Direct Proof: $\forall x \in\mathbb{ Z} , x^2-x \geq 0$ I'm currently in a Discrete Math course at my university (not a homework question) so I'm fairly new at proofs (only 1 week into the course). 
I'm a bit confused at the best approach to proving statements and furthermore, how to properly write them. I've split this question into cases because I'm not sure how to prove this in a simpler way. This is what I've done:
Proof: True. Assume $x \in \mathbb{ Z}$. 
Case 1: Assume $x = 0$. Then $x^2-x = 0$ which satisfies $x^2-x \geq 0$
Case 2: Assume $x > 0$. Let $x = 2$. Then $x^2-x = 2$ which satisfies $x^2-x \geq 0$
Case 3: Assume $x < 0$. Let $x = -2$. Then $x^2-x = 6$ which satisfies $x^2-x \geq 0$
Would this be considered a sufficient proof? 
EDIT: 
Sorry, I do know there is another method of solving this as I saw the answer that the practice problem provided. They solved it by completing the square and arrived at this:
$x^2-x = (x- \frac{1}{2})^2- \frac{1}{4} \geq - \frac{1}{4}$ 
and that was all that was there. Now, I understand the math but not why it's sufficient enough to qualify as a proof. Because to me it's no different than the original statement $x^2-x \geq 0$. 
 A: Hint: $$x^2-x=x(x-1)$$ so $$x^2-x\geq 0$$ if $$x\geq 1$$ or $$x\le 0$$
A: For any $n \in \mathbb{R}$, we have $n^2\geq0$. 
We also, know that if $n\leq 0$, then $(-n)\geq0$.
This implies $n^2-n \geq 0.$-------------$(*)$
Now, if $n\geq 1$.
Now, since, $n$ is positive, we can multiply $n$ to both sides. We get $n^2\geq n \implies n^2-n\geq 0.$--------$(**)$.
Since $\mathbb{Z} \subsetneq \mathbb{R}$ and it doesn't contain any number between $(0,1)$. Therefore, combining $(*)$ and $(**)$; It follows that $$n^2-n\geq 0 \enspace \forall \enspace n \in \enspace \mathbb{Z}$$ 
A: Showing $x^2-x = (x- \frac{1}{2})^2- \frac{1}{4} \geq - \frac{1}{4}$ is enough because:


*

*$x^2-x$ is an integer

*$x^2-x \geq \frac{-1}{4}$
Therefore, $x^2-x \geq 0$
But the "best answer" really depends on which tools you are "allowed" to use. With this I mean, should we consider ourselves familiar with rational and real numbers, or rather avoid them completely? With just integers, I'd try to proof by induction that $\forall x \in \mathbb{Z}, x^2 \geq x$. See:


*

*This statement is true for $x \leq 0$ as $x^2$ is always non-negative

*Assume it is also true for a given $x \geq 0$. Then, $(x+1)^2 = x^2 + 2*x +1 \geq x^2+1 \geq x+1 $
A: We want to show, that for every $x\in\mathbb{Z}$ it is $x^2-x\geq 0$. So $x(x-1)\geq 0$.
This product is greater then zero, if either every factor is positiv, or negativ.
The case, where one factor is zero is trivial.
So if $x>0$, then $x\geq 1$ and it is $x-1\geq 0$. So this holds. (both positiv)
If $x<0$, then $x-1<0$ obviously (both negativ).
If $x-1>0$, then $x>1$ (both positiv)
If $x-1<0$, then $x<1$ (both negativ)
[Keep in mind we 'made' the assumption, that neither factor is zero.]
A: Option:
Let $x \in \mathbb{Z}.$
Show $x^2\ge x$.
1)If $x \le 0$,  
$x^2 \ge 0$, and $x \le 0$, inequality is true.
2) If $x >0$,
$x^2 \ge x$ is equivalent to $x \ge 1$.  
(Division of inequality by $x >0$).
3)Altogether:
Inequality is valid for $x \in \mathbb{Z}$.
