At this point we can use everything we know about addition and order. (I will still mention everything I use from the previous sections).
Multiplication is defined as such:
Definition $2.3.1.$ (Multiplication of natural numbers). Let $m$ be a natural number.
To multiply zero to $m$, we define $0 \times m := 0$. Now suppose inductively that we
have defined how to multiply $n$ to $m$. Then we can multiply $n++$ to $m$ by defining
$(n++) \times m := (n \times m) + m$.
Where $n++$ is the successor of $n$.
We also proved that multiplication is commutative i.e. $n\times m=m \times n$ for all $n$,$m$ natural numbers.
I also call upon his second Peano axiom and a collorary
Axiom $2.3$. $0$ is not the successor of any natural number; i.e., we have $n++ = 0$ for every natural number $n$.
Corollary $2.2.9$. If $a$ and $b$ are natural numbers such that $a + b = 0$, then $a = 0$
and $b = 0$.
Lemma $2.3.3$ (Positive natural numbers have no zero divisors). Let $n$, $m$ be natural
numbers. Then $n\times m=0$ if and only if at least one of $n$, $m$ is equal to zero. In
particular, if $n$ and $m$ are both positive, then $nm$ is also positive.
Proof ($1$st try). We fix $m$ and induct on $n$. For $n=0$ we have to prove that $\forall m(0\times m=0\iff m=0)$ which is true by definition of multiplication. Now suppose the statement is true for $n$ i.e., $\forall n$,$m(n\times m =0 \iff [(n=0)\lor (m=0)])$. Since by Axiom $2.3$ we have $n++\neq 0$ the statement for $n++$ that we need to prove is $\forall n$,$m[(n++)\times m=0\iff m=0]$ which by the definition of multipliaction is equivalent to $\forall n$,$m[(n\times m)+m=0\iff m=0]$. First we show $(\implies)$ i.e., $\forall n$,$m[(n\times m)+m=0\implies m=0]$. Assume that $\forall n$,$m[(n\times m)+m=0]$ and suppose that $m\neq 0$ then by definition, $m$ is positive. Thus $(n\times m) +m=0$ yields a contradiction by Corollary $2.2.9$ so $m=0$. Now we prove $(\impliedby)$ i.e., $\forall n$,$m[m=0 \implies (n\times m)+m=0]$. Suppose that $m=0$ then $(n\times m)+m=(n\times 0)+0=(0\times n)=0$ by definition. This closes the induction. $\square$
I'm not very comfortable with logic symbols so let me know if something shouldn't be written as it is.
I'm unsure on this proof since Tao gave a hint to prove the second statement first and I didn't do that and after completing my proof I still don't see why he gave that hint. Which suggests that maybe I've missed something here.
The hint was: " prove the second statement first". The second statement being "In particular, if $n$ and $m$ are both positive, then $nm$ is also positive."
After a few comments I've realized how that previous proof was nonsense so here I will try to incorporate Tao's hint.
Proof ($2$nd try). We first prove the second statement. Suppose for the sake of contradiction that $nm=0$ and that $n$ and $m$ are both positive numbers. Since they are positive by Lemma $2.2.10$, $\exists !a$,$b$: $n=a++$ and $m=b++$. So $nm=0 \implies (a++)\times (b++)=0$. Which by definition can be written as $(a\times b++)+b++=0$ which by Corollary $2.2.9$ is a contradiction by Axiom $2.3$.
Now to prove the second statement; for $(\implies)$ if $n=0$ the statement is true by definition and commutativity of multiplication. For $(\impliedby)$ we take the contrapositive.
Best Answer
[[ Collecting my Comments + newer thoughts into this Answer ]]
(A) There may be various ways to Prove that , Tao is giving the "Hint" to Prove it in a Particular way.
(1) Prove that the Product of 2 Positive Integers is Positive Integer [[ $+ \cdot + = +$ ]] which should not be hard.
(2) Show that Product of 2 Positive Integers giving 0 is the Contraction to that.
(B) OP-Proof-1 is starting off with a flaw :
$\forall m(0\times m=0\iff m=0)$
This is wrong , because $ 0 \cdot 8 = 0 \iff 8 = 0 $ is wrong.
That makes the whole Induction invalid , in general.
I am not sure whether fixing that will help , here.
(C) OP-Proof-2 looks better & more to the Point.
(C1) It looks a little too long , it could be shortened.
(C2) User "Dr. Momo" has given Issues with this Proof. That should be rectified.
(D) My Suggestion , I am giving it in elaborate Detail :
(D2) Start with Product of 2 Positive Integers giving Positive Integer.
Proof : When we have $m$ & $n$ Positive , then we have $(m_1++) = m$ & $(n_1++) = n$ , hence we have $m \cdot n = (m_1++) \cdot (n_1++)$
$(m_1++) \cdot (n_1++) = m_1 \cdot (n_1++) + (n_1++)$
Here , $m_1 \cdot (n_1++) + (n_1++)$ cannot be 0 , because that will make Both terms 0 : In Particular , it will make $(n_1++) = 0$ , which can not occur , because O is not the Successor , according to Axiom.
In other words , Product of 2 Positive Integers is Positive Integer.
(D1) That immediately says that when Product of 2 Integers is 0 , at least 1 Integer must be 0.
Proof : When Both are not 0 (Positive) , Product is not 0 (Positive) , which we already Proved.
(E) That way was the way Tao wanted the Proof ; Alternative ways may Exist.
Particularly , Induction ways may Exist , though that was not necessary in my way.