# Prove that all numbers have a unique reciprocal

I'm learning logic for computer science and came across the question:

If $\ n$ is a real number, $\frac{1}{n}$ is the reciprocal of $\ n$. Prove that all numbers have a unique reciprocal.

I came up with the following method, but it seems so simple that I doubt it'll work:

$\frac{1}{n}=p$

Since we know that $\ n=ℝ$, and we assume that $\ p=$ the unique reciprocal, but will this work when proving all real numbers have unique reciprocals?

• I dont think your proof is correct. You should not assume that the reciprocal $p=1/n$ is unique, you need to show that. Try showing that if $n$ has two reciprocals, then they must be equal. – DKS Sep 7 '17 at 16:41
• Well, your proof will depend on what it is you're allowed to assume, which you haven't provided. Also, as written, this attempt is very problematic: It's circular since you're assuming your conclusion, and you write $n = \mathbb{R}$. – user296602 Sep 7 '17 at 16:41
• The notation $\frac 1n$ already assumes that it has a functional dependence on $n$, so if that is the definition I can't see there is anything to prove. I suspect they really meant to write: "If $n$ is a real number, then $a$ is a reciprocal of $n$ if $a\cdot n=1$" as a definition. – Henning Makholm Sep 7 '17 at 16:44
• I think it would be enough to prove that $f(x)=\frac1x$ is inversible. – u8y7541 Sep 7 '17 at 16:49

The way to prove that a certain property is uniquely satisfied, is to assume that two things satisfy and then prove that they are equal.

In this case, suppose that $p,q$ are both reciprocals of $n$. Then, $qn=1=pn$. We divide both sides by $n$ (which needs to be nonzero) to conclude that $p=q$.

Now, this doesn't prove existence; i.e. there might be no reciprocal for $n$ at all. Indeed this is true for $n=0$; without considering this exception the problem is false as written.

• Mention that $n \ne 0$ – I am Back Sep 7 '17 at 16:44
• @VidyanshuMishra Indeed, in general, $qn=pn$ does not imply $q=p$. – Hagen von Eitzen Sep 7 '17 at 16:46

Your conclusion is false. Not all $x\in\mathbb{R}$ has reciprocal. $0$ as no reciprocal.

To show unicity.

If $a\ne 0$ then $a^{-1}a=aa^{-1}=1$. If $a$ had another reciprocal $b$ such that $ab=1$ then left multiply by $a^{-1}$ you get $a^{-1}ab=a^{-1}$ hence $b=a^{-1}$. Same procedure for $ba=1$ right multiply etc

I assumed for granted associativity