Question

prove that seven is not a cube of rational number​

Answer 1

I would be inclined to mimic the classic Euclid proof that √ 2 2 is irrational. Assume that √ 7 7 is rational. That is, assume √ 7 = m n 7=mn for integers m and n, reduced to lowest terms. Then, cubing both sides, 7 = m 3 n 3 7=m3n3 so 7n3= m3. That tells us that m3 is a multiple of 7. Can you use that to prove that m itself must be a multiple of 7? Remember that in proving that if m2 is even, the m must be even, we have to show that the square of an odd number is always odd. Here, you will have to look at numbers that are not multiples of 7. Is it possible for the third power of a number that is not a multiple of 7 to be a multiple of 7? You will have to look 6 different cases