# Analysis Questions

• Feb 4th 2008, 10:05 PM
heathrowjohnny
Analysis Questions
1. Prove that there is no largest prime.

Proof: Assume by contradiction that there is a largest prime $\displaystyle n$. Then its divisors are $\displaystyle 1$ and $\displaystyle n$. Then how would I continue?

2. If $\displaystyle n$ is a positive integer, prove that the algebraic identity $\displaystyle a^{n} - b^{n} = (a-b) \sum_{0}^{n-1} a^{k} b^{n-1-k}$. So use induction on $\displaystyle n$?

3. If $\displaystyle 2^{n}-1$ is prime, prove that $\displaystyle n$ is prime. Then $\displaystyle 2^{n}$ is not prime. But this does not imply that $\displaystyle n$ is prime.
• Feb 4th 2008, 10:14 PM
Ryan J
for your first question: first imagine there is a largest prime number, X.

Think about how you could prove there is a larger one.

• Feb 4th 2008, 10:34 PM
earboth
Quote:

Originally Posted by heathrowjohnny
1. Prove that there is no largest prime.

Proof: Assume by contradiction that there is a largest prime $\displaystyle n$. Then its divisors are $\displaystyle 1$ and $\displaystyle n$. Then how would I continue?

...

Have a look here: http://www.mathhelpforum.com/math-he...t+prime+number
• Feb 5th 2008, 07:57 AM
ThePerfectHacker
Quote:

Originally Posted by heathrowjohnny
3. If $\displaystyle 2^{n}-1$ is prime, prove that $\displaystyle n$ is prime. Then $\displaystyle 2^{n}$ is not prime. But this does not imply that $\displaystyle n$ is prime.

Yes. If $\displaystyle n=pq$ where $\displaystyle p,q>1$ then $\displaystyle 2^n - 1 = (2^q)^p - 1$ and use the factorization trick above. Which means it cannot be prime.
• Feb 5th 2008, 09:08 AM
Jameson
Quote:

Originally Posted by earboth

OP - You're on the right track with assuming that there exists a largest prime and heading towards a contradiction. Call the largest prime $\displaystyle p_n$. Take the product $\displaystyle p_{1}p_{2}...p_n$. This is obviously not prime. What happens if you add 1 to the product? What divides $\displaystyle p_{1}p_{2}...p_n +1$?