How do I know $a+a+...+a$ $n$ times is $na$

Question

For a cardinal $a$ and a positive integer $n$

This seems almost trivial but I’m not sure how I can show it is true. Is it defined to be this way or must it be shown?

Answer 1

It is a consequence of distrbutive law that is since

$$n=\overbrace{1+1+\ldots+1}^{\text{n times}}$$

then

$$a \cdot n=a\cdot (1+1+\ldots+1)=\overbrace{a+a+\ldots+a}^{\text{n times}}$$

Answer 2

Proof by induction: For $n=1: na=a$ checks

Assume $n=k hold: a+a+a...a\ k\ times =ka$

So for $n=k+1: a+a+a..a\ k+1\ times=(k+1)$

$a+a+a...k+1\ times=a+a+a...a\ k\ times + a =ka + a =(k+1)a$

Therefore $a+a+a...a\ n\ times$ hold true.

Hence Proved