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

I have the following problem:

Prove that if a^{b} 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 2^{2} 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.

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

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Originally Posted by

**lamentofking** 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 2^{2} which is :

1.58740105197

First, $\displaystyle \left(\sqrt{2}\right)^3} =\sqrt{8}=2^{3/2} \ne\sqrt[3]{2^2}=\sqrt[3]{4}=2^{2/3}$. Second, you don't prove a claim by supposing that it is true. You need to show that *a* is irrational, so you can't assume it. On the other hand, you can assume that *a* is *not* irrational and show that it is impossible.

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

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Originally Posted by

**lamentofking** I have the following problem:

Prove that if a^{b} is irrational for any arbitrary positive integer b, then a

is irrational.

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.

**An example is never a proof.** So no that is not sufficient.

The statement is: If $\displaystyle a^b$ is irrational for every $\displaystyle b\in\mathbb{Z}^+$ then $\displaystyle a$ is irrational.

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

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**Plato** The statement is: If $\displaystyle a^b$ is irrational for every $\displaystyle b\in\mathbb{Z}^+$ then $\displaystyle a$ is irrational.

Arguably, the statement in the OP is stronger:

If $\displaystyle a^b$ is irrational for *some* $\displaystyle b\in\mathbb{Z}^+$ then $\displaystyle a$ is irrational.

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

But a variable cannot be irrational because it is a variable.

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

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**lamentofking** I have the following problem:

Prove that if a^{b} is irrational for **any arbitrary positive integer** b, then a

is irrational..

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Originally Posted by

**emakarov** Arguably, the statement in the OP is stronger:

If $\displaystyle a^b$ is irrational for *some* $\displaystyle b\in\mathbb{Z}^+$ then $\displaystyle a$ is irrational.

"any arbitrary positive integer" That means **some**?

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

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**Plato** "any arbitrary positive integer" That means **some**?

Well I thought in the world of math that if a statement was true then it was true for every case except where explicitly defined otherwise? But the question asks to prove that it's true in the sense that the statement is true now show why that is.

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

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**lamentofking** But a variable cannot be irrational because it is a variable.

You are saying this in response to what? In fact, unless you study formal logic, there is no distinction between a variable and its value. When you mention a variable, you refer to its value. Thus, we one says, "Let *a* be a rational (or irrational) number", one assumes that there is a rational (irrational) number and calls it *a*.

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Originally Posted by

**lamentofking** But the question asks to prove that it's true in the sense that the statement is true now show why that is.

You should prove it using the hint in post #2.

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**Plato** "any arbitrary positive integer" That means **some**?

It may. For example, "If you miss the deadline for any reason, you'll get only half the credit" means "(∃reason ...) ⇒ ..." or, equivalently, "∀reason (... ⇒ ...)", but not "(∀reason ...) ⇒ ...". I interpreted the statement the way I did because of this ambiguity of "any" and because a reasonable instructor would hardly ask to prove a statement for which a much stronger statement exists that can be proved equally easily.

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

If a is rational and b is an integer, a^b is rational. So a can't be rational.

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

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Originally Posted by

**emakarov** You are saying this in response to what? In fact, unless you study formal logic, there is no distinction between a variable and its value. When you mention a variable, you refer to its value. Thus, we one says, "Let *a* be a rational (or irrational) number", one assumes that there is a rational (irrational) number and calls it *a*.

You should prove it using the hint in post #2.

It may. For example, "If you miss the deadline for any reason, you'll get only half the credit" means "(∃reason ...) ⇒ ..." or, equivalently, "∀reason (... ⇒ ...)", but not "(∀reason ...) ⇒ ...". I interpreted the statement the way I did because of this ambiguity of "any" and because a reasonable instructor would hardly ask to prove a statement for which a much stronger statement exists that can be proved equally easily.

Can't I say that a will always be irrational regardless of what b equals as long as b does not equal 0?

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

I don't think you are understanding the key here. You can say it, but you haven't PROVEN it. What you need to suppose, is that, ALL YOU KNOW about the number a is that a^b is irrational for any natural number b. Based on this and based on ONLY THIS AND NOTHING ELSE, you need to establish that a must be irrational. You can't simply point out that "it would make sense" if your intended conclusion is the case, you need to show that it is the only possible case. For example, lets say someone asked you to prove that your friend Mike always told the truth. It is insufficient to point out an example of him telling the truth.

Now I think there might be a poor choice of words in your assigned problem or lost somewhere in communication. Your original post as literally interpreted is trivial. If a^b is irrational for EVERY natural b, then a^1 is irrational, and thus a is irrational. Q.E.D.

What I really think someone was trying to say, is that assume all you know is that a^b is irrational for SOME natural number b. Perhaps it's only one natural number or perhaps many or all, but all you know is that it's true for some b. The way to see this is to see what would happen if it WEREN'T the case that a is irrational. If a is rational, then any a^b must be rational, because it is simply a rational number multiplied by itself a finite number of times. This violates what you KNOW about a, specifically that a^b is irrational or some b. Therefore, it must be false that a is rational, hence a is irrational.

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

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**lamentofking** Can't I say that a will always be irrational regardless of what b equals as long as b does not equal 0?

What exactly is *a* here? And under what assumption is it irrational and why?

Your claim to prove is: For all real *a*, if there exists a positive integer b such that a^b is irrational, then *a* is irrational. Any proof must start with, "Fix an arbitrary real *a*". Then you need to prove an implication, so it is natural to assume the premise: "Assume there exists a positive integer b such that a^b is irrational". Then you need to prove that *a* is *not* rational. Negative statements are proved by contradiction, i.e., you say, "Assume that *a* *is* rational". Then see Hartlw's post.

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

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Originally Posted by

**lamentofking** I have the following problem:

Prove that if a^{b} is irrational for any arbitrary positive integer b, then a

is irrational.

This is an elementary problem in algebra. If a is rational, a^b is rational, so a has to be irrational.

What is the point of all the other posts? Why make something so simple so complicated?

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

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**Hartlw** What is the point of all the other posts? Why make something so simple so complicated?

Since this is posted in the Discrete Math forum, this problem is likely supposed to teach proofs (how to prove implications and universal statements, what are direct proofs and proofs by contradiction, etc.) and not algebra. So we were trying to explain the logic of this proof. Besides, there was a discussion about the exact language of the problem (some integer b vs all integer b, not that it matters for the proof).

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

If a belongs to S, then so does axa no matter how x is defined.

,ie,

rationalxrational=rational

complexcomplex=complex

.

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