# Thread: How can a be irrational if it's exponent is an integer?

1. ## Re: How can a be irrational if it's exponent is an integer?

So basically because the number is irrational, then that number times itself any amount of times will also yield an irrational product.

2. ## Re: How can a be irrational if it's exponent is an integer?

Originally Posted by lamentofking
So basically because the number is irrational, then that number times itself any amount of times will also yield an irrational product.
No. For example, $\sqrt{2}$ is an irrational number, but $\left(\sqrt{2}\right)^2$ is rational. However, any integer power of a rational number is rational.

You need to learn how to prove universal statement, implications and negations. In this case, you fix an arbitrary $a$ and assume that there exists a $b$ such that $a^b$ is irrational. Then you need to show that $a$ is not rational. You prove this by contradiction, i.e., you assume that $a$ is rational. Then $a^b$ is rational as well, contrary to the assumption.

3. ## Re: How can a be irrational if it's exponent is an integer?

Originally Posted by lamentofking
So basically because the number is irrational, then that number times itself any amount of times will also yield an irrational product.
sqrt2xsqrt2 is only defined in the set of real numbers, in which case sqrt2xsqrt2=2 only says realxreal is real. It’s a fine point of algebra. Google it. Your question is a good one.

4. ## Re: How can a be irrational if it's exponent is an integer?

Can't edit, so: You can't extrapolate from my post 15 that irrationalxirrational is irrational (which I believe you meant) because multiplication is not defined for the set of irrational numbers. Within the set of real numbers, emakorov is correct, you can have irrational times irrational is rational.

5. ## Re: How can a be irrational if it's exponent is an integer?

So in order to prove this statement I need to prove why a cannot be rational?

6. ## Re: How can a be irrational if it's exponent is an integer?

Originally Posted by lamentofking
So in order to prove this statement I need to prove why a cannot be rational?
Yes, that's what being irrational means.

7. ## Re: How can a be irrational if it's exponent is an integer?

Originally Posted by lamentofking
I have the following problem:

Prove that if ab is irrational for any arbitrary positive integer b, then a
is irrational.
I assume that's the statement you are referring to, in which case, as I stated much earlier:
If a were rational a^b would be rational. So a can't be rational, ie, it must be irrational since there are no other options in real numbers.

In my personal opinion, the issue has been confused by a lot of irrelevant discussion.

8. ## Re: How can a be irrational if it's exponent is an integer?

Why? a variable might be real that means rational / irrational and also it may be an imaginary one. it depends on the context were we are using it.

9. ## Re: How can a be irrational if it's exponent is an integer?

I agree this could be one approach.
let a be a rational number. that means a = c/d. now a^b = (c/d)^b = (c^b)/ (d^b) that means a^b is rational similarly proceed for when a is irrational.

10. ## Re: How can a be irrational if it's exponent is an integer?

Originally Posted by lamentofking
I have the following problem:

Prove that if ab is irrational for any arbitrary positive integer b, then a
is irrational.

I was thinking that if a was already an irrational number like the square root of 2. and b was equal to 3 then I would have the cubed root of 22 which is :

1.58740105197

Is this sufficient enough to prove a statement? I don't need to use any type of laws. Just need to prove if it's true/false.
Suppose $a$ is rational and that $a^b$ is irrational for every positive integer $b$. But $1$ is a positive integer and $a^1=a$ is rational - a contradiction. Hence if $a^b$ is irrational for every positive integer $b$ then $a$ cannot be rational and so must be irrational.

.

11. ## Re: How can a be irrational if it's exponent is an integer?

If a is rational, a^b can't be irrational. N/MxN/M=N^2/M^2

The context of the problem is assumed to be the set of real numbers.

rational number: a number which can be expressed as N/M where N and M are integers.
irrational number: a number which can't be expressed as N/M where N and M are integers, for ex: pi, sqrt2, e, ...

rational numbers and irrational numbers belong to the set of real numbers.

If a is rational, a^b is rational. If a is irrational, it does not follow that a^b is irrational. As emakorov pointed out sqrt2xsqrt2=2, or irrationalxirrational = rational. However, as we have pointed out countless times now, if a^b is irrational, a must be irrational (again, for if a were rational, a^b wouild be rational)

12. ## Re: How can a be irrational if it's exponent is an integer?

Thank you everyone. I do believe the problem has been made clear now. :-)

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