Why is multiplication treated differently to addition?

Question

I am a grade 11 student in South Africa. Just so you know, this is my first time posting here.

My understanding is that multiplication is simply a shorter way of writing addition problems. E.g. $$4\cdot 4 = 4+4+4+4 = 16$$

Also, subtraction is just another way of writing addition. E.g. $(+8)+(-7) = +1$

Also, aren't using exponents just a shorter way of writing multiplication. E.g. $$2^5 = 2\cdot 2\cdot 2\cdot 2\cdot 2 = 32$$

Because of this, does this mean that addition is the only fundamental mathematical function or am I completely on the wrong track?

As long as you only work with integers, you are right; for a arbitrary real numbers (e.g. $\pi$, $\sqrt(2)$), it's not possible to write $\sqrt(2)\cdot\pi$ as a (somewhat more complicated) addition. — , Jun 16 '15 at 10:45
Whether addition is "fundamental" or not depends (not surprisingly, really) on your choice of foundations. Try googling for "Peano axioms", for one of many possible foundations. There the "successor" (add one) operation is fundamental, and addition is defined in terms of that. Link: https://en.wikipedia.org/wiki/Peano_axioms — Mark Dickinson, Jun 16 '15 at 10:50
@Matt: If you work with integers, there's actually a more fundamental operation than addition, "successorship", that amounts to "taking the next integer". Loosely, for example, "adding $5$" to an integer $n$ means "taking the successor of $n$ five times". — Andrew D. Hwang, Jun 16 '15 at 10:51
Related posts: If multiplication is not repeated addition and Is there any way to define arithmetical multiplication as other thing than repeated addition? — Zev Chonoles, Jun 16 '15 at 10:53

score 1 · Answer 1 · answered Jun 16 '15 at 11:25

1

There is a more fundamental operation than addition. It is increment and addition is repeated increment. So I a = a +1 and 4+4= I I I I 4

answered Jun 16 '15 at 11:25

RowanS

1 Answers1