how could I prove for integers $a\neq b$ then $ab \neq 1$.

Attempt.Proof by contrapositive. Assume $ab=1$ Then the only cases are $(1)(1)=1$ and $(-1)(-1)=1$ Thus $a=b$ I have a feeling this is not right. Any ideas?

  • $\begingroup$ Do you think that $1$ has integer factors other than $\pm1$? $\endgroup$ – Lord Shark the Unknown Dec 2 '19 at 4:27
  • $\begingroup$ For $a,b \in \mathbb R$ ? $\endgroup$ – Zera Dec 2 '19 at 4:29
  • $\begingroup$ for a,b $\in \mathbb{Z}$ $\endgroup$ – user707991 Dec 2 '19 at 4:29
  • $\begingroup$ Is there any way I should use the division algorithm to approach this? I was thinking a proof by contradiction but could make no progress $\endgroup$ – user707991 Dec 2 '19 at 4:30
  • $\begingroup$ @lordsharktheunknown So should I say since the only integer factors are $1$ and $-1$ thus $(1)(1)=(-1)(-1)=1$ so $a=b$? $\endgroup$ – user707991 Dec 2 '19 at 4:32

Assume $ab=1$ for $a,b\in\mathbb{Z}$. Since $1$ is the multiplicative identity of $\mathbb{Z}$, we find that $a=b^{-1}$. If $a=1$, then $b=1$ as well. If $a=-1$, then we also have that $b=-1$. What are the multiplicative inverses of other elements of $\mathbb{Z}$? Particularly, are they integers? Conclude that $ab=1$ implies that $a=b=\pm1$ for $a,b\in\mathbb{Z}$.


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