Let $A$ be relation $[((a,b),(c,d)) \in \mathbb{N}^2 \times \mathbb{N}^2; (\exists k \in \mathbb{N}) (c=ka \land d=kb)]$
Prove that it is reflexive, antisymmetric, and transitive.
1.) Rexlexivity: $((a,b),(a,b)) \in A \to a=ka \land b=kb \to k=1 \land k=1$
It is reflexive because we get a tautology.
2.)Antisymmetricity: $((a,b),(c,d)) \in A \land ((c,d),(a,b)) \in A \implies (a,b)=(c,d)$
$(c=ka \land d=kb) \land (a=kc \land b=kd) \to (k=\frac{c}{a} \land k=\frac{d}{b}) \land (k=\frac{a}{c} \land k=\frac{b}{d})\to (\frac{c}{a}=\frac{d}{b})\land (\frac{a}{c}=\frac{b}{d})\to$
$ \to (\frac{a}{c}=\frac{b}{d})=(\frac{a}{c}=\frac{b}{d}) \to \frac{a}{c}=\frac{b}{d}$
This can only be true when $(a,b)=(c,d)$, so $a=c \land b=d$
3.) Transitivity: $((a,b),(c,d)) \in A \land ((c,d),(x,y)) \in A \implies ((a,b),(x,y)) \in A$
$(c=ka \land d=kb) \land (x=kc \land y=kd) \to (k=\frac{c}{a}\land k=\frac{d}{b}) \land (k=\frac{x}{c} \land k=\frac{y}{d}) \to (\frac{c}{a}=\frac{d}{b})\land(\frac{x}{c}=\frac{y}{d})\to$
$\to (\frac{c}{d}=\frac{a}{b})\land(\frac{x}{y}=\frac{c}{d})$ So $\frac{c}{d}=\frac{a}{b}=\frac{x}{y}$ which means that $\frac{a}{b}=\frac{x}{y}$ and from this we know that the relation is transitive.
In conclusion, the relation is reflexive, antisymmetric, and transitive. It doesn't have a maximum so it doesn't have a maximal element either. It does have a minimal element which is and the minimal element is $((0,0)(0,0))$
Is this correct?
Thanks!
For reflexivity, you have the right idea, but I think that you don't phrased it correctly. It should be something along these lines:
since $k=1$ is a natural number, and $a=1a$ and $b=1b$, then $((a,a),(a,a)) \in A$.
Thus, $A$ is reflexive. (Notice that you inverted the order of the reasoning; I'm not sure if it is wrong or just confusing.)
Now, for anti-symemtry, suppose $(a,b)A(c,d)$ and $(c,d)A(a,b)$.
Then there exist $k,k' \in \mathbb N$ such that $c=ka$, $d=kb$, $a=k'c$ and $b=k'd$.
Thus, $a=k'ka$, whence $k'k=1$, and it follows that $k=k'=1$.
Hence $a=c$ and $b=d$, and therefore $(a,b)=(c,d)$.
We conclude that $A$ is anti-symmetric.
For transitivity, suppose that $(a,b)A(c,d)$ and $(c,d)A(e,f)$.
Then there exist $k,k' \in \mathbb N$ such that $c=ka$ and $d=kb$, and $e=k'c$ and $f=k'd$.
Thus, $e=k'ka$ and $f= k'kb$, yielding $(a,b)A(e,f)$.
Hence, $A$ is transitive.
It follows that $(\mathbb N^2,A)$ is a partially ordered set.
It has a maximum element (thus, the only maximal element), which is $(0,0)$;
indeed, for every $(a,b) \in \mathbb N^2$, we get $0=0a$ and $0=0b$, so $(a,b)A(0,0)$ for all $(a,b) \in \mathbb N^2$.
Concerning minimal elements, it has infinitely many: for example, $(1,1)$ is a minimal element, and it is below precisely the elements of the form $(a,a)$, with $a \in \mathbb N$.
Otherwise, we get those of the form $(a,b)$ with $\gcd(a,b)=1$;
indeed, if $d|a$ and $d|b$ then $(a/d,b/d)A(a,b)$;
conversely if $(a,b)A(c,d)$, then $c=ka$ and $d=kb$, for some $k \in \mathbb N$; if $k=1$ this is just reflexivity; otherwise $k|c$ and $k|d$.