Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
5  
$(X+Y)^2 = X^2 + Y^2 + 2 X Y$. –  Ron Gordon Jan 31 '13 at 21:18
1  
$\exists a>0,b>0$ 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 2

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

share|improve this answer

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

share|improve this answer
    
$\sqrt{2+2} = 2 \neq 2\sqrt{2} = \sqrt{2} +\sqrt{2}$. –  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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.