Prove that square root of 6 is irrational by contradiction .

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Pignatpmv · Accepted Answer

Assume that       (    6  ) is rational.Then       (    6  )  =      p    q   where p and q are coprime integers.            6        2    =  6  =            p      2              q      2            p    2    =  6      q    2  therefore       p    2   is an even number since an even number multiplied by any other integer is also an even number. If       p    2   is even then p must also be even since if p were odd, an odd number multiplied by an odd number would also be odd.So we can replace p with 2k where k is an integer.  (  2  k      )    2    =  6      q    2    4      k    2    =  6      q    2    2      k    2    =  3      q    2  Now we see that   3      q    2   is even. For   3      q    2   to be even,       q    2   must be even since 3 is odd and an odd times an even number is even. And by the same argument above, if       q    2   is even then q is even.So both p and q are even which means both are divisible by 2. But that means they are not coprime, contradicting our assumption so sqrt(6) is not rational.

Prove that square root of 6 is irrational by contradiction .

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