0
$\begingroup$

Problem sounds easy enough - prove that if $m$ is in set of all natural numbers (let's call it $\mathbb N$) and so is $n$ than $m+n$ also must be there. Probably it should be done using induction. But how to do it formally? I mean I could tell that $\mathbb N$ is a set $\{1, 1+1, 1+1+1, \dotso\}$ and $m$ is $m$ ones and $n$ is $n$ ones so $m+n$ is $m+n$ ones so it's still in $\mathbf N$ but I doubt that my professor would accept such a proof. It's so intuitive that I don't even know where to start. What should I do?

$\endgroup$
  • 6
    $\begingroup$ What are your definitions of natural numbers and addition? Or what axioms are you using? $\endgroup$ – Henry Oct 9 '14 at 21:22
2
$\begingroup$

If you are using Peano axioms, induct on $n$ with $m$ fixed.

Note: $n^+$ is the sucessor of $n$.

Axiom 1. $0 \in \mathbf N$.

Axiom 2. if $n \in \mathbf N$ then $n^+ \in \mathbf N$.

Definition 3 (Adition). $m + 0 := m$ and $m + n^+ := (m + n)^+$.

Proof. Induct on $n$ (keeping $m$ fixed). Consider the case base $n = 0$. Since by hypothesis we have $m \in \mathbf N$, then $m + 0 = m \in \mathbf N$ by Definition 3. Now suppose inductively that $m + n \in \mathbf N$; we now have to show that $m + n^+ \in \mathbf N$. By Definition 3, we have $m + n^+ = (m + n)^+$. By Axiom 2, since $m + n \in \mathbf N$, we conclude $(m + n )^+ = m + n^+ \in \mathbf N$. Thus $m + n \in \mathbf N$ for every $m, n \in \mathbf N$. This close the induction.

$\endgroup$
0
$\begingroup$

There are different versions of the Peano's Axioms. The most commonly used version (with 5 axioms) does not define any add or multiplication function. You must construct these functions using the axioms of set theory. You can prove there exists a unique + function such that

$\forall a,b \in N:a+b\in N$

$\forall a \in N:a+0=a$

$\forall a,b \in N:[a+S(b)=S(a+b)]$

where $S:N\to N$ is the usual successor function defined in the axioms.

A first-order version (with 9+ axioms) gives you this theorem as an initial definition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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