Solution to Prove that using proof by contradiction. √2 + √6 < √15 - Sikademy
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Prove that using proof by contradiction. √2 + √6 < √15

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Assume the result is False; i.e. assume that \sqrt{6}+\sqrt{2}\geq\sqrt{15}.

Then


\sqrt{6}+\sqrt{2}\geq\sqrt{15}>0=>(\sqrt{6}+\sqrt{2})^2\geq(\sqrt{15})^2

=>6+2\sqrt{12}+2\geq15=>4\sqrt{3}\geq7

=>(4\sqrt{3})^2\geq(7)^2=>48\geq49,

which is contradiction.

Hence we have proved that \sqrt{6}+\sqrt{2}<\sqrt{15}.


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