# Consecutive prime numbers

Let's assume $$k$$ and $$n$$ are consecutive prime numbers, $$k \lt n$$.

An axiom: for any such $$k$$ and $$n$$, $$k^2 \gt n$$.

This seems "obviously" true to me, but could you please prove me wrong? Or if it is correct, could you please help me prove it?

• Since $k \ge 2$, it follows that the next prime after $k$ must be less than $2k$, which in turn is less than or equal to $k^2$. See Bertrand's postulate. – hardmath Oct 29 '14 at 18:37
• The problem statement vaguely resembles a famous unsolved problem, Legendre's Conjecture, that for each positive integer $n$, there exists a prime between $n^2$ and $(n+1)^2$. – hardmath Oct 29 '14 at 18:53

Primes occur no further intervals than $n$ and $2n$, and $n^2>2n$ for $n>3$
• done.${}{}{}{}$ – mookid Oct 29 '14 at 18:39