For all $x$ there is a $y$ such that if $x$ is non negative then $y^2 = x$
Is my logic correct in proving that statement is true ? Can provide an explanation of how to test this proof ?
$x=2$ then $y^2 = 2$
$x=2$ then $y = \sqrt 2$
Truth table for this scenario
$t , t$
This proves there is $y$ such that $y^2 = x$
On
Define a function, $$f:\mathbb{R_{\geq 0}} \to \mathbb{R_{\geq 0}} $$ $$ x \mapsto \sqrt{x}$$
Injective: $$\sqrt{x}=\sqrt{y} \Rightarrow x=y$$
Surjective: $$Dom(f)= \{x| x\geq 0\}$$ $$y \in Ran(f) \Rightarrow y \geq 0 \Rightarrow y \in Dom(f)$$ \
$\textbf{Comment}$: I left out details for you to fill in. Comment if things are still unclear. Note $y^2 \in Dom(f), \Rightarrow \exists x$ s.t $\ x = (y)^2 \Rightarrow \sqrt{x} = y.$
You cannot prove the statement by using only one value $x = 2$ to show that for that particular value $x$, there exists a $y$ such that $y^2 = x$.
Why not?
You can't stop at showing it's true for one $x$, or two $x$, or even a million values of $x$, because the statement is a claim about all $x\geq 0$. Proving a "for all" statement requires a different strategy than proving an "existence" statement. It is true that for each $x$, there exists a $y$, and you can show this existence of $y$ by providing one value (only one is required for existence), but you need to prove that for every $x\geq 0$, some $y$ exists so that $y^2 = x$.
We usually do this by picking an arbitrary $x \geq 0$. By not assigning it any particular value, we keep it arbitrary, so that what we then demonstrate about $x$ holds for every $x\geq 0$.
So we take $x \geq 0$. Then $(\sqrt x)^2 = x$, so we can put $y = \sqrt x$, and we're done, since no matter what the $x\geq 0$, $y = \sqrt x$ is defined, and hence exists.