# Proving Properties of Multiplication for Natural Numbers

• MHB
• evinda
In summary, the conversation discusses the definition of multiplication between natural numbers and the properties that need to be proven for each pair of numbers. These properties include commutativity, associativity, and the identities of multiplication. The conversation also includes a proof for one of these properties, using induction on one of the factors.
evinda
Gold Member
MHB
Hello! (Wave)

For each pair of natural numbers $m \in \omega, n \in \omega$ we define the multiplication between $m,n$ (as a function $\cdot: \omega \times \omega \to \omega$) like that:

$$m \cdot 0=0$$
$$m \cdot n'=m \cdot n+m$$

I want to show that for each $m \in \omega, n \in \omega, k \in \omega$ the following properties are satisfied:

• $m \cdot n=n \cdot m$
• $(m \cdot n) \cdot k=m \cdot (n \cdot k)$
• $(m+n) \cdot k=m\cdot k+n \cdot k$
• $1 \cdot n=n$
• If $k \neq 0$ and $n \cdot k=m \cdot k$ then $n=m$.
In order to prove the first sentence we have to show that $0 \cdot m=0$, $n' \cdot m=m \cdot n'$ and that $m \cdot n=n \cdot m$, right?In order to prove that $0 \cdot m=0$ I tried the following:

For $m=0: 0 \cdot 0=0 \checkmark$

We suppose a $m$ such that $0 \cdot m=0$.

We want to show that $0 \cdot m'=0$.

$0 \cdot m'=0 \cdot m+0 \overset{\text{induction hypothesis}}{=}0$.

Therefore, $0 \cdot m=0$, for any $m \in \omega$.Then we want to show that $n' \cdot m=m \cdot n'$.

For $m=0$: $n' \cdot 0 \overset{\text{definition}}{=}0 \overset{\text{previous sentence}}{=}0 \cdot n' \checkmark$We suppose a $m$ such that $n' \cdot m=m \cdot n'$ for all $n \in \omega$.We want to show that $n' \cdot m'=m' \cdot n'$.

$n' \cdot m'=n' \cdot m+n'=m \cdot n'+n'$How could we continue? (Thinking)

evinda said:
In order to prove the first sentence we have to show that $0 \cdot m=0$, $n' \cdot m=m \cdot n'$ and that $m \cdot n=n \cdot m$, right?
I would prove $0 \cdot m=0$ and
$n'm=nm+m\qquad(*)$
by induction on $m$. The equality (*) is used in the inductive step of proving $mn=nm$ by induction on $n$. Indeed,
$mn'\overset{\text{def}}{=}mn+m\overset{\text{IH}}{=}nm+m\overset{(*)}{=}n'm.$

Is it right like that?

We will show that $0 \cdot m=0$.
For $m=0: 0 \cdot 0 \overset{\text{definition}}{=}0$
We suppose a $m$ such that $0 \cdot m=0$.
We will show that $0 \cdot m'=0$.
$0 \cdot m'=0 \cdot m+0 \overset{\text{ind. hypothesis}}{=}0+0=0$


We will prove that $m \cdot n=n \cdot m$.

For $m=0: 0 \cdot n=0=n \cdot 0 \checkmark$

We suppose a $m$ such that $\forall n \in \omega: m \cdot n=n \cdot m$.

We will show $m' \cdot n=n \cdot m', \forall n \in \omega$
.
For $n=0: m' \cdot 0 \overset{\text{definition}}{=} 0 \overset{\text{above proposition}}{=} 0 \cdot m'$

Let $n$ such that $m' \cdot n=n \cdot m'$.

We will show that $m' \cdot n'=n' \cdot m'$.

$$m' \cdot n'=m' \cdot n+m' \overset{\text{ind. hypothesis}}{=}n \cdot m'+m'=n \cdot m+n+m' \overset{\star \star }{=} n \cdot m+n'+m \\ =(n \cdot m+m)+n'=n' \cdot m+n'=n' \cdot m'$$$$\star \star:$$

We will show that $n+m'=n'+m$.

For $m=0$: $n+m'=n+0'=(n+0)'=n'=n'+0=n'+m$

Let $m$ such that $n+m'=n'+m$.

We will show that $n+m''=n'+m'$.

$n+m''=(n+m')' \overset{\text{ind. hypothesis}}{=} (n'+m)' \overset{\text{definition}}{=} n'+m'$

evinda said:
We will show that $m' \cdot n'=n' \cdot m'$.

$$m' \cdot n'=m' \cdot n+m' \overset{\text{ind. hypothesis}}{=}n \cdot m'+m'=n \cdot m+n+m' \overset{\star \star }{=} n \cdot m+n'+m \\ =(n \cdot m+m)+n'=n' \cdot m+n'=n' \cdot m'$$
Note that $nm+m=n'm$ (second last equality) is not an instance of definition. You could justify it as follows.
\begin{align}
nm+m&=mn+m&&\text{external induction hypothesis (on }m)\\
&=mn'&&\text{definition}\\
&=n'm&&\text{external induction hypothesis}.
\end{align}

evinda said:
We will show that $n+m'=n'+m$.
You used other properties of addition, including commutativity and associativity. It is not clear why you are proving this particular fact and not others.

Evgeny.Makarov said:
Note that $nm+m=n'm$ (second last equality) is not an instance of definition. You could justify it as follows.
\begin{align}
nm+m&=mn+m&&\text{external induction hypothesis (on }m)\\
&=mn'&&\text{definition}\\
&=n'm&&\text{external induction hypothesis}.
\end{align}
I see... (Nod)

Evgeny.Makarov said:
You used other properties of addition, including commutativity and associativity. It is not clear why you are proving this particular fact and not others.

I didn't prove the other properties because they are given at the proposition of my other thread about addition. (Blush) I tried to show like that that $(m \cdot n) \cdot k=m \cdot (n \cdot k)$:
For $k=0: (m \cdot n) \cdot 0 \overset{definition}{=}0=m \cdot (n \cdot 0)=m \cdot 0=0$

Let $k$ such that $(m \cdot n) \cdot k=m \cdot (n \cdot k)$.We will show that $(m \cdot n) \cdot k'=m \cdot (n \cdot k')$.

$(m \cdot n) \cdot k' \overset{definition}{=} (m \cdot n)k +(m \cdot n) \overset{ind. hypothesis}{=} m \cdot (n \cdot k)+(m \cdot n) \overset{\star}{=} m \cdot (n \cdot k+n)\\=m \cdot (n \cdot k')$$\star:$$ We will show that$m \cdot (n \cdot k)+(m \cdot n)=m \cdot (n \cdot k+n)$For$m=0: 0 \cdot (n \cdot k)+(0 \cdot n)=0= 0 \cdot (n \cdot k+n) \checkmark$Let$m$such that$m \cdot (n \cdot k)+(m \cdot n)=m \cdot (n \cdot k+n), \forall n,k \in \omega$We will show that$m' \cdot (n \cdot k)+(m' \cdot n)=m' \cdot (n \cdot k+n)m' \cdot (n \cdot k)+(m' \cdot n)=m \cdot (n \cdot k)+(n \cdot k)+m \cdot n+n=m \cdot (n \cdot k)+m \cdot n+(n \cdot k)+n \overset{ind. hypothesis}{=} m \cdot (n \cdot k+n)+(n \cdot k)+n=m' \cdot (n \cdot k+n)\$

Am I right? (Smile)

## 1. What is the commutative property of multiplication?

The commutative property of multiplication states that the order in which numbers are multiplied does not affect the outcome. In other words, when multiplying two or more numbers, you can change the order of the factors and still get the same product.

## 2. What is the associative property of multiplication?

The associative property of multiplication states that the grouping of factors in a multiplication equation does not affect the outcome. This means that you can change the placement of parentheses in a multiplication equation and still get the same product.

## 3. How does the distributive property work in multiplication?

The distributive property in multiplication allows you to break down a multiplication equation into smaller, simpler equations. This is done by distributing one factor to all the other factors in the equation, and then adding the products together to get the final product.

## 4. What is the identity property of multiplication?

The identity property of multiplication states that when any number is multiplied by 1, the product is that number. This means that 1 is the identity element for multiplication, just like 0 is the identity element for addition.

## 5. How does the zero property of multiplication work?

The zero property of multiplication states that when any number is multiplied by 0, the product is 0. This means that 0 is the only number that, when multiplied with any other number, will result in 0.

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