How would you prove 7 is finite?

In his book “Infinity and the Mind”, Rudy Rucker challenges the reader to prove, without circularity, that the number 7 is finite.

So, for example, you can’t just count 7 steps and say “OK, I’m done” because that assumes the process of counting 7 steps is finite, which is circular. He reminds us that it’s possible to imagine an alien race that habitually counts up to infinity without even realizing it (we’re talking math, not physics here…yes it would take infinite time.)

As far as I can see, you can’t even use a Peano type system, starting with 0 and applying the successor function, because you’ll have to apply it 7 times…which begs the question.

Can anyone think of a simple proof for this?

Here’s his proof:

Assume that the numbers 4 and 3 are finite, and that the sum of two finite numbers is finite. Then 7 is finite.

But his proof is somewhat circular in that how do you know 3 and 4 is finite. Also, under his logic, what is 1+1+1+1+… since one is finite.

The definition of an infinite set is one that can be placed in a 1-to-1 bijection with a proper subset of itself. For example, I can show that the number of natural numbers is infinite since the perfect squares is a proper subset and t=s^2 is a bijection. Incidently, this would also prove the number of perfect squares is also infinite.

To show 7 is finite, it would sufice to look at all (2^7)-1 proper subsets of 7 objects (the power set minus the set {1, 2, 3, 4, 5, 6, 7}) and show that none have 7 elements.

7 is a real number. Real numbers are by definition finite. Ergo, 7 is finite.

Even if he meant “rational” when he said “infinite” it’s kind of in the definition. And then there’s that “proof”. I’m going to go out on a limb and guess that Infinity and the Mind is no Principia Mathematica.

I’m not sure I even understand the question. But wouldn’t counting to eight be proof that seven is not infinity and therefore must be a finite number?

If there is an infinite series that sums to 7, i.e. Σa[sub]n[/sub] = 7, and there exists a convergent majorant Σb[sub]n[/sub] to that series with |a[sub]n[/sub]| ≤ b[sub]n[/sub], then 7 is not infinite (because Σa[sub]n[/sub] isn’t divergent).

For instance, Σ[sup]7[/sup]/[sub]2[/sub][sup]n+1[/sup] = 7, [sup]7[/sup]/[sub]2[/sub][sup]n+1[/sup] < [sup]7[/sup]/[sub]2[/sub][sup]n[/sup], and Σ[sup]7[/sup]/[sub]2[/sub][sup]n[/sup] is convergent, as one easily sees using the convergence criterion of one’s choice (for instance, the quotient of two successive summands is 1/2).

Now if that’s not allowed since I used 7 (or rather, 7/2) in the series, then the proof in the OP is invalid, as well, since 4 = 7/2 + 1/2, and thus, to say ‘4 is finite’ it is necessary for 7/2, and thus 7, to be finite, as well.

I like SaintCads solution, but in this context, I’d simply define the property finite as zero is finite and any direct successor to a finite number is finite.

Now we can simply point out that 7 = s(s(s(s(s(s(s(0))))))) (in other words 7 is the successor of 6).

Now I may have used the successor function 7 times here, but since that’s not pertinent to the definition, there’s no circularity (if I hadn’t bothered to count the number of steps needed, that wouldn’t have invalidated the argument)

IANA expert, but if that’s the author’s idea of a proof, he’s far to clueless to be writing about math.

His “proof” isn’t exactly circular, but it does require assuming 3 & 4 are finite. Since 7 was arbitrarily chosen and couldn’t be assumed, why can 3 or 4 be assumed? His approach is effectively circular, even if reductionist.

This is the approach I’d reccommend too, and it fits under Peano arithmetic. Counting how many times you use an operation under any particular construction system isn’t necessary - it only matters whether you can formulate the construction or not.

My “proof”, then: Assume 7 is finite. Therefore 7 is finite. It’s just as watertight as his proof, and makes two fewer assumptions. :cool:

I disagree, if you mean counting up using finite numbers and ‘reaching’ infinity - that is not what infinity means. Infinite means beyond the numbers you can reach by counting up. It’s the mathematical equivalent of saying that if an alien race used enough rocket thrust, they could overcome relativity and transcend light speed ‘without even realizing it.’

Of course, it’s possibly to ‘count up to infinity’ by a different counting system, I can do it right now…

-1, 0, 1, aleph zero, aleph one, aleph two, aleph three…


But it’s less elegant. Making fewer assumptions is not the ideal thing in a math proof - you want to make OBVIOUS assumptions, and I think as a point of style you want to avoid assuming the objective if you possibly can. Even if you think something is obvious on the face of it, if you’re asked to prove it, and can do that by breaking it down into two things that are at least as obvious and taking those as assumptions, then you do that. :wink:

Rucker is a mathematics professor and a bit of a humorist in his popular writing, if I recall correctly. I’d be inclined to think that his proof was intended as a joke.

First define “7” and “finite” and then ask me the question. Surprisingly, there are several inequivalent definitions of “finite” once you avoid classical logic things like the axiom of choice and 2-valued logic.

The given proof depends on induction. Let P(n) be the proposition that n is finite. Assuming, as you must (if you want a proof of this) that the successor of a finite number is finite, then it is clear that n finite implies its successor is, and if you also assume P(0), you conclude that P(n) is always finite.

Always? That’s assuming that n is in the set of whole numbers.

You cannot use that proof, (or those assumptions,) to prove that any of the following entities are finite:
aleph null
aleph one
the letter q
the color green


With additional mechanisms for proving the finiteness of a given number x based on other finite numbers, we could address the first few from that list. I suppose this would do for a start:
Any number that can be derived using algebraic operations from finite numbers, (addition, subtraction, multiplication, division excluding division by zero, and roots of whatever degree,) is also finite.

That doesn’t handle pi though. :slight_smile:

I would take 7 marshmallows and eat them. If I did not explode then 7 must be a finite number.

That’s using your tummy! :slight_smile:

Given that we know that 3 + 3 = 7 for very large values of 3, it follows that there are infinite number of 7s. Or something.

Sets can be finite or infinite. Those terms don’t have meanings for individual numbers.

Maybe not, but it’s generally considered to be the best popular exposition you’re going to get of the mathematical notions of infinite numbers. I’d need to see the bit in the OP in context in order to get at the point the author (who is a well-regarded mathematician) is trying to make.

In order to write an actual proof, you need to know what set of axioms and definitions are being used. If you’re working with the Peano postulates, then the proof given by 4.66 and Hari Seldon is the way to go (SaintCad’s proof applies to a particular model of the Peano postulates, but lacks generality). On the other hand, if you’re working with the field axioms for the real numbers, you’re going to come up with something pretty close to what Sage Rat has (although you have to change “by definition” to “axiomatically” to be completely accurate).