i have the following function: $f:\mathbb{Z}\rightarrow\mathbb{Z}$ , $f(n)=n(n+1)$

Calculate: $f^{-1}({1})$, $f^{-1}({2})$, $f^{-1}(\mathbb{{N}})$ (Natural numbers)

for the first one i got the empty set, for the second one I got the solution $\{-2,1\}$.

However for the last one I couldn't really find a solution. If i put in an even number, the result is also even. If i put in an uneven number, I also get an even result. Therefore not every natural number is in the solution.

Where am I making a mistake here?

edit: can't really make it format right. my bad! but its supposed to be the set that contains all natural numbers, not only natural numbers

  • 3
    $\begingroup$ Hint: For $n\in \mathbb{Z}$, can $n(n+1)$ be negative? $\endgroup$ – Jason DeVito Sep 21 '20 at 15:07
  • $\begingroup$ oh you're right. n(n+1) is positive for all integers except 0 $\endgroup$ – 23408924 Sep 21 '20 at 15:17
  • $\begingroup$ And $-1{}{}{}{}$ $\endgroup$ – Kenta S Sep 21 '20 at 15:30

Think like this: $$f^{-1}(\mathbb{N}) = \bigcup_{a\in\mathbb{N}} f^{-1}{(a)}$$


  • $f^{-1}(1)=\emptyset$
  • $f^{-1}(2)=\{-2,1\}$
  • $f^{-1}(3)=\emptyset$
  • $f^{-1}(4)=\emptyset$
  • $f^{-1}(5)=\emptyset$
  • $f^{-1}(6)=\{-3,2\}$
  • $f^{-1}(1)=\emptyset$
  • $\vdots$
  • $f^{-1}(12)=\{-4,3\}$
  • $\vdots$

As you can see, most of the values will be empty, but numbers of the form $(n)(n+1)$ are non empty, and their values will give you the whole list of $\mathbb{Z}$ except for $\{-1,0\}$.

  • $\begingroup$ Thanks! so basically the solution is the set of numbers that can be written in the form of n(n+1). $\endgroup$ – 23408924 Sep 21 '20 at 18:50
  • 1
    $\begingroup$ @23408924 No!, the solution is all $\mathbb{Z}$ except for $-1$ and $0$. Remember that you were asked by inverse map $f^{-1}$, so you need all the numbers $n$, for which $n(n+1)$ is a natural number $\endgroup$ – Luis Felipe Sep 21 '20 at 18:55
  • $\begingroup$ oh yeah you're right I misunderstood. I think i need to reread some stuff. thank you very much! $\endgroup$ – 23408924 Sep 21 '20 at 18:56
  • $\begingroup$ Small correction: here, $f^{-1}$ is not the inverse map, but the inverse image (or preimage). It makes a difference, since $f$ is not bijective and hence an inverse map does not exist. In particular, the preimage $f^{-1}$ expects sets as an argument, and not elements of $\mathbb Z$ (or $\mathbb N$). $\endgroup$ – Marktmeister Sep 22 '20 at 5:59
  • $\begingroup$ @Marktmeister you are right, sorry for my vas english i will correct $\endgroup$ – Luis Felipe Sep 22 '20 at 9:43

Make a plot of the function $$f:\quad{\mathbb Z}\to{\mathbb Z}\qquad n\mapsto n(n+1)\ .$$ This plot consists of isolated points in the $(x,y)$-plane. Then determine the set $S\subset{\mathbb Z}$ of all $n\in{\mathbb Z}$ satisfying $f(n)\in{\mathbb N}$, i.e., $f(n)$ integer and $\geq1$.


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