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2015-04-14T20:13:12+05:30
(proof by contradiction)

Suppose \sqrt{5}+\sqrt{3} = \dfrac{a}{b} is rational.
Multiplying both sides by \sqrt{5}-\sqrt{3} gives
2=\dfrac{a}{b}(\sqrt{5}-\sqrt{3})\implies\sqrt{5}-\sqrt{3}=\dfrac{2b}{a}

Now (\sqrt{5}+\sqrt{3})+(\sqrt{5}-\sqrt{3})=\dfrac{a}{b}+\dfrac{2b}{a}
2\sqrt{5}=\dfrac{a^2+2b^2}{ab}
\sqrt{5}=\dfrac{a^2+2b^2}{2ab}

Which is a contradiction because \sqrt{5} is not rational. 

The only resolution to this contradiction is that the starting assumption is wrong. That is, \sqrt{5}+\sqrt{3} is irrational.
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Prove by taking a.
what a ? could you elaborate
2015-04-15T08:53:06+05:30
√3 +√5 let a is an rational number 
√3 +√5 = asquaring on both sides
(√3 +√5 )² = a²3+5+2√3 √5  = a²
8+2√15 = a²
a²-8 = 2√15
 \frac{ a^{2} -8}{2}  = √15
if a is rational then  \frac{ a^{2} -8}{2}  is also a rational
but √15 is a irrational
it is contradict to our assumption.
so √3 +√5 is not a rational
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