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#1 2024-05-07 10:32:10

mathxyz
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From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Irrational Number Raised To Irrational Number

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

Let A = sqrt{2}^(sqrt{2})

If A is rational, we are done. Why?

If A is irrational, we are done. Why?

Hint given in textbook: Consider A^(sqrt{2}).

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#2 2024-05-08 03:03:45

Bob
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Registered: 2010-06-20
Posts: 10,456

Re: Irrational Number Raised To Irrational Number

You'll be amused to hear that I've spent all day puzzling over this one. What had me stuck is how to show that A is irrational.

Then  after a pleasant walk in some public gardens it dawned on me you don't ever have to know this! The notes are telling you (me) to cover both cases


So here we go; first assume A is rational. In that case we have an irrational (root 2) to the power of an irrational is rational
Job done.

So what if A is irrational? Take the hint and evaluate A^(root 2)

Using (x^n)^m = x^(nm) we have

A^root2 =(root2^root2)^root2

= root2^(root2 times root2) =root2^2=2

So once again an irrational ^ an irrational comes out rational

Job done.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2024-05-08 09:41:42

mathxyz
Banned
From: Brooklyn, NY
Registered: 2024-02-24
Posts: 1,053

Re: Irrational Number Raised To Irrational Number

Bob wrote:

You'll be amused to hear that I've spent all day puzzling over this one. What had me stuck is how to show that A is irrational.

Then  after a pleasant walk in some public gardens it dawned on me you don't ever have to know this! The notes are telling you (me) to cover both cases


So here we go; first assume A is rational. In that case we have an irrational (root 2) to the power of an irrational is rational
Job done.

So what if A is irrational? Take the hint and evaluate A^(root 2)

Using (x^n)^m = x^(nm) we have

A^root2 =(root2^root2)^root2

= root2^(root2 times root2) =root2^2=2

So once again an irrational ^ an irrational comes out rational

Job done.

Bob

Wow! Great reply and more study notes for me.

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