For all $x \in \Bbb R$, if $x > 4$ then $x^2 > 9$ Can I just say $x > 4 \Rightarrow x^2 > 16$ and since $16$ is bigger than $9$, true?
 A: Yes of course that's correct indeed since $f: x \mapsto x^2$ is increasing for $x \gt 0,$
$$x>4 \implies x^2>16>9$$
What is not true is that for example
$$x^2>9 \implies x>3$$
A: Let $x \in \Bbb R$, if $x > 4$, then as $x \mapsto x^2$ is increasing over  $\Bbb R_+$,
$x^2 > 16 > 9$.
A: An alternative approach is $x>4>0\implies x^2>4x>4^2=16>9$.
A: Here is another take.
Write $x=3+y$,with $y>1$. Then $x^2=(3+y)^2=9+6y+y^2>9$, since $y>1>0$.
A: More or less.
You need some concept that $x^2$ is increasing for positive numbers so that if $0 < a < b$ then $a^2 < b^2$ (and so $x >4$ then $x^2 > 16 > 9$).
If you are doing calculus you can point out that for $f(x) = x^2$ then $f'(x) = 2x$ and if $x > 0$ then $f'(x) > 0$ so $f$ is increasing on all positives.
Or if you are doing ordered field axioms you can do via axiom: If $a < b$ and $c > 0$ then $ac < bc$.  And therefore if $0 < a < b$ then $a\cdot a < b\cdot a$ and $a^2\cdot b < b\cdot b^2$.  
So one way or another, yes, you can say it.  (But you should be prepared to defend it if anyone questions.)
....
And there are "tricks".
$x^2 - 9 = (x-3)(x+3)$ and $x - 3 > 4-3 = 1> 0$ and $x+3 > 4+3 = 7 > 0$ so $x^2 - 9 > 0$ so $x^ > 9$ and other cute little things like that.
