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Prove that √5 is a irrational no. |
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Answer» Step-by-step explanation: lets assume ROOT 5 is rational ... rand rational number cn be PRESENT in p/q form so ROOT5=p/q squaring both side root5^2=p^2/q^2 5=p^2/q^2 p^2in factor of 5 q^2is factor of 5 then 5=ak(any constant) squaring 5^2=a^2k^ 25=a^2k^2 25 is factor of a 25 is factor of k hence our assumption was wrong root5 is irrational |
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