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.
Thank you in advance!
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 $R^2=\{(a,c)\}\ne R$.