Odd divided by even is a fraction How can we prove that an odd number divided by an even number is a fraction? I started with odd $=2m+1$ and even $=2n$ and get left with with $(m+2)/n$.
 A: Your start is good (correct): an odd number can be represented as $2m + 1$, even number $2n$, for $m,n \in \mathbb{Z}$.
But then, to divide, take $$\frac {2m+1}{2n}=\frac{2m}{2n} + \frac {1}{2n} = \frac{m}{n} + \frac{1}{2n}.$$
Can you see why the right-most side of the equation cannot be whole (an integer)?
$$
\frac{2m+1}{2n} = k, \text{ where}\; k\in \mathbb{Z},$$ $$\text{ then} \; 2m+1 = 2kn.$$  Note that the remainder when the left-hand side ($2m+1$) is divided by $2$ is $1$, while the remainder when the right-hand side ($2kn$) is divided by $2$ is $0$. 
That's a contradiction.
A: A whole multiple of an even number is even, so if the quotient is whole and the denominator is even, the numerator would be even as well. Note that $\frac{2m+1}{2n}=\frac{m}{n}+\frac{1}{2n}\neq \frac{m}{n}+\frac{2}{n}= \frac{m+2}{n}$.
A: $$
\frac{2m+1}{2n} = \frac{m+\frac12}{n}
$$
So there's an error where you put $2$ where you need $1/2$.
However if
$$
\frac{2m+1}{2n} = a = \text{an integer}
$$
then $2m+1 = 2an$.  But the remainder when $2m+1$ is divided by $2$ is $1$, and the remainder when $2n$ is divided by $2$ is $0$.
A: Hint: Suppose to the contrary that $\dfrac{a}{b}=n$, where $a$, $b$, and $n$  are integers. Suppose also that $a$ is odd, while $b$ is even (and of course non-zero).  
Then $a=bn$. See whether you can show this is impossible. Here you will be using the fact that $a$ is odd and $b$ is even. 
A: Hint $\rm\ 2n\mid 2k+1\:\Rightarrow\ 2\,\mid\, 2k+1\,\ \Rightarrow\,\ 2\mid 1.\ $  Or, in terms of fractions,   
$\rm\quad j = \dfrac{2k\!+\!1}{2n}\in\Bbb Z\:\Rightarrow\: nj = k\!+\!\dfrac{1}{2}\in \Bbb Z\:\Rightarrow\: \dfrac{1}2\in\Bbb Z,\ $ a contradiction.
