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If our range such as in the question below is all the real numbers excluding $0$, to determine if a function is onto we must ask if all real numbers excluding $0$ can be mapped to at least one value of $x$ where $x$ is an element of the real numbers?

$$ \mathbb{R} − \{0\} \to \mathbb{R} \quad \text{defined by} \quad r(x) = \frac{6}{x} $$

So in the question above, all we'd need to do to prove that it isn't onto is find a value in our range that can't be mapped to a value of $x$, yes? The solution to the above question is

for any $y \in \mathbb{R} − \{0\}$ we have $r(6/y) = y$). So $r$ is not onto.

I'm struggling to understand what that means, is that saying that this function is one to one?

Thanks.

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  • $\begingroup$ $r(x) = 6/x$, right? $\endgroup$ – Luísa Borsato Jun 19 '16 at 13:57
  • $\begingroup$ yes, thank you I edited. $\endgroup$ – Perplexityy Jun 19 '16 at 14:01
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$r: \mathbb{R}- \{0\} \rightarrow \mathbb{R}- \{0\}$, $r(x) = \frac{6}{x}$

Let $y \in \mathbb{R}- \{0\}$. We need to decide if there is $x \in \mathbb{R}- \{0\}$ such that $r(x) = y$, that is, $6/x = y$. If you take $x = 6/y$, it may work.

For example: you have to decide if there is $x_0$ such that $r(x_0) = 2$. Take $x_0 = 6/2 = 3$. Then $r(3) = 6/3 = 2$, as we wanted.

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  • $\begingroup$ Actually, if the contradomain is $\mathbb{R}$, then $r$ is not onto, because there is no $x$ such $r(x) = 0$. $\endgroup$ – Luísa Borsato Jun 19 '16 at 14:06

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