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$\Bbb Z$ - integers
$\Bbb N$ - natural numbers (starting from 1)
$\Bbb R$ - real numbers

I believe the answer is the set of real numbers ($\Bbb R$), seeing as $b$ will not equal $0$ as the set of natural numbers start from $1$. Thoughts?

coldnumber
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Eddard
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1 Answers1

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The function is not surjective so the image is not $\Bbb R$; in fact, no irrational number is in im $f$, because a quotient of integers is never irrational.

If $a \in \Bbb Z$ and $b \in \Bbb N$, then $\dfrac{a-4}{7b} \in \Bbb Q$. This shows that im $f \subseteq \Bbb Q$.

On the other hand, any rational number $\frac cd$ is in the image of $f$:

Let $b \in \Bbb N$. Then

$$f\left(\frac {7cb}d +4, b\right)=\dfrac{\left(\frac {7cb}d+4\right)-4}{7b}=\frac cd$$

Hence, $\Bbb Q \subseteq $ im $f$.

This means that im $f$ is $\Bbb Q$.

coldnumber
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  • in f(7cb/d + 4, b) <-why is b left to it's own? – Eddard Aug 14 '15 at 00:18
  • I'm not sure of what you mean. I left it arbitrary because it doesn't actually matter what $b \in \Bbb N$ you choose because it cancels out (to check this just simplify the quotient in the answer). You can check that choosing $b=1$ or $b=2$ gives you the same image, or that $f\left(\frac {7c}d +4, 1\right)=f\left(\frac {14c}d +4, 2\right)$. – coldnumber Aug 14 '15 at 00:21
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    You're welcome! But I'm female, actually; it's not usually a good idea to try to guess the gender of the person behind the screen :) – coldnumber Aug 14 '15 at 00:36