Algebra Solutions: Determine whether or not a given set of rational numbers is additively closed

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

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.

We show that is a group under addition.
- $A_1$ is closed under addition as follows. If $\frac{a}{b}, \frac{c}{d} \in A_1$ , 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 $\frac{p}{q}$ 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 $\frac{a}{b} \in A_1$ , so that $\frac{0}{1}$ is an identity element under +.
- Given $\frac{a}{b} \in A_1$ , note that $\frac{-a}{b} \in A_1$ . 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 is a group under addition.
This set is not closed under addition since $\frac{1}{6} \in A_2$ but $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$ is not in $A_2$ . Hence $A_2$ is not a group under addition.
This set is not closed under addition since $\frac{1}{2} \in A_3$ but $\frac{1}{2} + \frac{1}{2} = 1$ is not in $A_3$ . Hence $A_3$ is not a group under addition.
This set is not closed under addition since $2, \frac{-3}{2} \in A_4$ 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 is a group under addition.
- $A_5$ is closed under addition as follows. Let $\frac{a}{1}, \frac{b}{1}, \frac{c}{2}, \frac{d}{2}$ be elements of $A_5$ in lowest terms. Then $\frac{a}{1} + \frac{b}{1} = \frac{a+b}{1}$ 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 $\frac{p}{q}$ in lowest terms with $q|2$ ; so $\frac{p}{q}$ 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 $\frac{a}{b} \in A_5$ , so that $\frac{0}{1}$ is an identity element under +.
- Given $\frac{a}{b} \in A_5$ , note that $\frac{-a}{b} \in A_5$ . 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 is a group under addition.
This set is not closed under addition since $\frac{1}{2}, \frac{1}{3} \in A_6$ but $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ is not in $A_6$ . Hence $A_6$ is not a group under addition.

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