Wikipedia:Reference desk/Archives/Mathematics/2008 June 29

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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)

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)

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

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

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

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

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)

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)

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)

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

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)

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)

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

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)

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

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)

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

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)

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)

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)

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