$0$ is finite. If $n$ is finite, then $n+1$ is finite. Hence, by induction, all numbers are finite. What is the catch?
Induction versus Natural Numbers
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0@AsafKaragila My bad, I didn't realize I was bumping them all. – 2013-03-23
5 Answers
You have to be careful with your wording in this type of question. There is more than one type of number. Usually, what is being referred to is clear from the context, but here it is not. Natural numbers (0,1,2,3,...,etc) are all finite, and standard induction only applies to these.
There are other types of numbers, such as cardinal numbers and ordinal numbers which can be infinite. The principle of induction does not apply to these. There is an extension of induction, called transfinite induction. On top of showing that the desired property holds for n+1 whenever it holds for n, transfinite induction also involves proving that the property holds in a limiting sense. The property of being finite does not survive this limit, so there is no problem with having infinite ordinal numbers.
Usually when someone says number without being clear about what they are referring to, they mean natural numbers, integers, real numbers or, possibly, complex numbers. In that case, there are no infinite numbers.
This is not the kind of thing which should cause problems in a serious mathematical discussion, but I've seen this ambiguity create confusion many times before on internet forums.
You might call each natural number finite (that's what induction applies to). Wait until you hear about these, though.
EDIT: As the comment to the question points out, the natural numbers are collectively infinite in number, if that's what you're referring to.
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5There are no infinite natural numbers. – 2010-08-21
The Axiom of Infinity may be relevant to your understanding, because it concerns the existence of an infinite set, which is also an infinite cardinal number. Without it all sets are finite and equivalent to natural numbers, for the reason that you state (induction is not powerful enough). Starting with Cantor's Theorem you can build even larger cardinals by taking the power set of the infinite set asserted to exist by the Axiom of Infinity. There is also a technique called transfinite induction which is used to prove propositions about ordinals and cardinals.
I'll approach this from a naive set theory standpoint, since it can easily be translated to other contexts. If we define zero as the empty set, and n+1 as (n)\union{n}, then we let the natural numbers be the set of all of these (yes, we need an axiom to guarantee its existence). In this case we define a set as finite iff it can be put in bijection with an element of the natural numbers. The key here is that 'finite' is just so defined that the statement that natural numbers (meaning the elements of the natural numbers) are finite is wholly trivial. This of course is only talking of finitude from the perspective of cardinallity.
Since finitude is often used to refer to other properties, being 'finite' means something distinct and often not logically related (e.g. as in points in an affine subspace of a projective space, as something being bounded, or as select elements of an algebraic structure), this question could be improved by specifying what you mean by 'finite'.
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0An infinite number is one which is not finite in the intuitive sense. As you describe, a number is finite in the intuitive sense, if it can be arrived at by "adding ones finitely many times". Anyway, I slightly changed my mind about these issues, because I learned that first-order logic can [define the set of non-standard natural numbers](http://math.stackexchange.com/a/202497/12490). It's only unable to define the negation of this property. – 2012-12-17
An interesting catch here is that the proposition "n is finite" cannot be expressed by a unary predicate "p(n)" in first-order logic. The first order induction schema (from the Peano axioms) only applies to unary predicates in first-order logic, hence you cannot prove that every natural number is finite, at least not with the first-order Peano axioms.
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0@AsafKaragila As long as the stystem can't prove that there are nonstandard models of PA. – 2012-05-10