Question

show that underroot 4 is irrational number​

Answer 1

Step-by-step explanation:

Say 4–√ is rational. Then 4–√ can be represented as ab, where a and b have no common factors.

So 4=a2b2 and 4b2=a2. Now a2 must be divisible by 4, but then so must a (fundamental theorem of arithmetic). So we have 4b2=(4k)2 and 4b2=16k2 or even b2=4k2, which implies that b=4n by the fundamental theorem. Now we have a contradiction (since can note that both a and b are divisible by 4 and we assumed they were coprime)

This proof is clearly false, yet I fail to see where it differs. Where does it do so?