any number not of the form $ n(n\minus{}1)$?

Let $ \frac{1}{b} \plus{} n \equal{} \frac{1}{\frac{p}{q}(\frac{p}{q}\minus{}1)} \equal{} \frac{q^2}{p*(p\minus{}q)}$.  Since the denominators are equal, then ... a long line of reasoning follows that may or may not lead to limiting the number of $ \frac{1}{b}$ that can be formed?

No, it's a very simple process. Start with $ \frac1n$ (induction). Add $ 1$ to get $ \frac{n\plus{}1}{n}$. Apply th transformation to get $ \frac{n^2}{n\plus{}1}$. Subtract $ n\minus{}1$ to get $ \frac1{n\plus{}1}$.

I have a question, regarding a solution which I realize is false but contains one minor error, and I'm wondering whether to expect any credit at all.

Suppose we have some rational number $ \frac{p}{q}$, with $ q \neq 1$.  If $ \frac{p}{q} \in S$, then there must be some $ \frac{a}{b} \in S$ with $ \frac{b^2}{a(a\minus{}b)} \equal{} \frac{p\plus{}nq}{q}$ for some $ n \in \mathbb{Z}$.  Choose $ q > 2$ a prime, and $ p \not\equiv 1\mod{q}$.  Write $ c \equal{} b\minus{}a$, so that $ b \equal{} a\plus{}c$.  Then we have $ q \equal{} \minus{}ac$ and $ (a\plus{}c^2 \equal{} p\plus{}nq$.

As $ q$ is a prime, we must have $ \{a,c\} \equal{} \{q,\minus{}1\}$ or $ \{a,c\} \equal{} \{\minus{}q,1\}$.  In either case $ (a\plus{}c)^2 \equal{} (q\minus{}1)^2$.  Hence we have $ (a\plus{}c)^2 \equal{} q^2\minus{}2q\plus{}1 \equal{} p\plus{}nq$, so that $ p \equiv 1\mod{q}$.  Contradiction; hence such $ \frac{p}{q}$ are not in $ S$.

The error, which is quite subtle, comes from the fact that we may not force $ q$ to be positive: that is, we may have $ q \equal{} ac$ and $ b^2 \equal{} \minus{}(p\plus{}nq)$, in which case we need $ p \not\equiv \minus{}1\mod{q}$ as well.

How much credit is to be expected?  Obviously, the argument is mostly sound, but there is a fine error made.  I realize that it shouldn't be more than two points, but am wondering whether I will get nothing or something.

[quote="Teki-Teki"]The error, which is quite subtle, comes from the fact that we may not force $ q$ to be positive: that is, we may have $ q \equal{} ac$ and $ b^2 \equal{} \minus{} (p \plus{} nq)$, in which case we need $ p \not\equiv \minus{} 1\mod{q}$ as well.[/quote]
In my opinion this was the whole point of the problem, so I would not expect too much.

I think $ \frac{2}{3} \not \in S$ is also true.

[quote="cefer"]I think $ \frac {2}{3} \not \in S$ is also true.[/quote]
[quote="jmerry"]That doesn't work. $ \frac23$ is reachable: $ 0\to 3\to\frac16\to\frac{\minus{}36}{5}\to\frac{4}{5}\to\frac{25}{\minus{}4}\to\frac34\to\frac{16}{\minus{}3}\to\frac23$.[/quote]
(The quote is from a PM.)

[quote="cefer"]I think $ \frac {2}{3} \not \in S$ is also true.[/quote]

we already know why have all $ \frac {1}{l}$ so try $ x\equal{}\frac {1}{4}$ get $ \frac {16}{\minus{}3}$ then adjust to  $ \frac {2}{3}$ (actually in addition to all $ \frac{1}{l}$ all $ \frac{\minus{}1}{l} \in S$ as well)

I think you can show $ \frac {3}{14}$ cannot be in $ S$, since $ 3$ is not a quadratic residue mod $ 7$.
(More detail: like in collegebookworm's solution, discriminant is a perfect square, so $ (p\minus{}nq)^2\plus{}4q(p\minus{}nq)\equal{}(p\minus{}nq)(p\minus{}nq\plus{}4q)$ is a perfect square. But if $ p\equal{}3$, $ q\equal{}14$ then $ \gcd(p\minus{}nq, p\minus{}nq\plus{}4q)\equal{}\gcd(3\minus{}14n, 4*14)\equal{}1$ so $ p\minus{}nq\equal{}3\minus{}14n$ is a perfect square, so $ 3$ is a quadratic residue mod $ 7$, a contradiction).

collegebookworm's solution is incorrect, as far as I can tell.

Maybe the last part is incorrect, but the part about discriminant being a square is correct.
Say you start with $ S_0 \equal{} {0}$ and define $ S_{i \plus{} 1}$ as everything you can get from $ S_{i}$ using exactly one operation (i.e. $ x \Rightarrow x \plus{} 1, x\Rightarrow x \minus{} 1, x \Rightarrow \frac {1}{x(x \minus{} 1)}$). Then if you assume $ \frac {3}{14}$ is in the set, you look at the last time the third operation was used and you will actually have $ \frac {p}{q} \equal{} \frac {1}{x(x \minus{} 1)} \plus{} n$ for some integer $ n$, since any composition of the first two operations is just adding/subtracting an integer. Also obviously $ x$ is rational.
From there you show discriminant must be a square; the specific choice of $ (3,14)$ allows to show $ \gcd$ must be $ 1$ and then reduce to the fact when $ p \minus{} nq$ is a perfect square.
In collegebookworm's solution I think the problem is he wrote the expression $ q^2n^2 \plus{} 2nq(p \plus{} 2q) \plus{} (p^2 \plus{} 4pq)$ is a square of a polynomial of degree $ 1$, which does not have to be true; it could just be a square for one specific value of $ n$. The point is to find $ p,q$ for which that expression is never a square.

The given conditions describe properties that elements of S satisfy.  However, there is no requirement that S be constructed using only those rules.  The set Q of all rational numbers satisfies these conditions, so S could be Q.  In other words, the conditions do not uniquely determine such a set S.

Thus a proof that says that 2/5, for example, *cannot* be in S is probably not a complete solution to this problem.  I believe that Albanian Eagle's solution is the only complete one I have seen posted so far.

Colin

I think what is usually meant is that $ \frac{2}{5}$ is not in the intersection of all such $ S$.

Yes, however, what if we do construct $ S$ according to those rules?
So we take $ S \equal{} \bigcup _{i \equal{} 0}^{\infty}S_{i}$; then $ S$ will satisfy the conditions of the problem however will not contain $ \frac {4}{13}$. I think this works now.

[quote="Kent Merryfield"]Let $ S$ be a set of rational numbers such that

(a) $ 0\in S;$

(b) If $ x\in S$ then $ x \plus{} 1\in S$ and $ x \minus{} 1\in S;$ and

(c) If $ x\in S$ and $ x\notin\{0,1\},$ then $ \frac {1}{x(x \minus{} 1)}\in S.$

Must $ S$ contain all rational numbers?[/quote]

From (a) and (b), all non-negative integers are $ \in S$.  From (3), if $ 2\in S$ then $ \frac{1}{2}\in S$, and $ \frac{1}{\left(\minus{}\frac{1}{2} \frac{1}{2}\right)}\equal{}\minus{}4\in S$.

Now, suppose we had $ \minus{}\frac{1}{4}$.  Is this an element of $ S$?  To find that, we need to find $ x|Z\plus{}\frac{1}{x\left(x\minus{}1\right)}\equal{}\minus{}\frac{1}{4}$, where $ Z$ is an integer.
\[ 4Z(x^2\minus{}x)\plus{}4\equal{}\minus{}x^2\plus{}x\]
\[ (4Z\plus{}1)x^2\minus{}(4Z\plus{}1)x\plus{}4\equal{}0\]
\[ x\equal{}\frac{(4Z\plus{}1)\pm \sqrt{16Z^2\minus{}56Z\minus{}15}}{8Z\plus{}2}\]
Evaluate the discriminant such that it equals 0.
\[ 16Z^2\minus{}56Z\minus{}15\equal{}0\]
\[ Z\equal{}\frac{56\pm \sqrt{3136\plus{}960}}{32}\equal{}\frac{56\pm \sqrt{4096}}{32}\equal{}\frac{56\pm 64}{32}\equal{}\frac{7}{4} \pm 2\notin \textbf{Z}\]
This is the maximum value the discriminant can take.  The only other integer discriminant that arises from $ 16Z^2\minus{}56Z\minus{}15\equal{}k^2, k\in \textbf{Z}$ is 0 at $ k\equal{}8$, which produces $ Z\equal{}\frac{7}{4}$.  Therefore, an x value that satisfies the equation does not exist, which means that $ \minus{}\frac{1}{4}\notin S$.  Therefore, it is not necessary that $ S$ contain all rational numbers.

$S$ will contain only numbers of the form $\frac{a}{b}$, where $b\in\mathbb N$, $\gcd(a,b)=1$ and $a$ or $-a$ is a quadratic residue modulo $b$. Hence for instance $\frac{2}{5}$ is not achievable.


linear algebra
matrix
linear algebra unsolved

How to prove that for a symmetric matrix A, if it have two eigenvalues a1 and a2 with corresponding eigenvectors v1 and v2, then v1 and v2 must be orthogonal to each other?

Of course we are assuming that $ a_1\ne a_2.$

We're given three things: $ Av_1\equal{}a_1v_1,Av_2\equal{}a_2v_2,$ and $ A^T\equal{}A.$

$ a_1v_1^Tv_2\equal{}a_1v_2^Tv_1$ (since any $ 1\times 1$ matrix is symmetric)

$ \equal{}v_2^T(a_1v_1)\equal{}v_2^TAv_1$

$ \equal{}(v_2^TAv_1)^T$ (again, since any $ 1\times 1$ is symmetric)

$ \equal{}v_1^TA^Tv_2\equal{}v_1^TAv_2$ (since $ A^T\equal{}A$)

$ \equal{}v_1^T(a_2v_2)\equal{}a_2v_1^Tv_2.$

So $ a_1v_1^Tv_2\equal{}a_2v_1^Tv_2$ which implies that either $ a_1\equal{}a_2$ or $ v_1^Tv_2\equal{}0.$ If we're given that $ a_1\ne a_2,$ then $ v_1$ and $ v_2$ must be orthogonal.

Note that the matrix product $ v_1^Tv_2$ is simply another way of writing the inner product $ \langle v_1,v_2\rangle$ or $ v_1\cdot v_2.$

----

This whole argument was written assuming you were talking about real symmetric matrices and real inner products. A very similar argument will work for Hermitian matrices and complex inner products, but we will have to make some careful adjustments to the notation.


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

Let $ K$ be a normal subgroup of order 2 in the group $ G$. Show that $ K$ lies in the centre of $ G$, that is $ kg \equal{} gk$ for all $ k \in K$ and $ g \in G$.

Describe a surjective homomorphism of the orthogonal group $ O(3)$ onto $ C_2$ and another onto the special orthogonal group $ SO(3)$.

For the first part I have that since $ geg^{\minus{}1} \equal{} e \quad \forall \quad g \in G$, $ e$ is conjugate to itself, so the other element $ k$ in $ K$ must be conjugate to itself. So $ \{e, k\}$ is a union of conjugacy classes of size 1 and is contained in the center of $ G$.

For the second part, can we map all elements $ A \in O(3)$ with $ \det (A) \equal{} 1$ to $ I$, and all elements $ B \in O(3)$ with $ \det (B) \equal{} \minus{}1$ to $ \minus{}I$? Since matrix multiplication is the operation concerned and $ \det(AB) \equal{} \det(A) \cdot \det(B) \quad \forall \quad A, B \in O(3)$, the conditions for homorphism are satisfied.

Any clues for the surjective homomorphism onto the special orthogonal group?

For the second problem, note that $ O_3\cong SO_3\times C_2$ by the map $ O_3\to SO_3\times C_2$: $ A\mapsto (\frac{1}{\det A}A, \det A)$. So surjective maps to $ SO_3$ and $ C_2$ are given by projection, i.e. $ A\mapsto\frac{1}{\det A} A$ and $ A\mapsto \det A$.

It's rather simple in hindsight  :maybe: Thanks!


search
logarithms
calculus
complex numbers
geometric series
real analysis
real analysis unsolved

Find all $ z \in \mathbb{C}$ such that $ \sum_{n\equal{}1}^\infty \frac{z^n}{n}$ converges.

[quote="WWW"]So what have you done so far?[/quote]
I am pretty sure that I can use the Abel's Test, and from that test usually the solution is $ | z | < 1$ (I used Mathematica to check for random complex numbers and it is true) . But I don't know exactly how to apply the Abel's Test.

Easy tests tell us: convergence when |z|<1 and divergence when |z|>1.  So the main part of the problem is to analyze the case |z|=1 .

[quote="Kent Merryfield"]I've posted here on this in detail, although it might have been several years ago and my own posts aren't always the easiest things to look for. Possible key words include "summation by parts."[/quote]

http://www.mathlinks.ro/viewtopic.php?search_id=1150242020&t=176509
Using Mathematica i got the same, that the series converges to $ \minus{}\log (1\minus{}z)$. Is there an easy way to prove it?

Thanks. That's one of my more complete explanations, so it's the right link.

General principle: as long as you stay strictly inside the radius of convergence, you can integrate and differentiate power series term by term as if they were polynomials.

Let $ f(z)\equal{}\sum_{n\equal{}1}^{\infty}\frac{z^n}{n}$ and suppose $ |z|<1.$

Then $ f'(z)\equal{}\sum_{n\equal{}1}^{\infty}z^{n\minus{}1}.$

That is a geometric series, and we know how to sum geometric series. $ f'(z)\equal{}\frac1{1\minus{}z}.$

Integrate that (with the appropriate knowledge of the calculus of the logarithm in the complex plane) to get that $ f(z)\equal{}\minus{}\log(1\minus{}z)\plus{}C.$ Then choose a branch of the logarithm such that $ \log(1)\equal{}0$ and evaluate the series at $ z\equal{}0$ to nail down that constant of integration. We then get that $ f(z)\equal{}\minus{}\log(1\minus{}z)$ for $ |z|<1.$

To extend that back to the appropriate values on the boundary circle, $ |z|\equal{}1$ but $ z\ne 1,$ quote Abel's Theorem.

[quote="Kent Merryfield"]Thanks. That's one of my more complete explanations, so it's the right link.
[/quote]
Thanks a lot for the extensive explanation.


inequalities
function
parameterization
analytic geometry
geometry
3D geometry

Let $ x$, $ y$, and $ z$ be non-negative real numbers such that
\[ 4^{\sqrt {5x \plus{} 9y \plus{} 4z}} \minus{} 68 \cdot 2^{\sqrt{5x \plus{} 9y \plus{} 4z}} \plus{} 256 \equal{} 0
\] What is the product of the minimum and maximum values of $ x \plus{} y \plus{} z$?

[hide="Solution"]Let $ y\equal{}2^{\sqrt{5x\plus{}9y\plus{}4z}}$. We have $ y^2\minus{}68y\plus{}256\equal{}0 \implies y\equal{}4, \; y\equal{}64$

Thus, $ \sqrt{5x\plus{}9y\plus{}4z} \equal{} 2$ or $ 6$, and $ 5x\plus{}9y\plus{}4z\equal{}4$ or $ 36$.

By elementary linear programming, we must only test the endpoints of the regions.

$ \blacktriangleright$ The minimum value occurs when $ 5x\plus{}9y\plus{}4z\equal{}4$, and $ (x,y,z)\equal{} \left(0, \; \frac{4}{9}, \; 0 \right) \implies x\plus{}y\plus{}z\equal{} \frac{4}{9}$

$ \blacktriangleright$ The maximum value occurs when $ 5x\plus{}9y\plus{}4z\equal{}36$, and $ (x,y,z) \equal{} (0,0,9) \implies x\plus{}y\plus{}z\equal{}9$

The product is $ \boxed{4}$

[/hide]

[quote="The Zuton Force"]By elementary linear programming, we must only test the endpoints of the regions.
[/quote]
Could you elaborate a little more on this? :P I'm not well-versed with linear programming

You don't need linear programming:

[hide]For the minimum, $ 9x\plus{}9y\plus{}9z\ge 5x\plus{}9y\plus{}4z\ge 4$, so $ x\plus{}y\plus{}z\ge 4/9$.

For the maximum, $ 4x\plus{}4y\plus{}4z\le 5x\plus{}9y\plus{}4z\le 36$, so $ x\plus{}y\plus{}z\le 9$.

We can see that both of these are achievable, so they are the actual min and max.[/hide]

[quote="aidan"]Does this mean to say that in constructing such inequalities it is alright to drop some terms or add them in? And does this prove that no better bound can be achieved?  :P[/quote]

I'm not sure exactly what you're asking, but for example $ 4x\plus{}4y\plus{}4z\le 5x\plus{}9y\plus{}4z$ is true because we're adding in non-negative terms ($ x$ and $ y$).  The bounds $ 4/9$ and $ 9$ are the best bounds because there are values that achieve these bounds, namely $ (0,4/9,0)$ and $ (0,0,9)$.  The reason we can get away with adding terms without ending up with weak bounds is because the variables we're adding are 0 in the equality case.  If you know the equality case you should always keep it in mind when trying to prove inequalities - if you make a step that changes the equality case, that step probably isn't going to work to prove the inequality.

[u]To elaborate on what I meant by "linear-programming"[/u]

Suppose you have a linear function of $ x,y,z$ of the form $ F(x,y,z) \equal{} a_1 x \plus{} a_2 y \plus{} a_3 z$. Now, we want to evaluate this along some line in 3-space. The line $ \ell$ can be parametrized in the form $ (x,y,z) \equal{} (b_1 t, b_2 t, b_3 t)$ with respect to some parameter $ t$. Therefore, if we want to evaluate the linear function along the line, we can write it in terms of the single paramter $ t$ as follows:

$ F_{\ell} (t) \equal{} F(x(t),y(t),z(t)) \equal{} a_1 (b_1 t) \plus{} a_2 (b_2 t) \plus{} a_3 (b_3 t) \equal{} (a_1 b_1 \plus{} a_2 b_2 \plus{} a_3 b_3) t$

So, as we move along $ \ell$ (as we increase or decrease), the function $ F_{\ell} (t)$ is either increasing with $ t$ or decreasing with $ t$, depending on the sign of $ a_1 b_1 \plus{} a_2 b_2 \plus{} a_3 b_3$. As a consequence, [b][color=red]this means that if we wish to evaluate a linear function on the coordinates over a line segment, one endpoint will produce the minimum and one will produce the maximum.[/color][/b] There is a qualifier to this, though: if $ a_1 b_1 \plus{} a_2 b_2 \plus{} a_3 b_3$ as shown in the original demonstration, is equal to $ 0$, then we have a constant value across the line. However, while optimization does not occur only at the endpoints, it does occur at the endpoints.

What does this mean? Suppose we have a convex n-D polytope shaped region in n-space. Clearly, optimization cannot occur in the interior of this, because we can draw a segment connecting two boundary points which contains this interior point, and optimization will occur on one of the two end-points; namely, optimization must occur on the boundary. Now, consider the (n-1)-D polytopes that make up the boundary, and execute the same argument. Continue doing so, and you will find that [b][color=blue]optimization, if possible must occur at one of the vertices[/color][/b].

[hide="A concrete example"]Say we have a cube. Take any point in the cube. We can create a line segment connecting two opposite faces that contains this interior point. Since optimization must occur on one of the endpoints, which are on the faces of the cube, optimization does not occur in the interior, but on the boundary (the collection of faces, edges, and vertices).

Now, on each face, any interior point of the face is also an interior point of a segment that connects two edges. Thus, optimization does not occur on the interior of the faces, but on the edges.

Now, the edge is a line segment, so optimization on the edge must occur at a vertex.
________________________________________________________________________[/hide]
In this case, we have two triangular regions. Each has 3 vertices, and we wish to optimize a linear function over these two triangles. The principle of linear programming implies that we only need to test the 6 vertices.

Thanks a lot guys! That was really helpful :)


linear algebra
matrix
vector
combinatorics unsolved
combinatorics

this nice problem is from a link that grobber introduced us.
There are 120 members in the Israeli parliament, and 40 committees. A committee can be assembled only when the number of committee members that are now present in the parliament hall, is odd (so that votes would be decisive). Show a way to assign the parliament members to committees, so that in all times, except for when the hall is completely empty, there would be at least one committee that can be assembled, or show that it is impossible.

Hi!
Can there be a person in more than one committee

If each person had to be in only one committee then we could just put an even number of members from each committee in the hall so the problem would be trivial. That's why I think it's crucial for some people to be members of more than one committee.

Actually, this appears to be rather easy, unless I'm making yet another one of my stupid mistakes :?

We construct a $40\times 120$ matrix $A=(a_{ij})_{i,j}$ s.t. $a_{ij}$ is $1$ if the $j$'th person is a member of the $i$'th committee and $0$ otherwise. We now work in $\mathbb F_2$, the field with $2$ elements. The problem asks for such a matrix $A$ that the system $A\cdot {\bf x}={\bf 0}$ has the null vector as its unique solution, where ${\bf x}$ is a column vector with $120$ entries, and ${\bf 0}$ is a column vector with $40$ entries.

We must thus construct a matrix $A$ which has rank $40$ (maximal rank, since it has $40$ rows). We can construct such a matrix by appending an all-$1$ $40\times 80$ matrix to the right of a $I_{40}$ (the identity $40\times 40$ matrix). It's clear that the rows of $A$ are independent (over $\mathbb F_2$), as we intended.

Am I missing something here? :?

Maybe I am the one who understood wrong, but it seems that the answer is completely opposite. I think $A$ must have rank $120$ not $40$ --  which is impossible.
For example, consider you example matrix $A$: if we get $x$ to be the column vector with $118$ first entries $0$ and only the last $2$ being $1$ then we obviously have $A$[b]x[/b]$=0$.
Am I right?

Actually, you are right. I don't know what the heck I was thinking :? So the answer is negative, then.

Edit: I forgot to say this wheh I replied, but I am [b]very[/b] embarrassed about this.  :blush:

Yes, as far as I can tell, the problem is solved now and the answer is negative: the members of the parliament cannot form committees so that they respect the conditions of the problem.

Now that I read the thread, I think that too.
Cool, one more that we will move soon into the solved problem section  :D

Pierre.

[quote="iura"]I think $A$ must have rank $120$ not $40$ --  which is impossible.[/quote]
I don't know why I didn't realize it from the beginning  :blush:


trigonometry
trig identities
Law of Cosines

In triangle ABC, AB = 7 units, BC = 8 units, and AC = 9 units.  A point D is located on side AC so that BD = 7 units.  Find AD.  Express your answer as a rational number.

My Answer: [hide]Let x be the side AD; in the triangle 7: 7+x is proportional to 8 : 16, so $\frac{7}{7+x} = \frac{8}{16}$, so $112 = 56 + 8x$, $56 = 8x$ and $x = 7$.[\hide][/hide]
Oops I got that wrong...

[hide="Solution"]Denote $x=AD.$ Double use of the Cosine Law:

From $\triangle ABC$:

$\cos A={{7^2+9^2-8^2}\over{2\cdot 7\cdot 9}}={11\over 21}$

From $\triangle ABD$:
\begin{eqnarray*}7^2=7^2+x^2-2\cdot 7\cdot x\cdot\cos A &\implies& 0=x^2-14\cdot{11\over 21}x\\ &\implies& x=14\cdot{11\over 21}={22\over 3}.\end{eqnarray*}[/hide]

[hide]

Stewart's Theorem: $AB^2\cdot CD+BC^2\cdot AD=AC(BD^2+AD\cdot CD)$

So $7^2\cdot (9-x)+8^2\cdot x=9(7^2+x(9-x))$, which gives $x=\frac{22}{3}$.

By the way, the theorem can be proven by using Law of Cosines twice, just like Farenhajt's solution. [/hide]


By the Power-mean inequality,

$4(p^3+q^3)\geq (p+q)^3$

Hence $8\geq (p+q)^3$, or $p+q\leq 2$

That's if you want it to hold for nonnegative reals

For the case of real numbers, say that $q<0$, so that we may rewrite as

$p^3-q^3=2$ and prove $p-q\leq 2$ (where $p,q>0$)

Then we get

$2=p^3-q^3=(p-q)(p^2+pq+q^2)$

Say $p-q>2$.

Then $p^2+pq+q^2<1$, so that $p<1$ and hence $p^3-q^3<1$, a contradiction.

Or :

$(p+q)^3 = 4*(p^3 + q^3) - 3*(p+q)*(p-q)^2$

If $p+q < 0$ then it's true obviously and else above equality makes it


floor function
number theory
greatest common divisor
analytic geometry
symmetry
number theory unsolved

Show that for all positive integers $m$ and $n$,

$\gcd(m,n)=m+n-mn+2\sum_{k=0}^{m-1}\lfloor \frac{kn}{m}\rfloor$

Consider the grid of the $(m+1)(n+1)$ integer points with coordinate $(x,y)$ such that $0 \leq x \leq m$ and $0 \leq y \leq n.$
Let $\Delta$ be the line with equation $y= \frac{nx}{m}$ which then goes through the origin and the upper-right vertex $(m,n)$.

Let $d = \gcd (m,n)$. Let $m=da, n=db.$
It is easy to verify that the integer points of the grid which belong to $\Delta$ are those of the form $(ka,kb)$ where $k=0,1,...,d.$ Thus there exactly $d+1$ integer points on the line.

Now, the number of integer points which are under or on the line and with $x=k$ is exactly $[ \frac{kn}{m}]+1.$ Summing over $k=0,1,...,m$ this leads to the total number of points of the grid which are under or on the line $\Delta.$
By symmetry there are as many points above or on the line. Only the points which belong to $\Delta$ are counted twice.
It follows that :
$2 \sum_{k=0}^{m}( [\frac{kn}{m}]+1) = (m+1)(n+1)+d+1$.
This is the desired result.

Pierre.