1. ## Divisors

Greetings

Prove that any integer $x>1$ has either a divisor $>1$ and $\leq\sqrt{x}$ or no divisor $>1$ and $\leq\sqrt{x}$ (In the latter case it is called a prime).
Prove this statement from the axioms governing the behavior of integers.

2. Originally Posted by I-Think
Greetings

Prove that any integer $x>1$ has either a divisor $>1$ and $\leq\sqrt{x}$ or no divisor $>1$ and $\leq\sqrt{x}$ (In the latter case it is called a prime).
Prove this statement from the axioms governing the behavior of integers.
Suppose $x$ has a proper divisor $d>1$, then there exists an integer $k>1$ such that:

$x=dk$

Now if $d \le \sqrt{x}$ there is nothing left to do, so assume $d>\sqrt{x}$, then:

$x=\sqrt{x}\sqrt{x}=dk>\sqrt{x}k$

which implies that $k<\sqrt{x}$ which is a factor $\le \sqrt{x}.$

Which leaves you tidy this up and sort out the details.

CB