Question

(a) Prove that the set $\mathit{S}$of rational numbers (in lowest terms) with odd denominators is a subring of

$\mathit{\mathbb{Q}}$.

Short answer

It is proved that $S$is a subring of$\mathbb{Q}$.

Step-by-step solution

Consider Theorem 3.2

Consider that $\frac{a}{b},\frac{c}{d}\in S$. Then,

$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$

Here, $2\cancel{|}bd$, so under addition, $S$is closed.

Suppose $\frac{a}{b},\frac{c}{d}\in S$then,

$\frac{a}{b}\frac{c}{d}=\frac{ac}{bd}$

Here,$2\cancel{|}bd$, so under multiplication, $S$is closed.

Show that S is subring of ℚ

It is known that $0=\frac{0}{1}$.Thus, $0\in S$

Also, if $\frac{a}{b}\in S$and $\frac{-a}{b}\in S$,Then,

$\frac{a}{b}+\frac{-a}{b}=0$

Therefore, $S$satisfied all the conditions of Theorem 3.2.

Hence, it is proved $S$is a subring of $\mathbb{Q}$.