Question

Prove or disprove that there is a rational number \(x\)and an irrational number \(y\)such that \({x^y}\)is irrational.

Short answer

For \(x\)as a rational number and \(y\)as irrational number, \({x^y}\)is an irrational number.

Step-by-step solution

Introduction

Consider \(x\)as the rational number and \(y\)as an irrational number.

Assume a rational number 2 as \(x\)and \(\sqrt 2 \)as \(y\).

That is,

\(\begin{aligned}x = 2\\y = \sqrt 2 \end{aligned}\)

Value of \({x^y}\)for \(x = 2\) and \(y = \sqrt 2 \)

Thus, the value of \({x^y}\)is calculated as,

\(\begin{aligned}{x^y} &= {2^{\sqrt 2 }}\\ &= {2^{1.41421356}}\\ &= 2.665144142690...\end{aligned}\)

It is irrational number.

Assumption and analysis for \(x = {2^{\sqrt 2 }}\) and \(y = \frac{{\sqrt 2 }}{4}\)

Assume \({2^{\sqrt 2 }}\) as a rational number and \(\frac{{\sqrt 2 }}{4}\) as an irrational number.

That is,

\(\begin{aligned}x &= {2^{\sqrt 2 }}\\y &= \frac{{\sqrt 2 }}{4}\end{aligned}\)

Calculate the value of \({x^y}\)as,

\(\begin{aligned}{x^y} &= {\left( {{2^{\sqrt 2 }}} \right)^{\frac{{\sqrt 2 }}{4}}}\\ &= {2^{\frac{{\sqrt 2 \left( {\sqrt 2 } \right)}}{4}}}\\ &= {2^{\frac{1}{2}}}\\ &= \sqrt 2 \end{aligned}\)

Here \(\sqrt 2 \)is irrational number.

Another example with\(x = 3\) and \(y = \sqrt 3 \)

Take another example.

Assume a rational number 3 as x and irrational number \(\sqrt 3 \)as y.

That is,

\(\begin{aligned}x &= 3\\y &= \sqrt 3 \end{aligned}\).

Thus, the value of \({x^y}\)is calculated as,

\(\begin{aligned}{x^y} &= {3^{\sqrt 3 }}\\ &= {3^{1.732058075688}}\\ &= 6.7049918538253100..\end{aligned}\)

It is irrational number.

Assumption and analysis for \(x = {3^{\sqrt 3 }}\) and \(y = \frac{{\sqrt 3 }}{6}\)

Assume \({3^{\sqrt 3 }}\)as a rational number and \(\frac{{\sqrt 3 }}{6}\)as an irrational number.

That is,

\(\begin{aligned}x &= {3^{\sqrt 3 }}\\y &= \frac{{\sqrt 3 }}{6}\end{aligned}\)

Calculate the value of \({x^y}\)as,

\(\begin{aligned}{x^y} &= {\left( {{3^{\sqrt 3 }}} \right)^{\frac{{\sqrt 3 }}{6}}}\\ &= {3^{\frac{{\sqrt 3 \left( {\sqrt 3 } \right)}}{6}}}\\ &= {3^{\frac{1}{2}}}\\ &= \sqrt 3 \end{aligned}\)

Here \(\sqrt 3 \)is irrational number.

Conclusion

Follow this procedure for n number of examples.

Therefore, there exists x as a rational number and y as irrational number such that \({x^y}\)is irrational.

Hence, for x as a rational number and y as irrational number, \({x^y}\) is an irrational number.