Recent content by Bingk

Prove that R(p,q) \leq \left(\stackrel{p+q-2}{p-1}\right) where p and q are positive integers I'm supposed to use induction on the inequality R(p,q) \leq R(p-1,q) + R(p,q-1) , but I'm having difficulty there. How do I go about doing this? I can show it's true for p=q=1. But, I can't...

Just to be sure ... if the person decided to do project 3 first, then 5, and nothing else (i.e. 35), you're saying that would be the same as doing 5 first then 3 (i.e 53 = 35)? If that's the case, then yes, I believe you have it correct, 32 outcomes. But, if you mean that you can do the...

How to say ... Hi ... I'm doing a small presentation and I was wondering how I would say the following summation: \sum_{0<i_1<...<i_n<p} \left(\frac{i_1}{3}\right) \frac{(-1)^{i_1}}{i_1 i_2 \cdot \cdot \cdot i_n} where \left(\frac{i_1}{3}\right) is the Legendre symbol, n is a positive odd...

I meant it would be tedious the way I did it :) ... Referring to thomas430's question if a similar method could be used to prove it :) Petek, that's great :) ... I didn't realize that LCM could be taken advantage of in that way :). In my defense, I was trying to explain how (p-1)/2 became...

Yeah, I admit that I oversimplified my example, just to help see what's going on ... For the case when n=p^2, this is what happens, 1<p<p^2 and 1<qp<p^2, where q is prime and q<p. so p and qp are still factors in (n-1)! so if we regroup it, (n-1)! = (n-1)*(n-2)*...*qp^2*...4*3*2 and we can...

b^2|b^3|a^2 so b^2|a^2 Can you show that if b^2|a^2 then b|a Here's how I did it: a and b are integers such that b^2|a^2 => mb^2 = a^2 Since the square root of a^2 = a, then we can take the sqrt(mb^2) = b*sqrt(m) = a. Since a and b are integers, sqrt(m) must also be an integer => b|a...

For your second part, if n is prime, then n cannot divide any number less than itself. In other words, it is not in the prime factorization of those numbers. So a product of those numbers will still not contain that prime number, and thus n will never divide (n-1)! For your first part, again...

Hi ... I suggest you wiki the Legendre symbol to get a better idea of it ... also wiki the proof for the Second Supplement to the Law of Quadratic Reciprocity. The idea is this, we know that even numbers are of the form 2n, and odd numbers 2n+1. We can extend this further by showing that 4n...

Couldn't you make use of prime decomposition? Take the prime decomposition of n, and let it's totient function be the product of the totient's of it's primes. Do the same for d, and since d|n, the prime decomposition of d is imbedded in n, so for n=d*a, a is just what's left to complete d so...

Actually, they are very strongly related, and in fact, number theory is sometimes taught with a group theory perspective ... A Classical Introduction to Modern Number Theory by K. Ireland and M. Rosen ... it's not an easy read, but it should give you a better idea of how they're related ... in...

Yes ramsey, that was basically my point :) Furry, I think it has to do with the fact that a^2|b^2 that's how I got a=1, but you're right, it's not that a has to equal 1, it's just that that's one of the possible solutions, the main point is that I didn't see anything wrong if the a's canceled...

I gave it some thought, and I kind of get what you're saying. I still think it's okay to claim it to be closure since we're dealing with integers only (divisibility, in this case, is defined for integers only ... because really, any number, except zero, divides any other number, it's just that...

Wouldn't it be closure?

1) "Since a<b are integers then sqrt(q) = m/n where m and n are coprime integers" ... that's what you said, so I'm wondering how you got "since a<b" ... and I got what you meant by coprime :) 2) b = ka + r ... if a>b, then k = 0 (my mistake :))and r = b ... and from "r = a(sqrt(q) - k)" which...

There's nothing wrong with trivial :) ... if you want, you can look at the prime factorization of a and b. So basically n contains all the prime factors missing from a to make it equal to b. If a does not divide b, it means that a contains at least one prime which is not in b, so no matter what...