Here is my solution not sure if it's quite right though.
To prove the theorem..
Assume n ≥ 2 where n is composite and all prime divisors of n are >√n.
Let n = 2
Since the only prime divisors of 2 is 1 or 2, then..
p = 1, or
p = 2
but If p = 1 then 1 > √2, Which is FALSE. This...
so if n = 2 then p^2 > 2. But there is no prime number that can be squared and divide 2. So there is a contradiction? I dunno... maybe it's bed time for me haha
Thanks for the hint!
Would I be correct then to assume the contrapositive of the statement then prove it to be true? my new assumption would be "If p does not divide n or p > √n Then n < 2 or n is prime."
Or is using the De Morgan's law here unnecessary?
Homework Statement
Prove the following Theorem.
Let n ε Z. If n ≥ 2 and n is composite, then there exists a prime p such that p divides n and p ≤ √n.
After proving this Theorem show that if 757 is not a prime, then it has a prime divisor p ≤ 23.
The Attempt at a Solution
I...
This is very new to me. I get what you mean about how it doesn't prove it for all integers, and I do understand how a = cx and b = cy. But I'm struggling to see where to go next. could I say that a*b = cx*cy = c^2xy? And then c^2xy must be divisible by c^2?
So here is the problem.
Prove that If the gcd(a,b) = c then c^2 divides ab
I know it looks very simple and it seems to be true, But I get the feeling I'm doing something wrong here in my proof. Would appreciate it if someone can explain if I'm on the right track or not.
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