Mathematics desk
< June 28	<< May \| June \| Jul >>	June 30 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

June 29

Very very hard problem

Prove that for every prime number greater than a googol, there exists a multiple of that prime number that is divisible by 7.

This problem is darn so hard that it is fit only for a university student. Instead our maths teacher forced us mere school students to do it. Oh yes, the teacher tell us that googol is a very large number.

When I went on the web to google googol, I despair when I found out just how large a googol is. How on earth am I going to prove this? I dont even know where to start.

Please help. 122.107.149.37 (talk) 07:11, 29 June 2008 (UTC)[reply]

If the question is just as you state it, then it must be an illustration of misdirection. Forget the googol part. Forget the prime number part. Find a multiple of 1 that is divisible by 7. Find a multiple of 2 that is divisible by 7. Show that for any whole number at all ~~(apart from 0)~~ there is a multiple of that number that is divisible by 7. Gandalf61 (talk) 07:40, 29 June 2008 (UTC)[reply]

Surely with any sensible definition, 0 is itself divisible by 7? Algebraist 09:34, 29 June 2008 (UTC)[reply]

Yes, indeed - brain not working. Gandalf61 (talk) 13:02, 29 June 2008 (UTC)[reply]

They're probably only working in the positive integers, and with positive integer multiples. Black Carrot (talk) 11:31, 29 June 2008 (UTC)[reply]

Since 0 isn't prime, who cares? "Prime number" generally refers to positive integers. --Tango (talk) 12:27, 29 June 2008 (UTC)[reply]

About limits

Hi, I asked a question here. Would you be so kind and take a care of it? Mozó (talk) 11:08, 29 June 2008 (UTC)[reply]

Quick question.

Hey guys. I was wondering if any of you could give me a quick proof of the following statement before I head off to work: the real part of any non-trivial zero of the Riemann zeta function is 1/2.

Come on folks, looks like kindergarten stuff. I'm off to get my coffee. Answer pronto! TIA. —Preceding unsigned comment added by 124.191.113.253 (talk) 11:24, 29 June 2008 (UTC)[reply]

It's trivial. The Plouffe's inversion of a Jordan-Eisanhimmenheimann space produces a havamorphism from pi to the unit n-dimensional sphere. The result follows. Black Carrot (talk) 11:33, 29 June 2008 (UTC)[reply]

In case this is a serious question - see Riemann hypothesis. --Tango (talk) 12:25, 29 June 2008 (UTC)[reply]

A tricky one-to-one correspondence problem

Hi all:

I've been trying to establish a one-to-one correspondence between the intervals [0, 1] and (0, 1), but so far this problem has proven surprisingly difficult.

I got as far as constructing the following function as a bijective map from (-∞, +∞) to (0, 1):

$f(x)={\frac {1}{\pi }}\arctan x+{\frac {1}{2}}$

and realizes that I just need to find a way to map [0, 1] to (-∞, +∞) to solve the problem. But so far it's been eluding me.

Any good ideas?

Thanks,

76.65.15.166 (talk) 20:17, 29 June 2008 (UTC)[reply]

This can't be done with 'nice' functions, since [0, 1] and (0, 1) aren't homeomorphic. If I was asked this question, I would take the lazy option and use the Cantor–Bernstein–Schroeder theorem, but you might want something more direct. Algebraist 21:15, 29 June 2008 (UTC)[reply]

One direct method would be to deal with the rationals and irrationals separately. Algebraist 21:17, 29 June 2008 (UTC)[reply]

Consider the sequence

0,1,{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},\dots

. You take every element of this list to the element two slots further down. The other points in

(0,1)

are mapped to themselves. This defines a simple bijection between

[0,1]

and

(0,1)

. Oded (talk) 21:23, 29 June 2008 (UTC)[reply]

Yes, that was essentially my idea. Algebraist 21:32, 29 June 2008 (UTC)[reply]

Thank you both OdedSchramm and Algebraist. While OdedSchramm provided the final, definitive answer. I would still like to show my understanding of Algebraist's indirect argument with the following exposition:

Forward injection: $f(x)={\frac {1}{2}}x+{\frac {1}{4}}$

Backward injection: trivially $f(x)=x$

Thus, by the Cantor–Bernstein–Schroeder theorem, there must exist a bijective relationship between the sets [0, 1] and (0, 1).

Am I correct according to your expert opinions, Algebraist? also submitting to OdedSchramm's expert scrutiny my above self-composed answer ;)

76.65.15.166 (talk) 21:37, 29 June 2008 (UTC)[reply]

That's perfectly correct. Oded (talk) 21:41, 29 June 2008 (UTC)[reply]

For bonus marks, unravel the proof of Cantor–Bernstein–Schroeder to get an explicit bijection from those injections. Algebraist 21:43, 29 June 2008 (UTC)[reply]

Unfortunately the margin of this fine encyclopedia is too small to contain the bonus solution. So I will try to show this some other time. ;) Thanks again Oded and Algebraist, really appreciate it! 76.65.15.166 (talk) 21:46, 29 June 2008 (UTC)[reply]

Of course the same way you can 'remove' any finite set of points — just skip appropriate number of initial elements in a sequence. You can also use similar method to get rid of a countable set of points, eg. to construct a $\mathbb {R} \to \mathbb {R} \setminus \mathbb {Z}$ bijection. Simply choose any countable subset $A\subset \mathbb {R} \setminus \mathbb {Z}$ , map its every element onto every other element, and use those freed elements to map points you want to remove. Technically: order $A$ into a sequence $(a_{n})_{n\in \mathbb {N} },$ order elements to be removed $\mathbb {Z}$ into a sequence $(b_{n})_{n\in \mathbb {N} }$ and use the function
$f(x)={\begin{cases}a_{2i}&\iff x=a_{i}\\a_{2i-1}&\iff x=b_{i}\\x&\mathrm {otherwise} \end{cases}}$
which is a desired bijection. --CiaPan (talk) 05:54, 1 July 2008 (UTC)[reply]

Thanks CiaPan for your answer to Algebraist's bonus question! 74.12.198.105 (talk) 21:34, 1 July 2008 (UTC)[reply]