Prove that a rational number minus an irrational number must be irrational.

Question

Please help with this homework problem I have! I don't know how to prove this.

Answer 1

You can answer this by contradiction. Assume it is rational and see what happens.

Answer 2

Hint. A rational number minus a rational number is a rational number.

Answer 3

$$x\equiv \frac{a}{b},y \,\mathrm{ irrational},z\equiv\frac{c}{d}$$ ($x,z \,$rational, $ y \,$irrational,$\{a,b,c,d\}\in\mathbb{Z}$) $x-y=z\implies \frac{a}{b}-y=\frac{c}{d}\implies ad-bdy=bc\implies -bdy=bc-ad\implies y=\frac{ad-bc}{bd}$. But we started with $\{a,b,c,d\} \in \mathbb{Q}\implies y\in\mathbb{Q}$, a contradiction!

Small proofs of rational #s are closed under multiplication, subtraction, and division: Multiplication: $\dfrac{a}{b}\dfrac{c}{d}=\dfrac{ac}{ef}\in \mathbb{Q}$ (Definition of a rational)

Division: $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\dfrac{a}{b}\dfrac{d}{c}=\dfrac{ad}{bc}$

Addition:$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad}{bd}+\dfrac{bc}{bd}=\dfrac{ad+bc}{bd}$ (Integers closed under addition:see https://proofwiki.org/wiki/Integer_Addition_is_Closed)

Subtraction: Proved by $c\mapsto -c$ in addition proof