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berekuk

Technically all numbers are comparable.

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What about imaginary numbers?

You can still compare them, e.g two numbers could or couldn't be equal. Equality is a form of comparison

wouldn't this make all things comparable?

Except apples and oranges apparently.

You can compare apples and oranges alphabetically.

As far as I can tell, it's not that you can't compare apples and oranges, it's just often misleading.

You don't even have to do that: Just well-order the universe and use the resulting ordering for your less-than operator.

1 != sqrt(-1). That's an example of a comparison between a real number and an imaginary number.

Technically incorrect: http://mrob.com/pub/math/largenum-2.html

That's a rather narrow definition of comparability. If nothing else said, one would understand the term with regards to the relation of any partially ordered set – not only the one implicitly defined on that page.

Would a tl;dr here be:

Is BB(26) - BB(25) greater than or less than BB(25)?

Google didn't help me. What's BB(#)?

Busy Beaver

I think people are downvoting you because they're assuming you're not serious. :(

Link: https://en.wikipedia.org/wiki/Busy_beaver#The_busy_beaver_fu...

Fun discussion to be found at: http://www.scottaaronson.com/writings/bignumbers.html

I am certain it is greater. BB(n) grows super-exponentially.

In fact I would be willing to bet serious money that BB(n+1)/BB(n) is greater than BB(n) if 3 < n.

(This is, of course, assuming that one assumes that BB(n) is well-defined. That is an interesting point of philosophy given the existence of Turing machines which can't be proven to not halt.)

> In fact I would be willing to bet serious money that BB(n+1)/BB(n) is greater than BB(n) if 3 < n.

BB(n) grows faster than any computable function. In order for BB(n+1)/BB(n) > BB(n) to hold, BB(n) merely has to grow faster than a sequence whose new terms are obtained by repeated squaring (like k^2ⁿ). That's computable, indeed primitive recursive, so BB(n) definitely grows dramatically faster than it.

https://en.wikipedia.org/wiki/Primitive_recursive_function

Edit: another way of looking at this is that the Ackermann function grows unbelievably faster than functions that easily satisfy the property you describe, and the Busy Beaver function grows unbelievably faster than the Ackermann function. Somehow putting it this way feels like an understatement, though!

It is unfortunately not that easy. Consider the following function, if n is even then f(n) = BB(n/2), else f(n) = BB(2n). Then f(n) grows faster than any computable function, but still if n is odd, then f(n+1) < f(n).

However you have encapsulated the reason why I would be confident of this result. :-)

I define a non-word "uncomparable" to mean something specific and rather arbitrary, but I do not intend it to be the antonym of the real word "comparable". I'll try to clarify this on the webpage you linked to. - Robert Munafo

Technically all nouns are comparable under equality.

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