I'm having trouble grasping a specific step in the following proof:
given three rationals $x,y,z$ if $x = z$ and $y = z$ then $x =y $
so it starts with defining x y and z to be a rational.
with $ x = z$ and $y = z $
so there exists integers a,b,c,d,e,f such that $x = \frac{a}{b}$ $y = \frac{c}{d}$ and $z = \frac{e}{f}$ ( b d and f cannot be 0)
so thus we know that since $x = z$ and $y =z$ then $af = eb$ and $de = cf$ due to the equality of rationals.
now this last step is something I don't quite understand. It says that by algebra, that we can multiply $af$ and $de$, along with $eb$ and $cf$ to make:
$afde = ebcf $
then we get $ad = bc$ which shows that $x = y$
How exactly is it valid to multiply af and de and eb and cf and equate each other? How is this a valid substitution?
Rationals are those numbers which can be expressed in p/q form where p,q are coprime.
Let x = $\frac{a}{b}$ and y = $\frac{c}{d} $ and z = $\frac{e}{f}$.
Now if $x = z$ then $a*f=b*e$ and if $y=z$ then $d*e = c*f$
Now we are dealing with $a*f,b*e,d*e,c*f$ which are all some integers/whole numbers.
Now to prove that $x=y$ it is enough to prove that $ad=bc$ So lets now multiply the equation $ a*f = b*e$ with $d*e$ on both sides.
So $a*f*d*e = b*e*d*e $ But we know $d*e=c*f$.
So $a*f*d*e = b*e*c*f $
Now cancelling the common terms $f,e$ we get $ a*d=b*c$ Hence we proved the claim.