How to prove that square-root of 'n'(a natural number) is irrational?
Asked by Vashi Vyas | 8th Sep, 2013, 07:39: PM
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 | 9th Sep, 2013, 12:43: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number