# Which is bigger, a^b or b^a?

1. Apr 13, 2005

### aa

Which is bigger, a^b or b^a? (set theory)

Hi!

Thanks for letting me join your physics forums!

Will anyone help me with a set theory question I have? I've been racking my brains over this for the last two hours with no progress.

Which is greater using ordinal exponentation: $$\omega^{\omega_1}$$ or $$\omega_1^{\omega}$$?

P.S. I know that $$\omega^{\omega_1}$$ equals the order type of $$\underbrace{ \omega \times \omega \times \omega \times ... }_{\omega_1 \ many \ times}$$, and $$\omega_1^{\omega}$$ equals the order type of $$\underbrace{ \omega_1 \times \omega_1 \times \omega_1 \times ... }_{\omega \ many \ times}$$, but I'm still stuck.

Last edited: Apr 14, 2005
2. Apr 14, 2005

### dextercioby

It depends,they can be equal even (for the a=b case and for 2 & 4,for example).So you can't formulate a general rule...

Daniel.

3. Apr 14, 2005

### aa

Yes, but this is referring to ordinal exponentiation, a la set theory.

I really appreciate any help you all can give.

Last edited: Apr 14, 2005
4. Apr 14, 2005

### Hurkyl

Staff Emeritus
What exactly is &omega;_1?

5. Apr 14, 2005

### aa

Sorry, I see what you mean. This is the problem as given to me on the homework, but I have edited my post to use aleph notation. Is it a well-defined question now?

6. Apr 14, 2005

### Hurkyl

Staff Emeritus
Ack, alephs are usually used for cardinals aren't they? They're not order types! And cardinal exponentiation is different than ordinal exponentiation!

You could just say that &omega;_1 was the first ordinal with cardinality greater than that of &omega;. :tongue2:

7. Apr 14, 2005

### aa

Hehe, alright, thanks. That's exactly what $$\omega_1$$ is supposed to be.

8. Apr 14, 2005

### Hurkyl

Staff Emeritus
Well, ω1 is a limit ordinal, so what does that tell you about ω^ω1?

9. Apr 14, 2005

### aa

$$\omega^{\omega_1} = sup\{\omega^\alpha : \alpha < \omega_1\}$$. I did think of that, but don't see how it helps.

10. Apr 14, 2005

### Hurkyl

Staff Emeritus
What's the cardinality of &omega;^&alpha; if &alpha; < &omega;_1?

11. Apr 14, 2005

### aa

It's $$\omega$$. So then $$\omega^{\omega_1}$$ is an ordinal with cardinality equal to the union of $$\omega_1$$ many countable ordinals, which means $$\omega^{\omega_1}$$ is an ordinal with cardinality $$\omega_1$$.

So we've shown that $$|\omega^{\omega_1}|$$ = $$\omega_1$$, but it could still be true that $$\omega^{\omega_1} > \omega_1$$.

Last edited: Apr 14, 2005
12. Apr 14, 2005

### Hurkyl

Staff Emeritus
Better yet, it's a specific ordinal with cardinality &omega;_1.

Don't think of it in terms of unions, think of it in terms of the ordering...

13. Apr 14, 2005

### aa

Oh schnap! You're right! $$\omega^{\omega_1} = \omega_1$$.

Boy, that's really surprising to me.

Thanks Hurkyl!!

That IS what you had in mind, right? I'm not sure what the ordering has to do with it, but I do think that it = $$\omega_1$$. Now, anyway.

14. Apr 14, 2005

### Hurkyl

Staff Emeritus
Well, I thought of it in terms of the ordering: ω_1 is an upper bound of that set, so we must have $\omega_1 \geq \omega^{\omega_1}$.

15. Apr 14, 2005

### aa

Hmm, let me think about it again for a sec...this is kinda tricky...

16. Apr 14, 2005

### Hurkyl

Staff Emeritus
You were right -- I didn't mean to suggest otherwise. Just offering how I arrived at that conclusion!

17. Apr 14, 2005

### aa

Yes, thank you, and I am interested in your thought process for arriving at this conclusion too. It's just that after reading your post, I tried to think about it again, and suddenly wasn't sure of what I was thinking before.

Let me try to get this straight once and for all.

Your reasoning Hurkyl, is that for any countable ordinal $$\alpha$$, $$\omega^{\omega_1} > \alpha$$ because $$\omega^{\omega_1} > \omega^{\alpha} > \alpha$$. We do need to use the fact that $$\omega^{\alpha} > \alpha$$ for all countable ordinals $$\alpha$$, right? That seems true, although a proof isn't immediately obvious to me.

But in any case, the basic idea is that $$\omega^{\omega_1} >$$ all countable ordinals, right?

18. Apr 14, 2005

### Hurkyl

Staff Emeritus
I know $\omega^{\omega_1}$ is bigger than any countable ordinal, because $|\omega^{\omega_1}| = |\omega_1|$. That's how I know that $\omega_1 \leq \omega^{\omega_1}$.

The other direction, $\omega_1 \geq \omega^{\omega_1}$, comes from the fact that $\omega_1$ is an upper bound of that set that defines $\omega^{\omega_1}$.

19. Apr 15, 2005

### aa

I have finished writing my solution and am ready to turn it in. Whew.

Thank you Hurkyl. I really appreciate your helping me through this problem.

20. Apr 21, 2005

### davidseed

going back to which is greater a^b or b^a. ( i can't use the special symbols)

then assume a>b (without loss of generality)
then
if b=1 then a^b =a b^a =1 so a^b is greater

there are some special cases for a,b, <=3 which I leave you to find

but for all other cases
b^a is greater e.g. 4^5 is greater than 5^4

I Haven't got a proof

this info comes from an Excel spreadsheet

but I think that the proof lies in putting
e=a-b

then
a^b = (b+e)^b = use binomial expansion
b^a = b^(b+e) = b^b * b^e

approach for proof