Is this proof of "$a\neq0\Rightarrow a^{-1}\neq0$" valid?

The statement I need to prove is the following:

$$a\neq0\Rightarrow a^{-1}\neq0$$, for all $$a$$ from $$\mathbb R$$.

I have tried to use the method of proof by contradiction and came up with the following:

Let $$a$$ be a random element from $$\mathbb R$$, such that $$a=b \Rightarrow a^{-1}=0$$, with $$b\neq 0$$.

$$a=b$$

$$\Rightarrow a\cdot b^{-1}=b\cdot b^{-1}$$

$$\Rightarrow a\cdot b^{-1}=1$$

$$\Rightarrow a\cdot b\cdot a^{-1}=1\cdot a^{-1}$$

$$\Rightarrow a\cdot a^{-1}\cdot b=a^{-1}\cdot 1$$

$$\Rightarrow 1\cdot b=a^{-1}\cdot 1$$

$$\Rightarrow b\cdot 1 = a^{-1}\cdot 1$$

$$\Rightarrow a^{-1}$$

Because $$b\neq 0$$, it contradicts the assumption we have taken at the beginning. Therefore,

$$a\neq 0 \Rightarrow a^{-1}\neq 0$$

I used the alebraical axioms to reason each step, but I was not sure if this is a valid proof as I am still not used to what makes proofs proofs. A confirmation or an explanation would be much appreciated. Thanks!

• What do you mean with "=> $a^{-1}$"? Oct 16 '21 at 7:33
• oh I meant to write $b=a^{-1}$ Oct 16 '21 at 7:41
• And how did you go from " $\Rightarrow a\cdot b^{-1}=1$" to "$\Rightarrow a\cdot b\cdot a^{-1}=1\cdot a^{-1}$"? Oct 16 '21 at 7:44
• I multiplied each side by $a^{-1}$ and as i am typing this i am seeing the typo. b is meant to be $b^{-1}$ Oct 16 '21 at 7:48

It's a little confusing to follow. There is no need to introduce a second variable $$b$$, and your second to last line is a little unclear $$\implies a^{-1}$$.
Instead you can argue like this, suppose $$a^{-1}=0$$, then $$1=a\cdot a^{-1}=a\cdot 0=0$$ which is a contradiction.
• There is no such rule in general. The logic is as follows, suppose $a$ is an arbitrary real number, and that $a^{-1}$ is its multiplicative inverse. If $a^{-1}$ also happens to be $0$, then we can show that a contradiction follows, namely that $0=1$. Therefore $a^{-1}$ cannot be $0$. Oct 16 '21 at 7:56