Which of the properties, Reflexive, Irreflexive, Symmetric, Asymmetric, Antisymmetric, Transitive, Linear, does F satisfy? Let  $S={(n,m) ∶n,m∈Z^+}$. Define the relation F on S by
${(n,m),(i,j)}∈F$ if and only if $nj=mi$.
In other words, let $F = {((n, m), (i, j)) ∈ S × S: nj = mi}$.
Proof F is reflexive:
Show that for all $(n, m) ∈ S, ((n, m), (n, m)) ∈F$.
Choose $(n_0, m_0) ∈ S$.
By the Commutative Property of Multiplication, we know that $n_0m_0 = m_0n_0$.
Hence, $((n_0, m_0), (n_0, m_0)) ∈F$.
Thus, for all $(n, m) ∈ S, ((n, m), (n, m)) ∈F$, and we have shown that F is reflexive.
Proof F is not irreflexive:
Show that $∃(n, m) ∈ S$ such that $((n, m), (n, m)) ∈F$.
Let $(n_0, m_0) = (1, 2) ∈S$.
Then, by the Commutative Property of Multiplication, $1(2) = 2(1)$.
Hence, $((1,2), (1,2))∈F$.
Thus, $∃(n, m) = (1, 2) ∈ S$ such that $((n, m), (n, m)) ∈F$.
Therefore, F is not irreflexive.
Proof F is symmetric:
Show that if $((n, m), (i, j)) ∈ F$, then $((i, j), (n, m)) ∈F$.
Choose $(((n_0, m_0), (i_0, j_0))  ∈ S × S$.
If $((n_0, m_0), (i_0, j_0)) ∈ F$, then $n_0j_0 = m_0i_0$.
By the Commutative Property of Multiplication, $j_0n_0 = n_0j_0 = m_0i_0  = i_0m_0$.
We can rewrite $j_0n_0 = i_0m_0$ as $i_0m_0 = j_0n_0$.
Thus, $((i_0, j_0), (n_0, m_0)) ∈ F$.
Hence, if $((n, m), (i, j)) ∈ F$, then $((i, j), (n, m)) ∈F$.
Therefore, F is symmetric.
Proof F is not asymmetric:
Let $n_0,m_0=1$.
Then $j=i$ and $i=j$.
Since $i=i$ and $j=j$,$∀i,j∈R,∃n,m=1$ such that if $j=i$ then $i=j,∀i,j∈R$.
Hence F is not asymmetric.
Proof F is antisymetric:
Choose $i=i_0$  and $j=j_0∈R$.
Then $nj_0=mi_0$ and $ni_0=mj_0$.  
F is transitive: 
Choose $l=l_0,  m=m_0$ and $n=n_0 ∈Z^+ $such that $l<m<n$.
Assume ${(l,m),(i,j)}∈F$ and therefore $lj=mi$.
Assume ${(m,n),(i,j)}∈F$ and therefore $mj=ni$.
We then need to show that ${(l,n),(i,j)}∈F$ and therefore $lj=ni$.  
F is linear:  
I got stuck on proving or disproving antisymmetry and transitivity and don't know where to start on proving or disproving linearity.
Also any checking of my other proofs would be greatly appreciated.
 A: Hint: Think of a pair $(m,n)$ as a $\frac{m}{n}$ , so that $((m,n),(i,j)) \in F$ iff $\frac{m}{n}=\frac{i}{j}$ , although $(m,0)$ and $(0,n)$ are some special cases. So $F$ is an equivalence relation: reflexive, symmetric, transitive, and not the rest of the properties.
But one still write $mj=ni$ while thinking $\frac{m}{n}=\frac{i}{j}$ 'cause there's no division in $Z$ at all.
More, $S/F$ gives classical introduction of $Q^+$.
A: Your proofs that the relation is reflexive, not irreflexivity, and symmetric are all good.
I have qualms with the not asymmetric proof - it doesn't read coherently as the others do, and it is overly complicated. All that is required is to exhibit an element in the relation which breaks asymmetry - in fact the example you used for the not irreflexive proof works just as well here.
Your attempt at the antisymmetric proof may be hindered by the fact that you are trying to prove the wrong thing - the relation is not antisymmetric, and you can see this with another counterexample (say the pair $((1,2),(2,4))$).
Transitivity is a bit trickier, but you can show it is also not true with a counterexample (try using the $(0,0)$ element of $S$ with things (and note that the $(0,0)$ element is related to every other element)).
I'm afraid I can't help you with the linear part though - I'm not sure what linear means in this context (If you comment with a definition I could probably update this part of the answer).
