1. ## Divisors

Greetings

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

$\displaystyle x=dk$

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

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

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

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

CB