Show that for all $n ∈ \mathbb Z$ , $n^2 ≥ n$. 
Show that for all $n ∈ \mathbb Z$ , $n^2 ≥ n$.

Hi, I'm trying to do this question. Does this mean I have to take any integers? Do I suppose its true and try to prove it. Is there a way to intuitively see if it's true or false even before beginning to prove.
I am new to this subject and here as well and I am sorry if my questions are not clear or are not the way I have to ask. 
Thanks.
 A: Yes, integers.
Intuition?  Well for $n\ge 1>0$ then $n*n\ge n*1$ by multiplication inequality axiom.  
And the only times that an integer isn't greater than or equal to one is if it is $0$ or negative.  (That's why this is only true for integers; other numbers can be between $0$ and $1$ and this just won't be true:  if $0 <x <1$ then $x*x <x*1$.)
If $n=0$ or $n <0$ can be thought of as  really just special case afterthoughts.  $0^2=0\ge 0$, and if $n <0$ then $n^2>0$ (does that need to be proven?) and so by transitivity  $n^2>0>n $.
A: For $n\leq0$ we have $n^2\geq0$, thus, $n\leq n^2$. For $n\geq1$ we can use induction; for the case $n=1$ we have the equality. Let us assume the statement is true for some $n=k$, then, for $n=k+1$ we have
$$n^2+2n+1\geq 3n+1$$
$$(n+1)^2\geq3n+1\geq n+1$$
Which completes the induction, notice that because $n>0, 3n+1>n+1$
A: Yes, to prove it in general you have to show it holds for any $n \in \Bbb Z$.
No, you don't want to "suppose it's true and try to prove it;" that is circular reasoning; you generally want your logic to proceed in a straight line.  
Okay, let's look at a little proof:
If $n = 0$, it's clearly true, right?
If $n \ne 0$, then
$\vert n \vert \ge 1; \tag 1$
also
$\vert n \vert \ge \vert n \vert; \tag 2$
if we multiply these two inequalities we obtain
$n^2 = \vert n \vert^2 \ge \vert n \vert; \tag 3$
also,
$\vert n \vert \ge n; \tag 4$
combining (3) and (4) yields
$n^2 \ge n, \tag 5$
as desired.
The question about intuition is almost too difficult to deal with in a small space, but this I know from experience:  the more little "obvious" facts you can prove, the more you develop an "intuitive nose" for sniffing out truth and falsehood; but of course, you really ultimately need to convert your intuition to logic.  I guess the key take-out here is this:  write it down.
A: The other answers focus on how to prove that theorem. I will focus on another part of your question:

Is there a way to intuitively see if it's true or false even before beginning to prove?

It's a good idea to shortly think about whether the theorem might be true before starting to prove it. 
One way to do so is by trying to find a counter-example to prove it wrong. So you just plug in some numbers, and see whether it is true for those. Try to choose numbers, that are somehow different from each other (e.g., positive numbers, negative numbers, small numbers, big numbers, $0$). 
If you find a number for which the theorem is wrong, congratulations, you just prove the theorem to be wrong. If the theorem is true for all numbers you have tried, the theorem might be true or there might be a counterexample that you haven't found. 
Now you need to prove the theorem as shown by the other answers. Only then you can be sure it is true.   
A: $$Let\space n\in Z$$
$$\text{Now, let us consider two functions such that,}$$
$$ \space  f_1(n)=n^2 \space and \space f_2(n)=n$$
$$ \text{Let me change value of n to n+1 , therefore we have,}$$
$$f_1(n+1)=(n+1)^2=n^2+2n+1; \space \space and \space f_2(n+1)=n+1;$$
$$\text{let us subtract }f_1 \text{from} f_2 $$
$$\phi=f_1(n+1)-f_2(n+1)=n^2+2n+1-(n+1)=n^2+n=n(n+1)$$
$$\phi=n^2+n=n(n+1)\space \text{is positive} \space \forall n \in[1,\infty)$$
$$\phi=n(n+1)=0 \space \space \text{if}\space n=0 \space or \space -1 $$
$$\text{suppose n is lesser than -1 then both n and n+1 is negative,}$$
$$\text{ therefore }\phi\space \text{is positive} $$
$$ \text{if }\phi \text{ is positive then }f_1(n+1)>f_2(n+1) \text{and if }\phi=0 \text{ then}\space f_1(n+1)=f_2(n+1) $$
$$\text{Therefore, } n^2 \ge n, \space \forall \space n\in Z$$
