2
$\begingroup$

Yes we can write numbers down and we all know 2+3=5 and 2x3=6. But how can we prove that all of these are true?

  • 0
    The tags are woefully inappropriate to the question, so please find some more appropriate ones. I like the question, though. – 2017-02-03
  • 0
    Sorry I have no idea what tags are appropriate for my question – 2017-02-04

2 Answers 2

3

This all come from Peano axioms Wikipedia : Peano axioms

Basically, you say : let there be $0$ and a successor operator named $S$ and let call $S(0)=1$. And similarly we define all integers like this $S(1)=2, S(2)=3, ...$

We define the addition recursively as :

$\begin{cases} a+0=a\\ a+S(b)=S(a+b) \end{cases}$

note that from this definition you can establish $S(a)=a+1$

And the multiplication as :

$\begin{cases} a*0=a\\ a*S(b)=a+a*b \end{cases}$

Have a look at the wikipedia page I linked for the details of why this leads to the usual addition and multiplication we know of.

  • Let's do it for $2+3$

$2+3=2+S(2)=S(2+2)=S(2+S(1))=S(S(2+1))=S(S(2+S(0))=S(S(S(2+0))=S(S(S(2)))=S(S(3))=S(4)=5$

  • And for $2*3$

$2*3=2*S(2)=2+2*2=2+2*S(1)=2+2+2*1=2+2+2*S(0)=2+2+2+2*0=2+2+2+0=6$

So it is not a question of proving anything, this is just some rigourous definition of addition and multiplication. From there we can develop shortcuts to do these calculations quickly and efficiently as every schoolboy knows.

0

Note that it is true if set on which binary operations $+$, and $*$ are defined on is unbounded.

For example, for group $G(\mathbb{Z}_5 = {0,1,2,3,4},+_5)$ then $2 + 3 = 0$.

Note, the same logic applies with days of the weeks which can be expressed as $G(\mathbb{Z}_7 = {Monday = 0, Tuesday = 1, Wednesday = 2, Thursday = 3, Friday = 4, Saturday = 5, Sunday = 6},+_7)$, then if it is Friday $=4$, so in $3$ days there will be $4 +_7 3 = 0$ which corresponds to Monday. Therefore, such statement make total sense.