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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.

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    $\begingroup$ I think $0$ is defined to be the additive identity. $\endgroup$ – Guy Mar 25 '14 at 20:06
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    $\begingroup$ When we define addition and multiplication, we pick the definition to ensure that the identities are what we want them to be. Usually $a + 0 = a$ is part of the definition of addition! (although $0 + a = a$ needs to be proven by some other means) $\endgroup$ – user14972 Mar 25 '14 at 20:06
  • $\begingroup$ @Hurkyl whatever the properties of $a+b=b+a$ and $a=b$,$b=c\implies a=c$ are called, aren't they axiomatic? Or do we need to prove? $\endgroup$ – Guy Mar 25 '14 at 20:08
  • $\begingroup$ @Sabyasachi: The three properties of an equality relation (reflexivity, symmetry, and transitivity) are axioms in PA. PA also includes that $a$ a natural number and $a=b$ implies $b$ a natural number (closure under $=$). $\endgroup$ – Eric Towers Mar 25 '14 at 20:28
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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.

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  • $\begingroup$ Can you prove the commutivity and the bits that follow "from definition"? What definitions are these, in particular? $\endgroup$ – kevin Mar 25 '14 at 20:13
  • $\begingroup$ I said "by" definition; when I say e.g. $a \times 0 := 0$ by definition I mean it is true because that's how we defined it. Note that multiplication may also equivalently be defined letting $0 \times a := 0$ and $s(a) \times b = a \times b + b$ $\endgroup$ – dani_s Mar 25 '14 at 20:25
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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.)

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  • $\begingroup$ Don't you actually need to prove $k + 0 = k$ in order to prove commutativity? $\endgroup$ – dani_s Mar 25 '14 at 20:27
  • $\begingroup$ Yes, that would probably be the first step in a proof of commutativity: $k+0=k$ by induction on $k$. So you could do this separately, as part of proving that $0$ is an additive inverse, if you prefer. $\endgroup$ – Jack Lee Mar 25 '14 at 20:57
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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.

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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.

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  • $\begingroup$ here you assumed that $1n=n$. you set out assuming what you had to prove. $\endgroup$ – Guy Mar 25 '14 at 20:05
  • $\begingroup$ @Sabyasachi: I disagree. I assume $n$ exists. This I can get with $n$ applications of $S$. Or do you choose to define multiplication perversely as somehting other than repeated application of $S$? $\endgroup$ – Eric Towers Mar 25 '14 at 20:06
  • $\begingroup$ No no. You didn't get my point. You said "the first multiple of $n$ is $n$". But the list of multiples is $\{0,1n,2n,\cdots\}$. How do you get your first statement? $\endgroup$ – Guy Mar 25 '14 at 20:09
  • $\begingroup$ @Sabyasachi: In the sequence $\{0, S0, SS0, \dots \}$ the first thing that is a repeated sequence of $n$ applications of $S$ is the one that is $n$ applications of $S$. $\endgroup$ – Eric Towers Mar 25 '14 at 20:20

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