Ordinal Exponentiation Comparison

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In summary: Ordinal numbers are those that are related to the natural numbers in a specific way. For example, the natural number 2 is greater than the ordinal number 1, but the ordinal number 3 is not greater than the natural number 2. ordinal numbers can also be thought of as being ordered in a specific way. So in general, a^b is greater than b^a, but there are some specific cases where this is not the case.
  • #1
aa
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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: [tex]\omega^{\omega_1}[/tex] or [tex]\omega_1^{\omega}[/tex]?

P.S. I know that [tex]\omega^{\omega_1}[/tex] equals the order type of [tex]\underbrace{ \omega \times \omega \times \omega \times ... }_{\omega_1 \ many \ times}[/tex], and [tex]\omega_1^{\omega}[/tex] equals the order type of [tex]\underbrace{ \omega_1 \times \omega_1 \times \omega_1 \times ... }_{\omega \ many \ times}[/tex], but I'm still stuck.
 
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  • #2
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
Yes, but this is referring to ordinal exponentiation, a la set theory.

I really appreciate any help you all can give.
 
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  • #4
What exactly is ω_1?
 
  • #5
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
Ack, alephs are usually used for cardinals aren't they? They're not order types! And cardinal exponentiation is different than ordinal exponentiation! :frown:

You could just say that ω_1 was the first ordinal with cardinality greater than that of ω. :-p
 
  • #7
Hehe, alright, thanks. That's exactly what [tex]\omega_1[/tex] is supposed to be.
 
  • #8
Well, ω1 is a limit ordinal, so what does that tell you about ω^ω1?
 
  • #9
Hurkyl said:
Well, ω1 is a limit ordinal, so what does that tell you about ω^ω1?
[tex]\omega^{\omega_1} = sup\{\omega^\alpha : \alpha < \omega_1\}[/tex]. I did think of that, but don't see how it helps.
 
  • #10
What's the cardinality of &omega;^&alpha; if &alpha; < &omega;_1?
 
  • #11
It's [tex]\omega[/tex]. So then [tex]\omega^{\omega_1}[/tex] is an ordinal with cardinality equal to the union of [tex]\omega_1[/tex] many countable ordinals, which means [tex]\omega^{\omega_1}[/tex] is an ordinal with cardinality [tex]\omega_1[/tex].

So we've shown that [tex]|\omega^{\omega_1}|[/tex] = [tex]\omega_1[/tex], but it could still be true that [tex]\omega^{\omega_1} > \omega_1[/tex].
 
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  • #12
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
Oh schnap! You're right! [tex]\omega^{\omega_1} = \omega_1[/tex].

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 = [tex]\omega_1[/tex]. Now, anyway.
 
  • #14
Well, I thought of it in terms of the ordering: ω_1 is an upper bound of that set, so we must have [itex]\omega_1 \geq \omega^{\omega_1}[/itex].
 
  • #15
Hmm, let me think about it again for a sec...this is kinda tricky...
 
  • #16
You were right -- I didn't mean to suggest otherwise. Just offering how I arrived at that conclusion!
 
  • #17
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 [tex]\alpha[/tex], [tex]\omega^{\omega_1} > \alpha[/tex] because [tex]\omega^{\omega_1} > \omega^{\alpha} > \alpha[/tex]. We do need to use the fact that [tex]\omega^{\alpha} > \alpha[/tex] for all countable ordinals [tex]\alpha[/tex], right? That seems true, although a proof isn't immediately obvious to me.

But in any case, the basic idea is that [tex]\omega^{\omega_1} > [/tex] all countable ordinals, right?
 
  • #18
I know [itex]\omega^{\omega_1}[/itex] is bigger than any countable ordinal, because [itex]|\omega^{\omega_1}| = |\omega_1|[/itex]. That's how I know that [itex]\omega_1 \leq \omega^{\omega_1}[/itex].

The other direction, [itex]\omega_1 \geq \omega^{\omega_1}[/itex], comes from the fact that [itex]\omega_1[/itex] is an upper bound of that set that defines [itex]\omega^{\omega_1}[/itex].
 
  • #19
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
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

start with e=1
(certainly true here)


then recast problem as b+e as b(1+x) where x=e/b ; 0<x<1

David
 
  • #21
The problem isn't about natural numbers -- it's about ordinal numbers.
 

FAQ: Ordinal Exponentiation Comparison

What do the variables "a" and "b" represent in this equation?

In this equation, "a" and "b" represent two different numbers that are being raised to powers. They can be any real numbers, but cannot be negative or zero.

How do you determine which is bigger between a^b and b^a?

In order to determine which is bigger between a^b and b^a, you can use logarithms. Take the log of both sides of the equation, and if a^b is larger, then log(a^b) will be larger than log(b^a). If b^a is larger, then log(b^a) will be larger than log(a^b).

Can you provide an example to illustrate this concept?

Sure. Let's take a=2 and b=3. Then a^b = 2^3 = 8, and b^a = 3^2 = 9. Since 9 is larger than 8, b^a is bigger than a^b.

Is there a general rule for determining which is bigger between a^b and b^a?

Yes, there is a general rule. If b > a, then b^a will be larger than a^b. If a > b, then a^b will be larger than b^a.

Can this concept be applied to any numbers or are there limitations?

This concept can be applied to any real numbers, as long as both a and b are positive. If one or both of the numbers are negative, then the concept does not apply.

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