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 &lt; y and y &lt; z then x &lt; z.

I have said this;

$ x + 0 &lt; y $

$ x - z + z &lt; y$

$ x - z &lt; y- z $

but $ y - z &lt; 0$

so $ x - z &lt; y- z $ implies $ x - z &lt; 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 &lt; y iff x - y is a negative rational number.

I think this is the information that was missing.
We have $$z-x=(z-y)+(y-x)$$
We know $z-y$ is a positive rational number, since $y&lt;z$.  We also know that $y-x$ is a positive rational number, since $x&lt;y$.  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 $x&lt;z$.
As Peter Franek pointed out in his comment, your proof is flawed.

You used if $x-z&lt;y-z$ and $y-z&lt;0$ then $x-z&lt;0$ 

This is the same as saying if $x&lt;y$ and $y&lt;z$ then $x&lt;z$. 

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 $x&lt;y$ and $y&lt;z$, 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.