# associative law

• July 14th 2008, 02:03 PM
particlejohn
associative law
Prove for any natural numbers $a,b,c$, we have $(a+b)+c = a+(b+c)$. So fix $a$ and $b$ and induct on $c$. For $c = 0$, $a+b = a+b$. Suppose inductively that $(a+b)+c = a+(b+c)$. We have to prove that $(a+b) + c++ = a+(b+c++)$. So $(a+b)+c++ = (a+b+c)++$. Also $a+(b+c++) = (a+b+c)++$. $\blacksquare$

Is this correct?
• July 14th 2008, 02:42 PM
ThePerfectHacker
Quote:

Originally Posted by particlejohn
Prove for any natural numbers $a,b,c$, we have $(a+b)+c = a+(b+c)$. So fix $a$ and $b$ and induct on $c$. For $c = 0$, $a+b = a+b$. Suppose inductively that $(a+b)+c = a+(b+c)$. We have to prove that $(a+b) + c++ = a+(b+c++)$. So $(a+b)+c++ = (a+b+c)++$. Also $a+(b+c++) = (a+b+c)++$. $\blacksquare$

Remember how we define addition. We define $+(a,0) = a$ and we denote it by $a+0 = a$. And $+(a,b+1) = +(a,b) + 1$. We denote $a+(b+1) = (a+b)+1$. Now use induction. Say $(a+b)+c = (a+b)+c$ is true. Then $(a+b)+(c+1) = ((a+b)+c)+1 = (a+(b+c))+1 = a + ((b+c)+1) = a+(b+(c+1))$.
• July 14th 2008, 02:46 PM
particlejohn
did I do that?
• July 14th 2008, 04:08 PM
Plato
Quote:

Originally Posted by particlejohn
did I do that?

The answer is “we don’t know”.
I am glad that TPH answered the way he did. I was about do the almost the same thing as he has done. But I would have used different notation. The fact is that there are almost as many notations used in Peano arithmetic as there are textbooks that address the topic. So the reason we don’t is we have not seen the notation you are using.