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 $b\neq 0$.


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

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

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

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

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

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

$\Rightarrow a^{-1}$

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!

  • 1
    $\begingroup$ What do you mean with "=> $a^{-1}$"? $\endgroup$
    – md2perpe
    Oct 16 at 7:33
  • $\begingroup$ oh I meant to write $b=a^{-1}$ $\endgroup$
    – Evank800
    Oct 16 at 7:41
  • $\begingroup$ And how did you go from " $\Rightarrow a\cdot b^{-1}=1$" to "$\Rightarrow a\cdot b\cdot a^{-1}=1\cdot a^{-1}$"? $\endgroup$
    – md2perpe
    Oct 16 at 7:44
  • $\begingroup$ I multiplied each side by $a^{-1}$ and as i am typing this i am seeing the typo. b is meant to be $b^{-1}$ $\endgroup$
    – Evank800
    Oct 16 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.

  • $\begingroup$ I see. That makes much more sense. Thank you. $\endgroup$
    – Evank800
    Oct 16 at 7:45
  • $\begingroup$ A trivial question that bothers me a little bit: when you have to prove the statement on the right from the one on the left, don't you have to tweak the left one and come to the right one? or is that not the case here? $\endgroup$
    – Evank800
    Oct 16 at 7:51
  • 1
    $\begingroup$ 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 $0$. $\endgroup$ Oct 16 at 7:56

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