Proving $<$ is transitive on $\mathbb{Q}$. I feel a little bit stupid asking this;
I am asked to prove that, for all rational numbers if, x < y and y < z then x < z.
I have said this;
$ x + 0 < y $
$ x - z + z < y$
$ x - z < y- z $
but $ y - z < 0$
so $ x - z < y- z $ implies $ x - z < 0 $.
I had a little search before posting this just to make sure its not a duplicate, if it is i will delete it right away, sorry and thanks in advance.
ORDERING OF THE RATIONALS:
Let x and y be rational numbers. We say that x > y iff x - y is a positive rational number, and x < y iff x - y is a negative rational number.
I think this is the information that was missing. 
 A: We have $$z-x=(z-y)+(y-x)$$
We know $z-y$ is a positive rational number, since $y<z$.  We also know that $y-x$ is a positive rational number, since $x<y$.  We now need some sort of property that the sum of two positive rational numbers is again a positive rational number.  With that property we know that $z-x$ is a positive rational number, and hence $x<z$.
A: As Peter Franek pointed out in his comment, your proof is flawed.
You used if $x-z<y-z$ and $y-z<0$ then $x-z<0$ 
This is the same as saying if $x<y$ and $y<z$ then $x<z$. 
All you have to do to see that is set $x'=x-z,y'=y-z,z'=0$
As for how to prove this, use the axioms of the real number system. 
By the axiom:
$x\leq y$ and $y\leq z$ imply $x\leq z$ for any real numbers $x,y,z$.
Now, say $x<y$ and $y<z$, it's the same as $x\leq y$ and $y\leq z$ and $x\neq y$ and $x\neq z$. 
This implies that $x\leq z$. What we have left to prove is that $y\neq z$
Assume $y=z$, then $x\leq y$ and $y\leq x$ and $x\neq y$, that is $x=x$ and $x\neq x$ which leads to a contradiction. Thus, $x\neq z$.
This ends the proof.
