Is the set of all rational numbers with odd denominators a subring of $\Bbb Q$? Is the set of all rational numbers with odd denominators a subring of $\Bbb Q$?(When the fraction is completely reduced)
I have tried to apply the subring test on this, and this means I want to show that it is closed under subtraction and multiplication.
To show it is closed under multiplication, I have done the following:
$$\frac{a}{2b+1}\frac{c}{2d+1},a,b,c,d\in\Bbb Z$$
$$=\frac{ac}{4bd+2b+2d+1}$$ Which still has odd denominator.
With regard to subtraction we have:
$$\frac{a}{2b+1}-\frac{c}{2d+1}$$
$$=\frac{a(2d+1)-c(2b+1)}{(2b+1)(2d+1)}=\frac{a(2d+1)-c(2b+1)}{4bd+2b+2d+1}$$
and since the denominator is odd, even if it isn't reduced, it will reduce to another odd denominator.
So we have closure under both of these operations it would seem. Is this sufficient?

Also if it were to show the set of all rational numbers with even demoninator, all I would need is $\frac{1}{2}-\frac{3}{2}=-\frac{2}{2}=-1$ to get a counter example right?
 A: Yes. Actually, it is the ring  $\mathbf Z_{(2)}$ – the localisation of $\mathbf Z$ at the prime ideal $(2)$. It has only one maximal ideal, namely  $\,2\mathbf Z_{(2)}$, and its residue field is isomorpphic to $\mathbf Z/2\mathbf Z$.
A: Your set can be described as the set $S$ of all rational numbers $q\in\mathbb{Q}$ such that $q=a/b$ for some integers $a$ and $b$, where $b$ is odd.
Indeed, as you observe, if $a/b$ is not the representation in minimal terms, the denominator will remain odd when we divide by the greatest common divisor with the numerator.
So you have to prove


*

*$0\in S$

*if $q_1\in S$ and $q_2\in S$, then $q_1-q_2\in S$

*if $q_1\in S$ and $q_2\in S$, then $q_1q_2\in S$

*$1\in S$ (if your definition of subring requires the identity)


or other sets of statements that imply $S$ is a subgroup with respect to addition and a subsemigroup (or submonoid) with respect to multiplication.
Your proof is basically right, but it's easier to observe that if $b$ and $d$ are odd, then $bd$ is odd as well. So, if $q_1=a/b$ and $q_2=c/d$ with $b$ and $d$ odd, then
$$
q_1-q_2=\frac{ad-bc}{bd},\qquad q_1q_2=\frac{ac}{bd}
$$
Thus both $q_1-q_2$ and $q_1q_2$ admit a representation as fractions with odd denominator.
As for your counterexample, you need to prove that assuming
$$
\frac{1}{2}=\frac{a}{b}
$$
with odd $b$ leads to a contradiction. The operations are irrelevant for this. You need to produce a rational number which is not in $S$.

Note. Students sometimes are confused by this subring, because they argue like
$$
\frac{1}{3}=\frac{2}{6}
$$
so $1/3\notin S$. This is incorrect: what's required is that a rational number can be written at least in one way as a quotient of an integer by an odd integer.
A: Yes, since $1$ is odd, and the product of two odd numbers is odd. You can generalize this in several ways, for instance, given a number $n$ you can consider fractions with denominators $\equiv 1$ $\!
\!\!\!\mod n$. Or you can take denominators that are relatively prime to $n$. In fact you will get the same subring of $\mathbb{Q}$. Note that in the first case you consider not only reduced fractions but possibly non-reduced. Say $n=6$ and we take $r = \frac{1}{5} = \frac{5}{25}$. 
