# properties of relations in set theory

I'm having a problem with the following questions (basically one question with several subquestions), here's the question and afterwards I'll write what I did.

Note: the questions are truth/false questions. I should only write if it's true or false.

$f:\mathbb{Q}-\{1\} \to \mathbb{Q}, g:\mathbb{Q}-\{1\}\to \mathbb{Q}-\{1\}$. they are defined as following: $f(x)=g(x)=\frac {x}{x-1}$ for every $x \in \mathbb{Q} - \{1\}$

a) $f$ is injective.

b) $g$ is surjective.

c) if $R$ is a relation over $A$ then $R= R^2$

d) the number of relations over $A=\{1,2,3\}$ in which there are exactly $2$ maximal variables is equal to the number of relations over $A$ in which there exists a bigger variable.

e) the number of non full relations over $A=\{1,2,3,4\}$ that has a single mininmal variable and single maximal variable is $8$.

What I tried to do:

a) $f$ is injective, since $f:\mathbb{Q}-\{1\}\to\mathbb{Q}$. i don't see how for a single $x$, $f(x)=\frac{x}{x-1}$ can get $2$ different values to disprove its injectivity.

b) $g$ is surjective by definition. If I understand it correctly, then $gg = g \circ g$ is surjective.

c) (It's under the same question on set theory, but it's not related to the previous details.) I think it's wrong. Not every relation $R$ is a relation over $A$ that applies $R = R^2$, i.e = not every $R = R \times R$.

d) I'm not sure how to solve it or how to write it mathmatically (even though I only need to write if it' correct or wrong). What do they mean by the numer of relations in which there are exacty two maximal values? I'm lost there.

e) if I counted it right, this sentence should hold true.

If I did mistakes, please help me and correct me. I've tried to understand and explain what I did and why.

• "$g$ is surjective by definition" is not a valid argument. You need to demonstrate this fact using the definition of surjective. – Christian Sykes Aug 13 '17 at 13:35
• To check whether $f$ is injective, you need to suppose $f(x)=f(y)$. Does this imply that $x=y$? You will have to make an argument if it does or provide a counterexample if it does not. – Christian Sykes Aug 13 '17 at 13:39
• "I don't see how for a single $x,f(x)=\frac{x}{x-1}$ can get $2$ different values..." on base of such (wrong) reasoning every function is injective. You must check wether for more than one $x$ the function can get the same value. – drhab Aug 13 '17 at 13:46
• thank you very much for the comments, but a stated those are truth/false questions. however i learn a lot from your elaborate explanations, so thank you for that! – BeginningMath Aug 13 '17 at 14:15
• Even if the homework is true/false, you should go through all the details anyway. The instructor is likely out to save themselves a bit of work on grading. Your goal should be to learn the material. – Christian Sykes Aug 13 '17 at 20:12

For a), write $\frac{x}{x-1}=\frac{y}{y-1}$. Then deduce that $(y-1)x=(x-1)y$ and $yx-x=xy-y$, which gives $x=y$. Therefore $f$ is injective because $f(x)=f(y)$ implies that $x=y$.
For a), we know that $g$ is surjective because for any $\frac{p}{q}\in\mathbb{Q}\setminus\{1\}$, we can solve $\frac{x}{x-1}=\frac{p}{q}$ to obtain $x=\frac{p}{p-q}\in\mathbb{Q}\setminus{1}$. (This is valid because $p\ne q$ and $q\ne 0$.) This gives that $g\circ g$ is surjective.
For c), $R^2$ is not the same as $R\times R$. The notation $R_1R_2$ for relations means we take the composite relation; i.e., $$R_1R_2 := \{(a,c)\in A\times A \mid (a,b)\in R_1\ \text{and}\ (b,c)\in R_2\ \text{for some b\in A}\}.$$ A possible counterexample would be $R:=\{(a,b),(b,c)\}$ on the set $\{a,b,c\}$. Then $R^2=\{(a,c)\}\ne R$.