We define a function $f : \mathbb Z/n\mathbb Z \to\mathbb Z/n\mathbb Z$ by$ f([l])=[7l^2]$.

Question

Let $n$ be a natural number and let $k$ be an integer number. Let $[x]$ be the relation class of $x$ (integer number) modulo $n$.

How now to prove this is a function?

score 1 · Answer 1 · answered Oct 15 '13 at 01:16

1

$\textbf{Hint:}$ You have to prove that the map given by $[l] \mapsto [7l^2]$ is well defined, i.e. that it does not depend on the choice of a representative for $[l]$.

answered Oct 15 '13 at 01:16

Arthur

5,524

Is this correct?
the domain X and codomain Y are given, so we only need to show that:

f is an relation from X to Y f⊆X×Y

The domain of f is Z/nZ, every element in X is related to some element of Y ∀x∈Z/nZ,∃y∈Z/nZ:(x,y)∈f No element of Z/nZ is related to more than one element of Z/nZ ∀x∈Z/nZ,∀y1,y2∈Z/nZ:(x,y1),(x,y2)∈f=>y1=y2
– arieenhenk Oct 15 '13 at 19:29

score 0 · Answer 2 · answered Oct 15 '13 at 01:17

To prove this is a function you need it to be well-defined. That is, there are multiple choices of $l$ for $[l]$; you need to prove that regardless of this choice, the function $f$ will have the same result. That is, you need to prove that if $[l_1]=[l_2]$, then $[7l_1^2]=[7l_2^2]$.

We define a function $f : \mathbb Z/n\mathbb Z \to\mathbb Z/n\mathbb Z$ by$ f([l])=[7l^2]$.

2 Answers2