How to prove or disprove $ \forall x \in \mathbb R, \exists y \in \mathbb R $ |x| = xy I think that the statement is true in general considering +1 or -1 for y. How can I prove it in proper notation.
Similarly I need to prove $ \exists y \in \mathbb R, \forall x \in \mathbb R st, x^2 +2x - 5 \leq y$
I did it as follows, is it wrong?
$$x^2 +2x - 5= (x+1)^2 -6 \leq -6 \forall x \in \mathbb R$$
 $$ \therefore \exists y \in \mathbb R, \forall x \in \mathbb R st,x^2 +2x - 5 \leq y$$ is true.
 A: $$|x|=\begin{cases}+x,\;x\ge0\\-x,\;x\le0\end{cases}$$
So,$$y=\frac{|x|}x=\begin{cases}+1,\;x>0\\-1,\;x>0\end{cases}$$
And for $x=0$, $|x|=xy\forall y\in\mathbb R$
$y$ can also be written as
$$y=\mathit{sgn}(x),\;\forall x\ne0\wedge y\in \mathbb{R},\;x=0$$

Actually, $$\exists y \in \mathbb R, \forall x \in \mathbb R \text{ such that } x^2 +2x - 5 \geq y$$
Since,
$$x^2 +2x - 5= (x+1)^2 -6 \ge -6 \qquad \forall x \in \mathbb R $$ 
Since,
$$(x+1)^2\ge 0 \qquad \forall x \in \mathbb R$$ 
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Here is another way to write down the proof for your first question.  I'm using a slightly different notation that I personally think is better for proofs like this.
(Note that this is not "proper notation", since I don't think that exists.  There are only different conventions, preferences, and opinions.  I happen to like those from EWD1300.)
For all $\;x \in \mathbb R\;$,
$$\calc
\langle \exists y : y \in \mathbb R : |x| = x \times y \rangle
\calcop{\equiv}{divide by $\;x\;$, with special case $\;x = 0\;$ using $\;|0| = 0 \times y\;$}
\langle \exists y : y \in \mathbb R : x = 0 \;\lor\; \frac{|x|}x = y \rangle
\calcop{\equiv}{logic: pull part not using $\;y\;$ out of $\;\exists y\;$}
x = 0 \;\lor\; \langle \exists y : y \in \mathbb R : \frac{|x|}x = y \rangle
\calcop{\equiv}{logic: one-point rule for $\;y\;$}
x = 0 \;\lor\; \frac{|x|}x \in \mathbb R
\calcop{\equiv}{arithmetic: right-hand side is always true (for $\;x \not= 0\;$); simplify}
\text{true}
\endcalc$$
