Problem

Determine which of the following sets are groups under addition:

1. the set of rational numbers (including \(0 = \frac{0}{1}\)) in lowest terms whose denominators are odd
2. the set of rational numbers in lowest terms whose denominators are even together with 0
3. the set of rational numbers of absolute value \(\lt\) 1
4. the set of rational numbers of absolute value \(\ge\) 1 together with 0
5. the set of rational numbers with denominators equal to 1 or 2
6. the set of rational numbers with denominators equal to 1, 2 or 3

Solution

a

Let \(S\) be the set as defined by (a).

For any \(\frac{a}{b}\) and \(\frac{c}{d} \in S\), \(\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}\). Since \(b\) and \(d\) are both odd, so is \(bd\) and thus the denominator of \(\frac{ad+bc}{bd}\) in lowest terms must be odd. Therefore, addition over \(S\) is a binary operation, or \(S\) is closed under addition.

• associativity: By the property of addition over real numbers, associativity holds.
• identity: \(\forall x \in S\), \(x + \frac{0}{1} = x = \frac{0}{1} + x\). Therefore, an identity exists.
• inverse: \(\forall x \in S\), \(-x \in S\) and \(x + (-x) = (-x) + x = 0\)

Therefore, \(S\) is a group under addition.

b

Let \(S\) be the set as defined by (b). Then, \(\frac{1}{2} \in S\). However, \(\frac{1}{2} + \frac{1}{2} = 1 = \frac{1}{1} \notin S\) because the denominator in lowest terms is not even.

Therefore, addition over \(S\) is not a binary operation, and thus \(S\) is not a group under addition, or \(S\) is not closed under addition.

c

Let \(S\) be the set as defined by (c). Then, \(\frac{1}{2} \in S\). However, \(\frac{1}{2} + \frac{1}{2} = 1 \notin S\).

Therefore, addition over \(S\) is not a binary operation, and thus \(S\) is not a group under addition, or \(S\) is not closed under addition.

d

Let \(S\) be the set as defined by (d). Then, \(-1 \in S\) and \(\frac{3}{2} \in S\). However, \(-1 + \frac{3}{2} = \frac{1}{2} \notin S\).

Therefore, addition over \(S\) is not a binary operation, and thus \(S\) is not a group under addition, or \(S\) is not closed under addition.

e

Let \(S\) be the set as defined by (e). \(\forall x \in \mathbb{Z}\), \(\frac{x}{1} = \frac{2x}{2}\). Therefore, \(S = \{\frac{x}{2} | x \in \mathbb{Z}\}\). It follows that for any \(x = \frac{a}{2}\) and \(y = \frac{b}{2}\) in \(S\), \(x + y = \frac{a+b}{2} \in S\). Therefore, addition over \(S\) is a binary operation, or \(S\) is closed under addition.

• associativity: By the property of addition over real numbers, associativity holds.
• identity: \(0 = \frac{0}{1} \in S\) and, by the property of addition over real numbers, it’s an identity.
• inverse: \(\forall x = \frac{a}{2} \in S\), \(-x = -\frac{a}{2} \in S\) and \((-x) + x = x + (-x) = 0\).

Therefore, \(S\) is a group under addition.

f

Let \(S\) be the set as defined by (f). Then \(a = \frac{1}{2} \in S\) and \(b = \frac{1}{3} \in S\). However, \(a + b = \frac{5}{6} \notin S\).

Therefore, addition over \(S\) is not a binary operation, and thus \(S\) is not a group under addition, or \(S\) is not closed under addition.

(a) and (e) are groups under addition.

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