Question 1042996
Since x and y are rational there exist
integers d and e such that {{{x = d/e}}}
and integers f and g such that {{{y = f/g}}}
Notice that
{{{x = d/e}}} = {{{x = (d/e)(g/g)}}} = {{{x = dg/eg}}}
and
{{{y = f/g}}} = {{{y = (f/g)(e/e)}}} = {{{y = fe/eg}}}
So we have {{{x = dg/eg}}} and {{{y = fe/eg}}}
Since the product of two integers is an integer,
we can set c = eg where c is an integer
set a = dg where a is an integer
set b = fe where b is an integer
Substituting we have
{{{x = a/c}}} and {{{y = b/c}}}