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.
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It is indeed not true that $\sqrt{x}+\sqrt{y}=\sqrt{x+y}$ in general, but it is true that $\sqrt{x+y}\leq \sqrt{x}+\sqrt{y}$ for $x,y\in \mathbb{R}^+$, which is easy to prove. – projectilemotion Nov 02 '17 at 09:18
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$\sqrt x + \sqrt y\neq \sqrt{x+y}$ for the same reason as $(x+y)^2\neq x^2+y^2$. – Ennar Nov 02 '17 at 09:22
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You recieved 2 answers to your question. If any of them is what you need, you should accept it by clicking the grey tick mark below the question score. Otherwise, explain what is still missing in the answers! – 5xum Nov 02 '17 at 10:18
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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 $\sqrt{1+1}=\sqrt2<2$.
José Carlos Santos
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remember what $^2$ is:
$$u^2=u\cdot u\\\text{so: }\left(\sqrt{a}\cdot\sqrt{b}\right)^2=\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{a}\cdot\sqrt{b}=\sqrt{a}\cdot\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{b}=a\cdot b=\sqrt{ab}^2\implies\sqrt x \cdot\sqrt y = \sqrt {xy}\\\text{and you can do: }\left(\sqrt{a}+\sqrt{b}\right)^2=\left(\sqrt{a}+\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)=\sqrt{a}^2+2\sqrt{a}\cdot\sqrt{b}+\sqrt{b}^2{=a+2\sqrt{a}\cdot\sqrt{b}+b}=\sqrt{a+2\sqrt{a}\cdot\sqrt{b}+b}^2\ne\sqrt{a+b}^2\implies\sqrt{a}+\sqrt{b}\ne\sqrt{a+b}$$
ℋolo
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