Problem

Decide which of the following are subrings of \(\mathbb{Q}\):

1. the set of all rational numbers with odd denominators (when written in lowest terms)
2. the set of all rational numbers with even denominators (when written in lowest terms)
3. the set of nonnegative rational numbers
4. the set of squares of rational numbers
5. the set of all rational numbers with odd numerators (when written in lowest terms)
6. the set of all rational numbers with even numerators (when written in lowest terms)

Solution

a

Let \(S\) be the set of all rational numbers with odd denominators (when written in lowest terms). Let \(x = \frac{a}{b}, y = \frac{c}{d}\) be arbitrary elements of \(S\) in lowest terms.

First, we show that \(S\) is a group.

• closure: \(x + y = \frac{ad + bc}{bd} \in S\) because \(bd\) is odd.
• associativity: addition in \(S\) is associative by the property of that in \(\mathbb{Q}\).
• identity: the identity \(0 = \frac{0}{1}\) exists in \(S\).
• inverse: for any \(x = \frac{a}{b}\), \(-x = -\frac{a}{b}\) exists in \(S\).

Next, because both \(b\) and \(d\) are odd, so is \(bd\). As a result, the denominator of \(\frac{ac}{bd}\) in lowest terms must be odd. That is, \(xy = \frac{ac}{bd} \in S\).

Therefore, \(S\) is a subring of \(\mathbb{Q}\).

b

Let \(S\) be the set of all rational numbers with even denominators (when written in lowest terms).

\(0\) is the additive identity in \(\mathbb{Q}\) but \(0 = \frac{0}{1} \notin S\). Therefore, \(S\) is not a group.

Thus, \(S\) is not a subring of \(\mathbb{Q}\).

c

Let \(S\) be the set of nonnegative rational numbers.

\(1 \in S\) but there is no \(x \in S\) such that \(1 + x = 0\) because \(-1 \notin S\). Therefore, \(S\) is not a group.

Thus, \(S\) is not a subring of \(\mathbb{Q}\).

d

Let \(S\) be the set of squares of rational numbers.

\(1 \in S\) but there is no \(x \in S\) such that \(1 + x = 0\) because \(-1 \notin S\). Therefore, \(S\) is not a group.

Thus, \(S\) is not a subring of \(\mathbb{Q}\).

e

Let \(S\) be the set of all rational numbers with odd numerators (when written in lowest terms).

\(0\) is the additive identity in \(\mathbb{Q}\) but \(0 = \frac{0}{1} \notin S\). Therefore, \(S\) is not a group.

Thus, \(S\) is not a subring of \(\mathbb{Q}\).

f

Let \(S\) be the set of all rational numbers with even numerators (when written in lowest terms). Let \(x = \frac{a}{b}, y = \frac{c}{d}\) be arbitrary elements of \(S\) in lowest terms. Note that since these are expressed in lowest terms, \(a\) and \(c\) are even while \(b\) and \(d\) are odd.

First, we show that \(S\) is a group.

• closure: \(x + y = \frac{ad + bc}{bd} \in S\) because \(ad + bc\) is even while \(bd\) is odd, and thus the numerator must be even in lowest terms.
• associativity: addition in \(S\) is associative by the property of that in \(\mathbb{Q}\).
• identity: the identity \(0 = \frac{0}{1}\) exists in \(S\).
• inverse: for any \(x = \frac{a}{b}\), \(-x = -\frac{a}{b}\) exists in \(S\).

Next, \(ac\) is even while \(bd\) is odd. As a result, the numerator of \(\frac{ac}{bd}\) in lowest terms must be even. That is, \(xy = \frac{ac}{bd} \in S\).

Therefore, \(S\) is a subring of \(\mathbb{Q}\).

(a) and (f) are subrings of \(\mathbb{Q}\). Others are not.

$$\tag*{$\blacksquare$}$$