# If a, b are irrational, then is $a^b$ irrational?

## Homework Statement

True or false and why: If a and b are irrational, then $a^b$ is irrational.

## Homework Equations

None, but the relevant example provided in the text is the proof of irrationality of $\sqrt{2}$

## The Attempt at a Solution

Attempt proof by contradiction. Say $a^b$ is rational, then we can express
$a^b = \frac{c}{d}$
for some integer c and d. Without loss of generality, we can choose c and d to express the fractions in lowest terms.
Then, $a^b d = c$ which implies that $a^b$ must be an integer. This seems to me like a suspicious statement. The problem reduces to whether or not we can have $a^b = c$ where c is an integer. How would I show this possible/ impossible, if indeed this is the correct path to a solution?

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Mark44
Mentor

## Homework Statement

True or false and why: If a and b are irrational, then $a^b$ is irrational.

## Homework Equations

None, but the relevant example provided in the text is the proof of irrationality of $\sqrt{2}$

## The Attempt at a Solution

Attempt proof by contradiction. Say $a^b$ is rational, then we can express
$a^b = \frac{c}{d}$
for some integer c and d. Without loss of generality, we can choose c and d to express the fractions in lowest terms.
Then, $a^b d = c$ which implies that $a^b$ must be an integer. This seems to me like a suspicious statement. The problem reduces to whether or not we can have $a^b = c$ where c is an integer. How would I show this possible/ impossible, if indeed this is the correct path to a solution?
Can you raise e (which is irrational) to some irrational power, but still get a rational number?

Can you raise e (which is irrational) to some irrational power, but still get a rational number?
Yes, if we raise it to the natural log of some rational number. $e^ {\ln{ \frac{c}{d} }} = \frac{c}{d}$.

Applying this to the original statement, we would have $b = \log_{a} \frac{c}{d}$. Do we have to prove that $\log_{a} \frac{c}{d}$ is irrational to finish this? My attempt to do so: assume it is rational. Then $\log_{a} \frac{c}{d} = \frac{e}{f}$ where e, f are integers such that fraction is in lowest terms. Then $\frac{c}{d} = a^{\frac{e}{f}}$. I'm tempted to say that raising an irrational number to a rational power (other than 0) will result in an irrational and hence a contradiction, but I'm not sure how to prove. Where do I go from here?

Mark44
Mentor
More simply $e^{\ln(2)} = 2$. Surely $\ln(2)$ is irrational...

Math_QED
Homework Helper
2019 Award

## Homework Statement

True or false and why: If a and b are irrational, then $a^b$ is irrational.

## Homework Equations

None, but the relevant example provided in the text is the proof of irrationality of $\sqrt{2}$

## The Attempt at a Solution

Attempt proof by contradiction. Say $a^b$ is rational, then we can express
$a^b = \frac{c}{d}$
for some integer c and d. Without loss of generality, we can choose c and d to express the fractions in lowest terms.
Then, $a^b d = c$ which implies that $a^b$ must be an integer. This seems to me like a suspicious statement. The problem reduces to whether or not we can have $a^b = c$ where c is an integer. How would I show this possible/ impossible, if indeed this is the correct path to a solution?
As your text refers to the irrationality of $\sqrt{2}$

Consider: $\sqrt{2}^{\sqrt{2}}$. If this is rational, we are done. Otherwise, when it is irrational, we have $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{2} = 2$. Thus, surely any of these two numbers will be the number you seek. Note that this proof does not show which one (But I believe it's the second one)

Math_QED
Homework Helper
2019 Award
More simply $e^{\ln(2)} = 2$. Surely $\ln(2)$ is irrational...
Why is ln(2) irrational?

fresh_42
Mentor
Just as a remark on the subject in case you're interested in: Theorem of Gelfond-Schneider

If $a$ and $b$ are algebraic numbers (e.g. rational) with $a \notin \{0,1\}$ and $b$ irrational, then any value of $a^b$ is a transcendental number (which are irrational).

Isaac0427
Gold Member
My favorite example of this (maybe a stretch because it uses imaginary numbers) is simply @Math_QED s profile picture.

$e^{i\pi}=-1$

Even though it is imaginary, I still believe that $i\pi$ is irrational, because it is i*3.14159265.... (it still goes on forever).

mfb
Mentor
As the original question is answered, but the flaw in the attempt didn't get pointed out:

## The Attempt at a Solution

Attempt proof by contradiction. Say $a^b$ is rational, then we can express
$a^b = \frac{c}{d}$
for some integer c and d. Without loss of generality, we can choose c and d to express the fractions in lowest terms.
Then, $a^b d = c$ which implies that $a^b$ must be an integer.
It does not. Ignoring the details of the exponent, imagine $a^b = \frac 5 3$. Therefore, $a^b \cdot 3 = 5$. Does that imply $a^b$ has to be an integer?

Mark44
Mentor
Even though it is imaginary, I still believe that iπ is irrational, because it is i*3.14159265.... (it still goes on forever).
The irrational numbers are a subset of the reals. $i\pi$ is a pure-real complex number.

Mark44
Mentor
mfb
Mentor
Mark44
Mentor
That number wouldn't help for the original question here - 2log_2(3) is rational, but 2 is rational as well.
I wasn't giving it as another possible counterexample, but rather in answer to Isaac's question. Proving that $\log_2(3)$ is irrational is pretty easy, but proving that $\ln(2)$ is irrational would be quite a bit harder.

fresh_42
Mentor
Most of the numbers produced by the natural log function are themselves transcendental, a subcategory or the irrationals, with the other subcategory being algebraic numbers.
This is also interesting in this context (and the first time I heard of / read about it):
https://en.wikipedia.org/wiki/Transcendental_number_theory

LCKurtz
As your text refers to the irrationality of $\sqrt{2}$
Consider: $\sqrt{2}^{\sqrt{2}}$. If this is rational, we are done. Otherwise, when it is irrational, we have $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{2} = 2$. Thus, surely any of these two numbers will be the number you seek. Note that this proof does not show which one (But I believe it's the second one)
Nice! Best argument of the bunch because it only uses the irrationality of $\sqrt 2$.