Prove that root 6 is a …

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Yogita Ingle 5 years, 11 months ago
Let √6 be a rational number , then
√6 = p÷q , where p,q are integers , q not = 0 and p,q have no common factors ( except 1 )
=> 6 = p² ÷ q²
=> p² = 2q² ................(i)
As 2 divides 6q² , so 2 divides p² but 2 is a prime number
=> 2 divides p
Let p = 2m , where m is an integer .
Substituting this value of p in (i) , we get
(2m)² = 6q²
=> 2m² = 3q²
As 2 divides 2m² , 2 divides 3q²
=> 2 divides 3 or 2 divide q²
But 2 does not divide 3 , therefore , 2 divides q²
=> 2 divides q
Thus , p and q have a common factor 2 . This contradicts that p and q have no common factors ( except 1 ).
Hence , our supposition is wrong . Therefore , √6 is an irrational number.
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