# Determine whether or not a given set of rational numbers is additively closed

**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.6**

Determine which of the following sets are groups under addition:

- The set $A_1$ of rational numbers in lowest terms whose denominators are odd. (Including 0 = 0/1.)
- The set $A_2$ of rational numbers in lowest terms whose denominators are even. (Including 0 = 0/2.)
- The set $A_3$ of rational numbers of absolute value less than 1.
- The set $A_4$ of rational numbers of absolute value at least 1 together with 0.
- The set $A_5$ of rational numbers in lowest terms with denominator 1 or 2.
- The set $A_6$ of rational numbers in lowest terms with denominator 1, 2, or 3.

Solution:

**(1)** We show that $A_1$ is a group under addition. $A_1$ is closed under addition as follows. If $\dfrac{a}{b}$, $\dfrac{c}{d} \in A_1$, then $b$ and $d$ are odd. Then $bd$ is odd. Now $$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}.$$ This fraction is equal to some other fraction $\dfrac{p}{q}$ in lowest terms, such that $q|bd$. Since $bd$ is odd, $q$ must also be odd. So $A_1$ is closed under addition. Addition on $A_1$ is associative since addition on all of $\mathbb{Q}$ is associative. We have $$\frac{0}{1} + \frac{a}{b} = \frac{a}{b} = \frac{a}{b} + \frac{0}{1}$$ for all $\dfrac{a}{b} \in A_1$, so that $\dfrac{0}{1}$ is an identity element under +. Given $\dfrac{a}{b} \in A_1$, note that $\dfrac{-a}{b} \in A_1$. Since $$\frac{a}{b} + \frac{-a}{b} = \frac{0}{b} = \frac{0}{1}$$ and $$\frac{-a}{b} + \frac{a}{b} = \frac{0}{b} = \frac{0}{1},$$ every element of $A_1$ has an additive inverse. Thus $A_1$ is a group under addition.

**(2)** This set is not closed under addition since $\dfrac{1}{6} \in A_2$ but $$\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$$ is not in $A_2$. Hence $A_2$ is not a group under addition.

**(3)** This set is not closed under addition since $\dfrac{1}{2} \in A_3$ but $$\frac{1}{2} + \frac{1}{2} = 1$$ is not in $A_3$. Hence $A_3$ is not a group under addition.

**(4)** This set is not closed under addition since 2, $\dfrac{-3}{2} \in A_4$ but $$2 + \frac{-3}{2} = \frac{1}{2}$$ is not in $A_4$. Hence $A_4$ is not a group under addition. We show that $A_5$ is a group under addition.

**(5)** $A_5$ is closed under addition as follows. Let $\dfrac{a}{1}$, $\dfrac{b}{1}$, $\dfrac{c}{2}$, $\dfrac{d}{2}$ be elements of $A_5$ in lowest terms. Then $\dfrac{a}{1} + \dfrac{b}{1} = \dfrac{a+b}{1}$ is in $A_5$, $$\frac{c}{2} + \frac{d}{2} = \frac{2c + 2d}{4} = \frac{c+d}{2}$$ is in $A_5$, and $$\frac{a}{1} + \frac{c}{2} = \frac{2a+d}{2}$$ is equal to some fraction $\dfrac{p}{q}$ in lowest terms with $q|2$; so $\dfrac{p}{q}$ is in $A_5$. So $A_5$ is closed under addition. Addition on $A_5$ is associative since addition on all of $\mathbb{Q}$ is associative. We have $$\frac{0}{1} + \frac{a}{b} = \frac{a}{b} = \frac{a}{b} + \frac{0}{1}$$ for all $\dfrac{a}{b} \in A_5$, so that $\dfrac{0}{1}$ is an identity element under +. Given $\dfrac{a}{b} \in A_5$, note that $\dfrac{-a}{b} \in A_5$. Since $$\frac{a}{b} + \frac{-a}{b} = \frac{0}{b} = \frac{0}{1}$$ and $$\frac{-a}{b} + \frac{a}{b} = \frac{0}{b} = \frac{0}{1},$$ every element of $A_5$ has an additive inverse. Hence $A_5 $is a group under addition.

**(6)** This set is not closed under addition since $\dfrac{1}{2}$, $\dfrac{1}{3} \in A_6$ but $$\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$$ is not in $A_6$. Hence $A_6$ is not a group under addition.