Question

prove that by the method of contradiction any prime no. is odd

Answer 1

Answer:

Let us assume that the square root of the prime number p is rational. Hence we can write p–√=ab. (In their lowest form.) Then p=a2b2, and so pb2=a2.

Hence p divides a2, so p divides a. Substitute a by pk. Find out that p divides b. Hence this is a contradiction as they should be relatively prime, i.e., gcd(a,b)=1.

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