Question

Prove that if \({x^3}\) is irrational, then x is irrational.

Short answer

It is shown that if \({x^3}\) is irrational, then \(x\) is irrational.

Step-by-step solution

Important property

If a is a rational number, then there exist integers b and c such that \(a = \frac{b}{c}\).

Prove that if \({x^3}\) is irrational, then x is irrational

The given statement can be proved by contradiction.

\({x^3}\)is irrational and thus there do not exist integers \(v\) and \(w\) such that \({x^3} = \frac{v}{w}\).

Assume that \(x\) is rational then there exist \(y\) and \(z\) such that \(x = \frac{y}{z}\).

Take the 3th power of each side of the equation \(x = \frac{y}{z}\).

\(\begin{array}{c}{x^3} = {\left( {\frac{y}{z}} \right)^3}\\ = \frac{{{y^3}}}{{{z^3}}}\end{array}\)

Since \(y\) and \(z\) are integers, \({y^3}\) and \({z^3}\) are also integers and thus, it is shown that \({x^3}\) is rational. However, \({x^3}\) is known to be irrational and thus, the contradiction is obtained. Therefore, the assumption that \(x\) is rational is then not correct and thus, \(x\) is irrational.

Therefore, it is shown that if \({x^3}\) is irrational, then \(x\) is irrational.