A clean proof of $x^2 \geq x$, for any integer $x$ I am trying to prove that $x^2 \geq x$ for any integer $x$.
Since we know that for any number $n$, $n^2 \geq 0$ we conclude that if $x \leq 0$ the proposition will hold.
Next we must prove that the proposition holds for $x > 0$.
First we note that $x^2=xx$, and apply this to the inequality in the proposition.
$xx \geq x$.
Divide each side by $x$ and arrive at the solution set $x \geq 1$. Since there is no integer $p$; $0 < p < 1$ the proposition must be true.
The problem I am having is the second part for $x > 0$. I feel like my argument is circular. If anyone has any advice I'd appreciate it!
 A: If $x\geq 1$ then $x>0$. Multiply both sides of $x\geq 1$ by $x$ to get $x^2\geq x$. 
A: Since $x \geq 1$, you have $x^{2} = x \cdot x \geq x \cdot 1 = x$.
A: $$x^2 \geq x \iff x^2 - x \geq 0 \iff (x-1/2)^2 \geq 1/4$$
And $(x-1/2)^2$ is minimal when $x = 0, 1$. 

Alternatively: 
$x \geq 1 \implies x^2 \geq x$.
$x \leq 0 \implies x \leq 1 \implies x^2 \geq x$
These two cases cover all integers.
A: Rewrite
$$x(x-1)\ge0.$$
This is true as $x$ and $x-1$ have the same sign, except when they straddle zero. But in this case, one of them must be zero.
A: If $n\geq 0$, $n^2 - n = n(n-1)\geq 0$ if $n\geq 0$ and $n\geq 1$. Since $n$ is an integer greater or equal than zero, when $n$ is equal to zero, it is certainly true because you have $0=0$. If $n>0$, then, automatically it is also $n\geq 1$ and the relation is again proved true. Done.
A: If you re looking for this result with integers you might as well consider the natural numbers and do induction (why?). Since for negative integers, it follows immediately that $x^2 \geq x$ since if $x<0$ then $x^2 >0$. 
A: Axiomatically we are given $x>0$ and $y>0 \implies xy > 0$.
We can prove Corollary 1 if $0 < x$ and $a < b$ then $x*a < x*b$.  (Proof:  $a < b \implies 0 < b - a \implies 0 < x(b-a) = xb - xa \implies xa < xb$.)
We can prove Corollary 2 if $0 > x$ and $a < b$ then $x*a > x*b$.  (Proof:  if $x < 0$ then $0 = x -x < 0 -x = -x$.  So $-x*a < -x*b$ so $0 < xa - xb$ and $xb < xa$.)
Now let $x \in \mathbb Z$.
Case 1:  $x = 0$.  Then $x^2 = 0 = x$.
Case 2: $x > 0$.   Then $x \ge 1$ and  by Corollary  1 $ \implies x^2 = x*x > 1*x = x$.
Case 3: $x < 0$. Then by Corollary 2 $x^2 = x*x > 0*0 = 0$.  So $x^2 > 0 > x$.
That's it.
A: $0\ne x\in Z\implies x^2=(-x)^2=|x|^2\geq 1$ $\implies x^2=|x|^2=|x|(|x|-1)+|x|\geq $ $\geq |x|(0)+|x|=|x| \geq x.$
$0=x\implies x^2=x.$
A: With no division:
$$\begin{align}
x^2
&=x\cdot x\\
&=x\cdot(1+x-1)\\
&=x\cdot1 + x\cdot(x - 1)\\
&=x + \text{product of two numbers each $\geq0$}\\
&=x + \text{some number $\geq0$}\\
&\geq x
\end{align}$$
A: By completing the square, you can rewrite
$$(2x-1)^2\ge1.$$
As the squared number is odd, the inequality holds.
