Prove that cube root of 7 is an irrational number?

Answer 1

The proof is by the method of reductio ad absurdum. We start by assuming that cuberoot(7) is rational. That means that it can be expressed in the form p/q where p and q are co-prime integers. Thus cuberoot(7) = p/q. This can be simplified to 7*q^3 = p^3 Now 7 divides the left hand side (LHS) so it must divide the right hand side (RHS). That is, 7 must divide p^3 and since 7 is a prime, 7 must divide p. That is p = 7*r for some integer r. Then substituting for p gives, 7*q^3 = (7*r)^3 = 343*r^3 Dividing both sides by 7 gives q^3 = 49*r^3. But now 7 divides 49 so 7 divides the RHS. Therefore it must divide the LHS. That is, 7 must divide q^3 and since 7 is a prime, 7 must divide q. But then we have 7 dividing p as well as q which contradicts the requirement that p and q are co-prime. The contradiction implies that cuberoot(7) cannot be rational.