Question

Prove that \(\sqrt[3]{2}\) is irrational.

Short answer

\(\sqrt[3]{2}\) is irrational

Step-by-step solution

Introduction

Any real value which cannot be represented as the quotient of two integral values is said to be an irrational value.

Proof using contradiction

Suppose that \(\sqrt[3]{2}\)is rational.

Then by definition, there exist integral values\(p\,\,and\,\,q\)with\(q \ne 0\)such that\(\sqrt[3]{2} = \frac{p}{q}\) , where \(p\,\,and\,\,q\)have no common factors, i.e.\(gcd\left( {p,q} \right) = 1\)

Now\(\sqrt[3]{2} = \frac{p}{q}\)

Cubing both sides,

Conclusion

Both\(p\,\,and\,\,q\)are even.

Thus, 2 divides both\(p\,\,and\,\,q\).

Thus,\(p\,\,and\,\,q\)have common factors.

That is the statement \(\sqrt[3]{2}\)is irrational is true.

Therefore, proved that \(\sqrt[3]{2}\)is irrational.