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!
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.