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How to prove that square-root of 'n'(a natural number) is irrational?

Asked by Vashi Vyas 8th September 2013, 7:39 PM
Answered by Expert
Answer :

Assume x=m/n where mbelongs to Z and nbelongs to N

this implies x=sqaure root(m)/ square root(n)

which means that for an integer to have a rational root it's root must be the ratio of two integer roots. For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity.

From contradiction Hence Proved

Answered by Expert 9th September 2013, 12:43 PM
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