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Answer: one pair $(a=\frac12$ and $b= -1)$


$ab=\dfrac{a}{b}\Rightarrow ab^2=a \Rightarrow b^2=\dfrac aa=1$ so $b= 1$ or $-1$.

$a+b=ab$ becomes       $a+1=a$        or $a-1=-a$
                                    imposible            $\Rightarrow 2a=1\\
\Rightarrow \ a=\tfrac12$

So $a=\frac12, b=-1$ is the only pair.

(check: $\frac12\times-1=-\frac12, \quad \frac12\div-1=\frac1{-2}=-\frac12, \quad \frac12+-1=-\frac12$)



This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.