Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

One of the axioms out there is: if n and m are integers and they have the same next integer then n=m

But why is it considered to be an axiom since I can easily prove it:

Have the same next integer means: n+1=m+1 <=> n=m


share|cite|improve this question
You have just restated the axiom, you have not proved it. ;) – fkraiem Jan 31 '14 at 11:46
@fkraiem No I did prove it since in the axiom we have if n and m are integers such that they have the same next integer which in algebra means: m+1=n+1 then I just did elementary algebra adding to both sides the same number to prove it – user125180 Jan 31 '14 at 11:58
The elementary algebra you used depends on this axiom. You may have not noticed that you were using the axiom because it seems obvious that you're steps were valid. Axioms are statements which seem to be obviously true but defy proof using the other available axioms. Its supposed to be self evident but it cannot be proven this way. – Spencer Jan 31 '14 at 12:08
up vote 9 down vote accepted

If you leave out this axiom you can satisfy Peano's other axioms with a set that is not the natural numbers.

Consider the set containing $0,a_1,a_2$ with $0\neq a_1,a_2$.

We can define

  • $a_1=0+1$,
  • $a_2=a_1+1$,
  • $a_1=a_2+1$

This set satisfies all of Peano's axioms except that in this case successor of two numbers being equal does not imply that the numbers are equal ($0$ and $a_2$ have the same successor).

The axiom is then in some sense really a definition. If you leave it out you get a theory of arithmetic that is different from the usual one. This could be compared to leaving out the parallel line postulate to get non-Euclidean geometries.

share|cite|improve this answer
Here is a list of Peano's axioms for anyone who is interested: – Spencer Jan 31 '14 at 15:31

The original axiom is part of a very small set of axioms needed to derive the usual properties of addition and multiplication. Instead of that one axiom you took all the usual properties of addition as given.

There are multiple sets of axioms that can be used to define the natural numbers. We have largely settled on the Peano axioms as the simplest definition of the natural numbers. Along with a formal definition of equality, Peano's axioms define the natural numbers as a set $ N $ with a function $ S: N \rightarrow N $ and an element $ 0 \in N $ such that:

  • $ \forall n \in N, S(n) \neq 0 $
  • $ S(a) = S(b) \Rightarrow a = b $
  • for any set A, if $ 0 \in A $ and $ a \in A \Rightarrow S(a)\in A $, then $ A \supset N $

Given these axioms, we can define addition recursively as:

  • $ n + 0 = n $
  • $ n + S(a) = S(n + a) $

and define multiplication as:

  • $ n * 0 = 0 $
  • $ n * S(a) = n * a + n $

Given this foundation, we can prove that addition and multiplication are commutative and associative and that multiplication distributes over addition. That is left as an exercise for the reader.

share|cite|improve this answer

To address the questioner's proof, the proof was stated like this:

Have the same next integer means: n+1=m+1 <=> n=m

However, this assumes that the principle trying to prove is true. Therefore the proof is begging the question.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.