# Determination of Functions, 1:1, and Inverse

For the following relations, I need to answer: 1) Is it a function? If not, explain why and stop. Otherwise, 2) What are its domain and image, 3) Is the function 1:1. If not, explain why and stop. Otherwise, 4) What is its inverse function.

i) $\{(x,y): x,y \in \mathbb Z, x|y\}$

ii) $\{(x,y): x,y \in \mathbb N, x|y\text{ and }y|x\}$

iii) $\{(x,y): x,y \in \mathbb N, \binom{x}{y}=1\}$

I am self teaching myself discrete mathematics and these questions which seems quite simple got me a bit confused.

My attempt: i) 1) Function because we cant have the pairs $(x,y)$ and $(x,z)$ where $x$ can be same numbers with two different outputs. 2) $\mathrm{dom}(f) = \mathbb{Z}$ and $\mathrm{im}(f) = \mathbb{Z}$. 3) Not 1:1. For questions ii and iii, I am totally lost.

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I think that you are thinking the second component like the image of the first. I may mistake, but the first doesn't seem a function, it contains all the pairs of the form $(1,*)$. The second is the identity map on integers. For the third, I think this also not be a function because if we admit that $0!=1$, than it contains the pairs $(1,0)$ and $(1,1)$. – Lorban Apr 14 '13 at 19:13

i) $2|6 \quad \text{and} \quad 2|8 \quad$ so i) can not be function because you have two pairs (2,6),(2,8) with the same x

ii) Can you complete next statement? : $\forall x,y \in N: x|y \,\,\,\, \& \,\,\,\ y|x \implies x \,\,\color{red}?\,\, y$

iii) $\forall x,y\in N: {x \choose y}=1 \Leftrightarrow (y = 0 \quad \text{or} \quad y = x)$. From Pascal triangle you can see that first and the last element in the each line are equal. It means, iii) is not function

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Hint: Instead of messing with $\mathbb{Z}$ and $\mathbb{N}$ try considering the set of integers $$S=\{ x \in \mathbb{Z} \mid 1\leq x \leq 10 \}.$$ The rephrased questions using $S$ would be:

i) $\{(x,y) \mid x,y \in S, x|y \}$,

ii) $\{(x,y) \mid x,y \in S, x|y \text{ and } y|x \}$,

iii) $\{(x,y) \mid x,y \in S, { x \choose y}=1 \}$.

Now you can very explicitly write the sets to get a feeling for what the original questions are asking, how to tackle the problems, and whether the sets are functions.

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