STEP II 2003 question 3 solution

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Prove that the cube root of any irrational number is an irrational number

Solution

First, recall that a rational number is one that can be expressed in the form p/q, where p and q are integers. An irrational number is a number that cannot be expressed in such a way. Now take an irrational number x. Suppose that the cube root of x is rational, say p/q for some integers p and q. Then x = (p^3)/(q^3) which is a contradiction since x is irrational. So the cube root of x must be irrational.

Let u(n) = 5^(1/(3^n)). Given that the cube root of 5 is irrational, prove by induction that u(n) is an irrational number for every positive integer n

u(1) = the cube root of 5, which we are told is an irrational number. Now suppose u(k) is irrational for some positive integer k. We need to show that u(k+1) is irrational. But u(k+1) is equal to the cube root of u(k), so from the first part of the question we can deduce that u(k+1) is irrational, as required. So, by the principle of induction, u(n) is an irrational number for every positive integer n.

Hence, or otherwise, give an example of an infinite sequence of irrational numbers which converges to a given integer m.

Look at the sequence u(n). As n gets bigger, 1/(3^n) converges to 0, so that u(n) converges to 1. Now note that if x is irrational and m is an integer, then x+m-1 is irrational. This is because if x+m-1 were rational, then we could write x+m-1=p/q for some integers p and q. But then x = p/q + 1 - m = (p+q(1-m))/q which is a contradiction since x is irrational. Observe also that if w(n) is an infinite sequence converging to a then w(n) + b is an infinite sequence converging to a+b.

So for m an integer, u(n) + m - 1 is an infinite sequence of irrational numbers converging to m.