• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • Thread starter QuietMind
  • Start date
  • #1
42
2

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?
 

Answers and Replies

  • #2
33,152
4,836

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?
 
  • #3
42
2
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?
 
  • #4
33,152
4,836
More simply ##e^{\ln(2)} = 2##. Surely ##\ln(2)## is irrational...
 
  • #5
Math_QED
Science Advisor
Homework Helper
2019 Award
1,364
491

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)
 
  • #6
Math_QED
Science Advisor
Homework Helper
2019 Award
1,364
491
More simply ##e^{\ln(2)} = 2##. Surely ##\ln(2)## is irrational...
Why is ln(2) irrational?
 
  • #7
12,476
8,880
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).
 
  • #9
627
112
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).
 
  • #10
33,888
9,606
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?
 
  • #11
33,152
4,836
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.
 
  • #12
33,152
4,836
  • #13
33,888
9,606
  • #14
33,152
4,836
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.
 
  • #15
12,476
8,880
  • #16
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,515
733
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##.
 

Related Threads for: If a, b are irrational, then is ##a^b## irrational?

Replies
11
Views
4K
Replies
18
Views
3K
Replies
7
Views
1K
Replies
7
Views
2K
Replies
26
Views
17K
Replies
5
Views
2K
Replies
0
Views
2K
Top