Chapter 4.7, Problem 8ES

Fill in the blanks for the following proof that the difference of any rational number and any irralional number is irrational.

Proof (by contradiction]: Suppose not. That is, suppose that there exist (a) x and (b) y such that $x-y$ is rational. By definition of rational, there exist integers a, b, c, and d with $b\ne 0$ and $d\ne 0$ so that $x=$(c) and $x-y=$ (d) . By substitution,
   $\frac{a}{b}-y=\frac{c}{d}$

Adding y and subtracting $\frac{c}{d}$ on both sides gives
   $y=$ (e) by substitution
   $=\frac{ad}{bd}-\frac{bc}{bd}$

   $=\frac{ad-bc}{bd}$ by algebra.

Now both $ad-bc$ and bd are integers became products and diffrences of (f) are (g) . And $bd\ne 0$ by the (h) . Hence y is a ratio of integers with a nonzero denominator, and thus y is (i) by de?nition of rational. We therefore have both that y is irrational and that y is rational, which is a contradiction. [Thus the supposition is false and the statement to be proved is true.]