Successor of 0 is 1

Question

Using the Peano axioms as the foundation for arithmetic (but further elementary structure can be developed), where S is the successor operation and 0 is an element of what we will call the set of natural numbers, how does one prove that for an element defined as 1=S(0), 1 is also the multiplicative identity? Multiplication is defined as iterated addition, which is defined as iterated succession. In fact, first, can you tell me why 0 is the additive identity? I usually learn it as axiomatic, but I am thinking that, in this case, we probably can prove it from the definition of addition from S.

Answer 1

We define multiplication recursively: we let $a \times 0 := 0$ and $a \times s(b) := a \times b + a$.

$1$ is the multiplicative identity because $$a \times 1 = a \times s(0) = a \times 0 + a = 0 + a = a$$

where $a \times 0 = 0$ by definition and $0 + a = a$ can be shown by induction. Also from the commutativity (which requires a proof) follows $1 \times a = a \times 1$.

As for $0$ being the addivite identity, $a + 0 = a$ is true by definition, but as I said $0 + a = a$ requires a proof.

Answer 2

In Peano arithmetic, both addition and multiplication are defined recursively. For example, you can use the following recursive definition of addition for each fixed natural number $k$: $$0+k=k,\quad S(n)+k = S(n+k).$$ From this it follows immediately that $0$ is a (left) additive identity. (You also have to prove -- by induction of course -- that addition is commutative, from which it follows that $0$ is also a right additive identity.)

Then multiplication is defined similarly: for each natural number $k$, $$0\centerdot k = 0,\quad S(n)\centerdot k = n\centerdot k + k.$$ To show that $1$ is a (left) multiplicative identity, just use the definitions: $$1\centerdot k = S(0)\centerdot k = 0\centerdot k + k = 0 + k = k.$$ (And of course, as before, you prove commutativity and thereby show that it's also a right inverse.)

Answer 3

I usually learn it as axiomatic, but I am thinking that, in this case, we probably can prove it from the definition of addition from S.

You are right. Usually, these days, Peano's axioms do not define addition and multiplication, just a successor function $S$.

Using only the successor function and the axioms of set theory, you can construct i.e. prove the existence of the addition and multiplication functions. It's kind of tricky for the beginner. I won't go into detail, but for both functions, you construct the set of ordered triples $N^3$, select the required subset of $N^3$, and prove that subset is a function. For the addition function $+$, you eventually obtain:

A1: $\forall a,b\in N: a+b\in N$

A2: $\forall a\in N: a+0=x$

A3: $\forall a,b\in N: a+S(b)=S(x+y)$

It can be shown that addition is commutative and associative (another long and convoluted proof).

For the multiplication function ($\times$), you obtain:

M1: $\forall a,b\in N: a\times b\in N$

M2: $\forall a\in N: a\times 0=0$

M3: $\forall a,b\in N: [a\times S(b)=a\times b + a )$

It can also be shown that multiplication is commutative and associative.

Let $x\in N$. Specifying $a=x$ and $b=0$ in M3 , we obtain:

$x\times S(0)=x\times 0 + x$

Applying M2, A2 and the commutativity of addition, we have

$x\times S(0)=x$

Thus, $S(0)$ is the right multiplicative identity.

From the commutativity of multiplication, $S(0)$ is also the left multiplicative identity.

Answer 4

The simplest answer is that $n$ is the first multiple of $n$. The sequence of multiples of $n$ is $\{0, 1n, 2n, \dots \}$, so $1$ must be the multiplicative identity.