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I'm told to find the domain, codomain, and range of the function $f : \mathbb{Z} \to \mathbb{Z}$ which is defined as $f = \{ (x, 4x + 5) : x \in \mathbb{Z} \}$.

I can determine the first two, but I'm struggling to define the range. I tested out a few values of $x$, such as:

$f(-2) = -3$

$f(-1) = 1$

$f(0) = 5$

$f(1) = 9$

$f(2) = 13$

I don't really see a pattern here, though. They're not primes, and they're not just all odd integers, but rather a subset of odd integers. I also can't factor anything from $4x + 5$ to say, for instance, that it's the set of all multiples of $n \in \mathbb{Z}$.

How can I express the range for this function? Any intuition/procedure one can use to arrive at an answer would be appreciated, as I don't see one myself.

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  • $\begingroup$ Well, you could just say the range is the set of integers which are $5$ more than a multiple of $4$. Since $5=4+1$ it would probably be more typical to say the range was the set of integers which are $1$ more than a multiple of $4$. $\endgroup$ – lulu Jun 15 '17 at 23:03
  • $\begingroup$ It's exactly half the odd numbers! It's all the multiples of 4 but with 5 added to them. $\endgroup$ – mdave16 Jun 15 '17 at 23:04
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    $\begingroup$ Or you could write it as $\{ y \in \mathbb{Z} : y \equiv 1 \pmod{4} \}$ if you're familiar with that notation. $\endgroup$ – Daniel Schepler Jun 15 '17 at 23:04
  • $\begingroup$ It is the set $$4\mathbb Z+5$$ $\endgroup$ – Piquito Jun 15 '17 at 23:06
  • $\begingroup$ @DanielSchepler I'm familiar with the modulo operator, but I'm having trouble interpreting what the set you defined means. $\endgroup$ – AleksandrH Jun 15 '17 at 23:08
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For range of the function, you can write -

$4x+5=4(x+1)+1=4k+1$ Since $x \in \mathbb Z \, \implies k \in \mathbb Z$

Thus, range of the given function is all the integers of the form $4k+1$.

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  • $\begingroup$ Thank you! It says I can accept an answer only in 11 minutes. Question: couldn't I also write it as, say: $4x + 5 = 4(x - 2) + 13 = 4k + 13$ for some $k \in Z$, and any other such variations? Edit: I guess I could, seeing as how no matter how you write it, $k$ still remains an arbitrary integer. $\endgroup$ – AleksandrH Jun 15 '17 at 23:06
  • $\begingroup$ Welcome @AleksandrH, for your question - Yes, definitely you can write many variations, and can also answer accordingly, for example : Range of this function is all the integers of the form $4k+13$, and it's absolutely correct, but that seems to be a bit awkward, since we do not write a remainder greater than the divisor itself. For example - question "What is the remainder when 53 is divided by 10" so, here both 13 and 3 are correct answers, but we will generally answer 3. $\endgroup$ – Jaideep Khare Jun 15 '17 at 23:11
  • $\begingroup$ Thank you! That makes sense. $\endgroup$ – AleksandrH Jun 15 '17 at 23:16

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