## Multiplication

After explaining addition and its properties, it’s time to explain another operation: multiplication.

It can be defined by repeated addition:

Or recursively by:

0*n = 0

S(m)*n = (m*n) + n

Once again we’ll use mathematical induction to prove what we want.

First of all, I will prove that both definitions are equivalent. We know that

Therefore:

Now let’s define the properties of multiplication.

**Commutativity**

First, as before, we need to prove that numbers commute with 0.

*Commutativity with 0*

It is evident that 0*0 = 0 = 0*0, so that property is valid at 0. If we can prove that being valid at k implies being valid at S(k), then it will be valid for all natural numbers. Thus, we need k*0 = 0*k to imply S(k)*0 = 0*S(k)

1. S(k)*0 = (k*0) + 0 [Definition of multiplication]

2. (k*0) + 0 = k*0 [Definition of addition]

3. k*0 = 0*k [Premise]

4. 0*k = 0 [Definition of multiplication]

5. 0*S(k) = 0 [Definition of multiplication]

And so it’s verified that multiplication commutes with 0.

*Commutativity with Successor*

Now let’s move on to prove that k*S(n) = k + (k*n). First, by definition 0*S(n) = 0 = 0 + (0*n) so that’s valid at k = 0. Now let’s find out whether validity at k implies validity at S(k).

1. S(k)*S(n) = (k*S(n))+S(n) [Definition of multiplication]

2. (k*S(n))+S(n) = k + (k*n) + S(n) [Premise]

3. k + (k*n) + S(n) = (k*n) + k + S(n) [Commutativity of addition]

4. (k*n) + k + S(n) = (k*n) + S(k + n) [Definition of addition]

5. (k*n) + S(k + n) = (k*n) + S(n + k) [Commutativity of addition]

6. (k*n) + S(n + k) = (k*n) + n + S(k) [Definition of addition]

7. (k*n) + n + S(k) = (S(k)*n) + S(k) [Definition of multiplication]

8. (S(k)*n) + S(k) = S(k) + (S(k)*n) [Commutativity of addition]

Since the property is valid at 0 and validity at k implies validity at S(k), the property is valid for all natural numbers.

*Commutativity*

Now we’re ready to prove commutativity in general using Peano’s Axioms and the definition of multiplication. We already know that 0*m = 0 = m*0, so we want to prove that if k*m = m*k then S(k)*m = m*S(k).

1. S(k)*m = (k*m) + m [Definition of multiplication]

2. (k*m) + m = (m*k) + m [Premise]

3. (m*k) + m = m + (m*k) [Commutativity of addition]

4. m + (m*k) = m*S(k) [Commutativity with Successor]

So Commutativity is proved for all natural numbers. Q.E.D.

**Distributivity over Addition**

We are now going to prove that k*(m+n) = k*m + k*n. First, for k = 0 that is trivially true:

1. 0*(m+n) = 0 [Definition of multiplication]

2. 0*m + 0*n = 0 + 0 = 0 [Definitions of multiplication and addition]

Now let’s prove that k*(m + n) = k*m + k*n implies S(k)*(m + n) = S(k)*m + S(k)*n.

1. S(k)*(m+n) = k*(m + n) + (m + n) [Definition of multiplication]

2. k*(m + n) + (m + n) = (k*m) + (k*n) + (m + n) [Premise]

3. (k*m) + (k*n) + (m + n) = (k*m + m) + (k*n + n) [Commutativity and associativity of addition]

4. (k*m + m) + (k*n + n) = S(k)*m + S(k)*n [Definition of addition]

So multiplication distributes over addition for all natural numbers. Q.E.D.

**Associativity**

We mean to prove that (m*n)*p = m*(n*p).

*Associativity with 0*

First we prove that (m*n)*0 = m*(n*0).

1. (m*n)*0 = 0*(m*n) [Commutativity]

2. 0*(m*n) = 0 [Definition of multiplication]

3. m*(n*0) = m*(0*n) [Commutativity]

4. m*(0*n) = m*0 [Definition of multiplication]

5. m*0 = 0*m [Commutativity]

6. 0*m = 0 [Definition of multiplication]

Since (m*n)*0 = 0 = m*(n*0), commutativity will guarantee that multiplication always associates with 0.

*Associativity*

Now we need (m*n)*p = m*(n*p) to imply (m*n)*S(p) = m*(n*S(p)).

1. (m*n)*S(p) = (m*n)*p + (m*n) [Definition of multiplication and commutativity]

2. (m*n)*p + (m*n) = m*(n*p) + (m*n) [Premise]

3. m*(n*p) + (m*n) = m*(n*p + n) [Distributivity over addition]

4. m*(n*p + n) = m*(n*S(p)) [Definition of multiplication and commutativity]

And so multiplication is associative for all natural numbers. Q.E.D.

### Notes

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it’s really upsetting me that I have neither ever been shown a proof of this nor attempted to prove it for myself there...

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