I am given these 5 questions about the relations
- $xRy$ if $x\ge y^2$ (on real numbers), and
- $aRb$ if $a|b$ (on $\mathbb N$).
a. Find the domain and range of R.
b. Prove or disprove that R is reflexive.
c. Prove or disprove that R is symmetric.
d. Prove or disprove that R is transitive.
e. Prove or disprove that R is antisymmetric.
I have no idea how to do a, because I haven't done it without being given some sort of set written out. for b, would I be able to say that x is reflexive because it is greater than or equal to itself, and also a because a divides itself?
I'm also not sure how to do c or e for both relations.
Any help is greatly appreciated.
Recall that a relation $R$ on a set $X$ is a subset of $X\times X$, i.e. it contains elements of the form $(a,b)$, $a,b\in X$. The domain is defined as $\{a\mid \exists b\in X\; (a,b)\in X\}$, and the range is $\{b\mid \exists a\in X\; (a,b)\in X\}$. Less formally, the domain is the set of $x$-coordinates of the relation, and the range is the set of $y$-coordinates of the relation.
We'll look at your first relation first, then your second after. For part a) if you think about it geometrically, the relation defines the set of points in $\mathbb{R}^2$ to the right the parabola $x=y^2$ (including the boundary). So we would expect the range to be all real numbers, and the domain to be all nonnegative real numbers. Sure enough, $(y^2,y)\in R$ for all $y\in\mathbb{R}$, so the range is simply $\mathbb{R}$. If $(x,y)\in R$, then $x\geq y^2\geq0$, which implies the domain is a subset of $\mathbb{R}_{\geq0}$, and if $x\in\mathbb{R}_{\geq0}$, then $(x,\sqrt{x})\in R$, proving that the range is equal to the nonnegative reals.
The statement that $R$ is reflexive amounts to $x\geq x^2$ for all values of $x$. However we know that this is not always true, take $x=2$ as an example. So this relation is not reflexive.
$R$ is also not symmetric. $(1,0)\in R$, but $(0,1)\notin R$.
$R$ is also not transitive. $\left(\frac{1}{16},\frac{1}{4}\right)\in R$ and $\left(\frac{1}{4},\frac{1}{2}\right)\in R$, however $\left(\frac{1}{16},\frac{1}{2}\right)\notin R$, since $\frac{1}{16}\not\geq\left(\frac{1}{2}\right)^2$.
$R$ is also not antisymmetric. $\left(\frac{1}{3},\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{3}\right)\in R$, but $\frac{1}{3}\neq\frac{1}{2}$.
For the second relation, since $a\mid a$ for all $a\in\mathbb{N}$, then both the domain and range of $R$ are $\mathbb{N}$.
For this same reason, $R$ is reflexive.
$R$ is not symmetric, for example $1\mid 2$, but $2\nmid 1$.
$R$ is transitive, because $a\mid b$ and $b\mid c$ implies that $a\mid c$. The reason for this, is that $a\mid b$ and $b\mid c$ imply that there exist integer $k_1,k_2$ such that $b=k_1 a$ and $c=k_2 b$. But then $(c=k_1k_2) a$, so $a\mid c$.
$R$ is also antisymmetric: for natural numbers $a,b$, if $a\mid b$ and $b\mid a$, then $a=b$. The reason for this, is that if $a\mid b$ and $b\mid a$ then there exist $k_1,k_2\in\mathbb{N}$ such that $k_1a=b$ and $k_2b=a$. This means that
\begin{align*} k_1(k_2b) &= b \\ \implies k_1k_2 &= 1 \\ \implies k_1,k_2 &= 1 \\ \implies a &= b. \end{align*}