I know it's probably a well known problem, but my intention to post/publish this question (and of course my approach) here is to verify my intuition about about the proof .Thus I done it, my way and honestly I don't know whether I had copied someone's approach or not.
To prove that there is no largest prime, my approach is as follows:
Suppose there is a largest prime $p\in \mathcal P$ (set of all primes) i.e. $p=\sup\mathcal P$.
Since there are prime/primes $> 2$, let there be a largest prime $p>2$.
Now every prime except 2 is an odd number, and the difference between two odd numbers is always even. So for every $p_i \in \mathcal P/\{2\}$ there exists a $k \in \Bbb N$ such that $p_i+2k \in \mathcal P$. Which implies $p+2k\in \mathcal P$ for some $k \in \Bbb N$.
Which is a contradiction. Because $p+2k>p$, but we claimed $p$ is the largest member of $\mathcal P$.
So there exists no largest prime number. □
Please let me know my faults, if any, and suggest improvements, if any.
Best Answer
This is not correct, because you haven't justified why for every odd prime $p_i$ there is a $k\in\mathbb N$ such that $p_i+2k\in\mathcal P$. It is true that if there is a prime larger than an odd prime, then their difference must be even. However, how do you know that for each odd prime there is a larger prime?—that is exactly what you are trying to prove.