# Proof of the Sum of Square Roots

I have a question about a problem I encountered:

$\exists$ a,b $\epsilon$ $\mathbb{R}$+ such that $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$

Any tips for going about solving this?

I tried:

$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$

$a+b=a+b$

I have a feeling this isn't a legal operation...

-
$(X+Y)^2 = X^2 + Y^2 + 2 X Y$. – Ron Gordon Jan 31 '13 at 21:18
$\exists a>0,b>0$ with $\sqrt{a+b} = \sqrt{a}+\sqrt{b}$? The problem is there is no such $a$ and $b$. – Anon Jan 31 '13 at 21:20

## 2 Answers

$\textbf{Hint:}$ Suppose such $a,b\in \mathbb{R}^+$ do exist, then square both sides of $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$.

-

i think you just need to find the value of $a$ and $b$ so that the conditions are met but we know that only $a=0$ or $b=0$ such that conditions met is $0\in\mathbb{R^+}$? if yes then the statement is true

-
$\sqrt{2+2} = 2 \neq 2\sqrt{2} = \sqrt{2} +\sqrt{2}$. – Git Gud Jan 31 '13 at 21:26
ahh,i'm sorry haha – A Ricko Maulidar Jan 31 '13 at 21:34
i think the conditions only met when $a=b=0$ isn't it? – A Ricko Maulidar Jan 31 '13 at 21:35
@A Ricko Maulidar No, the conditions are met if, and only if, $a=0 \vee b=0$. But notice that the OP wants solutions in $\mathbb{R}^+$, so there are none. – Git Gud Jan 31 '13 at 21:35
yes,that's what i mean haha :p i'm sorry :) thankyou for the correction :) – A Ricko Maulidar Jan 31 '13 at 21:39