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SO the question is, is $f(x)=1/x$ an injective, surjective, bijective or none of the above function? The domain is all real numbers except 0 and the range is all real numbers.

It's injective because $f(x)=f(y)$
$1/x=1/y$
$y=x$

I also think that it's surjective in the given domain, as $x=1/y$
$f(y)=1/1/y$
$f(y)=y$

Thus it's surjective and by addition it's also bijective. Am I correct?

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2 Answers

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The first part showing that it is injective is good, though you could be a little more descriptive. Like:

Let $x,y \in \mathbb{R} - \{0\}$ such that $f(x) = f(y)$. This implies that $1/x = 1/y$. Rearranging, we have that $x = y$. Thus, $f$ is injective.

The part showing that $f$ is surjective is a little jumbled. For one, it doesn't make sense to say that $f(y) = (1/y)^{-1}$. Because by definition $f(y) = 1/y$ and these two expressions are generally not the same.

In general, to show that a function is surjective, you pick an element $y$ in the codomain and show that the function maps some element in the domain to $y$. But there is an issue when you try to do this for the function $f$ in this case. Can you spot a real number $y$ that is not in the range of $f(x) = 1/x$? If you are having trouble, just look at the graph of $f(x)$.

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It is indeed injective but not surjective.Because the range is $\mathbb{R} $ not $\mathbb{R} *$

$$f: \mathbb{R}* \to \mathbb{R} \\ \quad \ x\mapsto \frac 1 x$$ $f$ is not surjective because $\forall x \in \mathbb{R}*, \ f(x)\neq 0$ $$g: \mathbb{R}* \to \mathbb{R} *\\ \quad \ x\mapsto \frac 1 x$$ $g$ however is bijective.

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