Irrational Raised to Irrational

[nycmath]

Precalculus
David Cohen
Edition 3
Chapter 1, Section 1.1

66. Can an Irrational number raised to an irrational power yield an answer that is rational?

(a) Let A = (1/2)^(1/2)

Now, either A is rational or A is irrational. If A is rational, we are done. Why?

(b) If A is irrational, we are done. Why?

Hint: Consider A^(1/2)

[Math Tutor]

Yes. Let [tex]A=\left(\sqrt{2}\right)^{\sqrt{2}}[/tex] - irrational base, irrational exponent.

(a) If A is rational, we are done: A itself is the example.

(b) If A is irrational, raise it to [tex]\sqrt{2}[/tex]:

[tex]A^{\sqrt{2}}=\left(\sqrt{2}\right)^{\sqrt{2}\cdot\sqrt{2}}=\left(\sqrt{2}\right)^{2}=2[/tex]

Irrational base A, irrational exponent, rational answer. Done again.

A is one or the other, so such a pair exists - though the proof never tells us which case is the real one.

[Eigenvalue]

Did you mean sqrt(2)? 0.5 is rational as it can be written as 1/2

[nycmath]

Did you mean sqrt(2)? 0.5 is rational as it can be written as 1/2

Let sqrt = square root

A = sqrt{2}^(sqrt{2})

This is the value of A for this example.

[nycmath]

Yes. Let [tex]A=\left(\sqrt{2}\right)^{\sqrt{2}}[/tex] - irrational base, irrational exponent.

(a) If A is rational, we are done: A itself is the example.

(b) If A is irrational, raise it to [tex]\sqrt{2}[/tex]:

[tex]A^{\sqrt{2}}=\left(\sqrt{2}\right)^{\sqrt{2}\cdot\sqrt{2}}=\left(\sqrt{2}\right)^{2}=2[/tex]

Irrational base A, irrational exponent, rational answer. Done again.

A is one or the other, so such a pair exists - though the proof never tells us which case is the real one.

Thank you for another detailed and to the point reply.