Let $f: \mathbb{Z}^2 \to \mathbb{Z}^2$ be defined as $f(m, n) = (m + n, 2m − 5n)$ . Is $f$ a bijection, i.e., one-to-one and onto?
Since my function is mapped on the domain consisting of all integers I was wondering if it is valid to have $m$ and $n$ be two non-integers that form an integer. For example is $f(0,1) = (1/7 + -1/7, 2(1/7) - 5(-1/7))$ valid or do $m$ and $n$ have to be integers even before they are used in the equation?
Thanks!!
Let $f : X \to Y$. We say $f$ is surjective (onto) if for every $y \in Y$, there is $x \in X$ such that $f(x) = y$.
Note that $x$ must be in $X$, the domain of $f$. Here, $f$ has domain $\mathbb{Z}^2$ so you need $(m, n) \in \mathbb{Z}^2$.
From my reading of your first sentence, since the domain of $f$ is $\mathbb{Z}^2$, $m$ and $n$ have to be integers, yes.
So, if you're looking for $m$ and $n$ which map to $(0,1)$, for example, you wouldn't consider numbers like $\frac{1}{7}$ or $-\frac{1}{7}$ as possibilities for $m$ or $n$.
The function is clearly one-one since if $f(m_1,n_1)=f(m_2,n_2)$, then $(m_1+n_1,2m_1-5n_1)=(m_2+n_2,2m_2-5n_2)$ which implies $m_1=m_2$ and $n_1=n_2$. Since $f$ maps a finite set to itself and it is one-one it is onto. Hence $f$ is bijective.
Since your domain is $Z_2$ you need $(m,n) \in Z_2$
