Irrational Number Raised To Irrational Power

[nycmathguy]

Set 1.1
Question 60

See attachment.

I look for questions that create a discussion. This one certainly pushes one to think a bit more clearly. Work out (a) and (b).

NOTE: No need to work out every question posted tonight. You can take your time and spread your math work for several days if too busy.

 

[MathLover1]

60.
Can an irrational number raised to an irrational power yield an answer that is rational?

a.
let A=(sqrt(2))^(sqrt(2))

Let’s consider the number (sqrt(2))^(sqrt(2)) . This number is either rational or irrational. Let’s examine each case.

Case 1: rational or irrational
Recall that sqrt(2) is irrational. So if sqrt(2) ^(sqrt(2)) is rational, then we have proven that it’s possible to raise an irrational number to an irrational power and get a rational value. Done!

It turns out that it’s the second. The Gelfond–Schneider theorem tells us that for any two non-zero algebraic numbers a and b with a ≠ 1 and b irrational, the number a^b is irrational.
So sqrt(2) ^(sqrt(2)) is in fact irrational.

b.

Case 2: (sqrt(2))^(sqrt(2)) is irrational.

In this case, (sqrt(2))^(sqrt(2)) and sqrt(2) are both irrational numbers. So what if we raise the (sqrt(2))^(sqrt(2)) to the power of sqrt(2) ?

(sqrt(2))^(sqrt(2)) ^(sqrt(2)) ) = sqrt(2) ^ (sqrt(2)*sqrt (2)) = (sqrt( 2)) ^ 2 = 2 which is rational

 

[nycmathguy]

60.
Can an irrational number raised to an irrational power yield an answer that is rational?

a.
let A=(sqrt(2))^(sqrt(2))

Let’s consider the number (sqrt(2))^(sqrt(2)) . This number is either rational or irrational. Let’s examine each case.

Case 1: rational or irrational
Recall that sqrt(2) is irrational. So if sqrt(2) ^(sqrt(2)) is rational, then we have proven that it’s possible to raise an irrational number to an irrational power and get a rational value. Done!

It turns out that it’s the second. The Gelfond–Schneider theorem tells us that for any two non-zero algebraic numbers a and b with a ≠ 1 and b irrational, the number a^b is irrational.
So sqrt(2) ^(sqrt(2)) is in fact irrational.

b.

Case 2: (sqrt(2))^(sqrt(2)) is irrational.

In this case, (sqrt(2))^(sqrt(2)) and sqrt(2) are both irrational numbers. So what if we raise the (sqrt(2))^(sqrt(2)) to the power of sqrt(2) ?

(sqrt(2))^(sqrt(2)) ^(sqrt(2)) ) = sqrt(2) ^ (sqrt(2)*sqrt (2)) = (sqrt( 2)) ^ 2 = 2 which is rational

Great reply. Can you check out my thread Using Triangle Inequality?