To prove cube root of 26 is irrational?

Answer 1

The proof is by the method of reductio ad absurdum. We start by assuming that cuberoot of 26, cbrt(26), is rational. That means that the cube root can be expressed in the form p/q where p and q are co-prime integers. That is, cbrt(26) = p/q.Therefore, p^3/q^3 = 26 which can also be expressed as 26*q^3 = p^3 Now 26 = 2*13 so 2 divides the left hand side (LHS) and therefore it must divide the right hand side (RHS). That is, 2 must divide p^3 and since 2 is a prime, 2 must divide p. That is p = 2*r for some integer r. Then substituting for p gives, 26*q^3 = (2*r)^3 = 8*r^3 Dividing both sides by 2 gives 13*q^3 = 4*r^3. But now 2 divides the RHS so it must divide the LHS. That is, 2 must divide q^3 and since 2 is a prime, 2 must divide q. But then we have 2 dividing p as well as q which contradicts the requirement that p and q are co-prime. The contradiction implies that cbrt(26) cannot be rational.