Why is $\sqrt x \times \sqrt y = \sqrt {xy}$

Question

Sorry I do not know latex. Why is $\sqrt x \times \sqrt y = \sqrt {xy}$? It also applies for division, but why not addition and subtraction? i.e., why is $\sqrt x + \sqrt y$ not equal to $\sqrt {x+y}$? Thank you for editing and answering.

Answer 1

You were told in the comments where to find an answer to your first question. Now, why is it not true that $\sqrt{x+y}=\sqrt x+\sqrt y$? Because, for instance, $\sqrt1+\sqrt1=2$, whereas $\sqrt{1+1}=\sqrt2<2$.

Answer 2

remember what $^2$ is:

$$u^2=u\cdot u\\\text{so: }\left(\sqrt{a}\cdot\sqrt{b}\right)^2=\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{a}\cdot\sqrt{b}=\sqrt{a}\cdot\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{b}=a\cdot b=\sqrt{ab}^2\implies\sqrt x \cdot\sqrt y = \sqrt {xy}\\\text{and you can do: }\left(\sqrt{a}+\sqrt{b}\right)^2=\left(\sqrt{a}+\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)=\sqrt{a}^2+2\sqrt{a}\cdot\sqrt{b}+\sqrt{b}^2{=a+2\sqrt{a}\cdot\sqrt{b}+b}=\sqrt{a+2\sqrt{a}\cdot\sqrt{b}+b}^2\ne\sqrt{a+b}^2\implies\sqrt{a}+\sqrt{b}\ne\sqrt{a+b}$$