Why is $f(n) = n + 6 \mod{1729}$ not injective?

Question

The exercise was to determine if the function $f : \mathbb{Z} \rightarrow \mathbb{Z}_{1729}, f(n) = n + 6 \mod{1729}$ is injective or not.

My thinking was that $f(-7) = 1728\\ f(-6) = 0\\ f(-5) = 1\\ \vdots\\ f(0) = 6\\ f(1) = 7\\ \vdots\\ f(1722)=1728\\ f(1723) = 0$

and so on, we will get all numbers in $\mathbb{Z}_{1729}$.

However, this turned out to be wrong, but the solution doesn't say why, though. Why is this function not injective?

Answer 1

You seem to be thinking surjective, meaning that the image of $\mathbb{Z}$ under $f$ is all of $\mathbb{Z}_{1729}$. Injective means there would not be two different integers, say $n,m$ such that $f(n)=f(m)$. Equivalently, if $f(n)=f(m)$ then we must have $n=m$.

This cannot be the case for $f$ since it repeats at least every 1729 integers. For example, $$ f(0)=0+6 \equiv 0 \mod 1729 $$ $$ f(0+1729)=f(1729)=1729 \equiv 0 \mod 1729 $$ In fact, this will work for any integer $n$: $f(n+1729)=f(n)$. So $f$ cannot be injective.

Answer 2

There is no injective function from an infinite set to a finite one.

Answer 3

You are definitely confusing surjective with injection.

SURjective means "we will get all the numbers in $\mathbb Z_{1729}$".

INjective means "we will never get any number in $\mathbb Z_{1729}$ more than once".

$f(-6) = 0$ and $f(1723) = 0$ is enough to prove we get some numbers at least twice.

So $f$ is NOT injective.

.......

Injective means if $f: X\to Y$ then for any $k \in Y$ there is at most one $a \in X$ so that $f(a) = k$.

Surjcetive means if $f: X \to Y$ then for every $k \in Y$ there is at least one $a \in X$ so that $f(a) = k$.

So if $f(a) = k$ has solutions for every $k$ it is surjective. If for any $k$, $f(a) = k$ does not have a solution it is not surjective.

And if $f(a) = k$ has a unique or no solution for every $k$ it is injective. If for any$ $k$, $f(a) = $k\in \mathbb Z_{1729}$ more than one solution it is not injective.

For any $k \in \mathbb Z_{1729}$ then $f(k - 6 + n\cdot 1729)=k$.

So for any $k$ there is more than one solution. So it is not injective.

And for every $k$ there is at least one solution (infinitely many in fact). So it is surjective.