Question:
Prove √2k is irrational where k is an odd integer.
My attempt:
Proof by contradiction:
Now, assume √2k is rational. Then, √2k=ab where a,b∈Z, b not equal 0 and a,b have no common factors.
√2k=ab⟹2k=a2b2⟹(b2)(2k)=a2⟹2|a2⟹2|a, since 2 is prime⟹∃c∈Z such that a=2c
Then, (2k)(b2)=a2⟹(2k)(b2)=4c2⟹kb2=2c2⟹2|kb2⟹2|b2⟹2|b, since 2 is prime.
So, 2|a and 2|b , a contradiction.
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