Is this proof of $ab = 0$ correct?

Question

I have to prove the following theorem(Apostol's Calculus I, exercise 1 page 19):

If $ab = 0$ then either $a = 0$ or $b = 0$.

My attempt to solve it was: $ab = 0$ can be rewritten as $ab = a0$, because $a0 = 0$(already been proved). So we now can cut a on both sides(already been proved too), so we have $b = 0$. Also, we could rewrite the original equation as $ab = b0$ and b on both sides of equation and turn it into $a = 0$.

I think my proof covers the basic steps but I don't think it's asserting anything.

Answer 1

This is almost perfect but you missing a bit:

My attempt to solve it was: $ab=0$ can be rewritten as $ab=a0$, because $a0=0$(already been proved). So we now can cut a on both sides(already been proved too), so we have $b=0$.

Here you need to add "assuming that $a\ne 0$" to be able to justify the last part.

Same thing with:

Also, we could rewrite the original equation as $ab=b0$ and b on both sides of equation and turn it into $a=0$.

You need to first assume that $b\ne 0$.

Now you left with the case that $a=b=0$, in which case it is trivial.

---

As @Theo point out we can do it using only $2$ cases:

Either $a\ne 0$ hence we can do the cancellation: $ab=a0\implies b=0$

Or $a=0$, in this case we are done.

Answer 2

Proof: Let $a,b \in \mathbb{R}$ with $ab = 0$.

Then, if $a \neq 0$, we know there exists $a^{-1} \in \mathbb{R}$ such that $a\cdot a^{-1}= 1$. Thus,

$$ab = 0 \implies a^{-1} (ab) = a^{-1} 0 = 0$$But,$$\begin{align*} a^{-1} (ab) = 0 &\implies (a^{-1}a)b = 0 \\ &\implies 1b = 0\\&\implies b = 0. \end{align*}$$

Answer 3

Consider the contrapositive of the statement:

if $a\neq 0$ and $b\neq0$, then $ab\neq0$. Which is trivial.

Answer 4

Correct me if wrong.

Proof by contradiction

Assume:

$ab= 0$ implies $a \not =0$ and $b \not = 0$.

If $a \not =0$, and $b \not =0$, $a, b$ have inverses $a^{-1},b^{-1}$.

$(b^{-1}a^{-1})(ab)=b^{-1}(a^{-1}a)b=$

$ b^{-1}(1b)= b^{-1}b=1;$

By assumption we have $ab=0$, hence

$(b^{-1}a^{-1})(ab)= (b^{-1}a^{-1})0=0.$

Hence $1=0$, a contradiction.