Prove that square root of 6 is irrational by contradiction .

atgnybo4fq

atgnybo4fq

Answered

2022-11-22

Prove that square root of 6 is irrational by contradiction .

Answer & Explanation

Pignatpmv

Pignatpmv

Expert

2022-11-23Added 22 answers

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.

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