Prove Lemma 2.3.3 Let $n,m$ be natural numbers. Then $n \cdot m = 0$ if and only if at least one of $n,m$ is equal to $zero$.
Helpful/Required definitions and such:
Definition 2.3.1 (Multiplication of natural numbers). $\,$ Let
$m$ be a natural number. To multiply $zero$ to $m$, we define $0 \cdot m := 0$. Now suppose inductively that we have defined how to multiply $n$ to $m$. Then we can multiply $n+1$ to $m$ by defining $(n+1) \cdot m := (n \cdot m) + m$.
Lemma 2.3.2 (Multiplication is commutative). $\,$ Let $n,m$ be natural numbers. Then $n \cdot m = m \cdot n$.
All assumptions when using addition of natural numbers has already been proven.
My attempt –
Proof. Let $n,m$ be natural numbers. Then by definition of a natural number, we have only these cases for $n,m$:
(i) $\,$$n=0,\,m=0$
(ii) $\,$$n=0,\,m \gt 0$
(iii) $\,$$n \gt 0,\,m=0$
(iv) $\,$$n \gt 0,\,m \gt 0$
Suppose at least one of $n,m$ is equal to $zero$, i.e. we concern ourselves with only properties (i), (ii), and (iii) from the enumeration above. (i) implies $n \cdot m = 0 \cdot 0 = 0$, $\,$(ii) implies $n \cdot m = 0 \cdot m = 0$, and (iii) implies $n \cdot m = n \cdot 0 = 0$, all by definition of multiplication for natural numbers. This exhausts the required cases, and so we have proven that if at least one of $n,m$ is equal to $zero$, then the product of $n,m$ equals $zero$.
Conversely, let's consider the product of $n,m$. We will show $n \cdot m \neq 0$ when both $n$ and $m$ are positive, that is property (iv). Suppose $n \gt 0,\,m \gt 0$ and consider the sum $0+\sum_{i=1}^n m_{i}$, where each $m_{i} = m$. Let's show this is equivalent to $n \cdot m$ and then show $0+\sum_{i=1}^n m_{i}$ must be positive.
First consider the base case when $n=1$.
Then $n=1 \Rightarrow 0+\sum_{i=1}^1 m_{i} = 0+m=m$.
Also, $n=1 \Rightarrow n \cdot m = 1 \cdot m = m$.
Thus, $0+\sum_{i=1}^n m_{i} = n \cdot m$ is true when $n=1$, proving the base case.
Now suppose we have shown inductively that $0+\sum_{i=1}^n m_{i} =n \cdot m$. Consider the sum $0+\sum_{i=1}^{n+1} m_{i}$. This equals $0+ m_{n+1} + \sum_{i=1}^{n} m_{i}$. By the inductive hypothesis, we can rewrite this sum as $0+ m_{n+1} + n \cdot m$. Furthermore, $m_{n+1} = m$ by construction. Then we have $0+ m_{n+1} + n \cdot m = 0+ m + n \cdot m = 0 + (n+1) \cdot m = (n+1) \cdot m$ by Definition 2.3.1 and Lemma 2.3.2. Thus, because we have shown $0+\sum_{i=1}^{n+1} m_{i} = (n+1) \cdot m$, then $0+\sum_{i=1}^{n} m_{i} = n \cdot m$, closing the induction.
Taking a closer look at the sum $0+\sum_{i=1}^{n} m_{i}$, we can infer that $0+\sum_{i=1}^{n} m_{i} \gt 0$, since each $m_{i} \gt 0$ by our hypothesis. But $0+\sum_{i=1}^{n} m_{i} \gt 0 \Rightarrow n \cdot m \gt 0$. Thus, we have shown if $n \gt 0,\,m \gt 0$, then $n \cdot m \gt 0$. That is, we have shown if $n \cdot m = 0$, then at least one of $n,m$ must be equal to $zero$.
Have I proven this correctly? And if I did, could this be a shorter proof. Thank you!
Best Answer
Simpler:
Every number can be written either as $0$ or as $k + 1$ for some $k$. Note that $(k + 1) \cdot (j + 1) = k \cdot (j + 1) + (j + 1) \geq j + 1 > j \geq 0$. So two numbers which can be written as $k + 1$ and $j + 1$ have $(k + 1) \cdot (j + 1) > 0$.