# 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 $(m+2)/n$.

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You mean "divided by a nonzero even number", and you probably mean "is not an integer" instead of "is a fraction". To me $\frac 21$ and $\frac{21}7$ are fractions, even though they are also integers. –  Marc van Leeuwen Nov 25 '13 at 14:07

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.

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THANKYOU!! you just made it seem so simple :) –  james Nov 20 '12 at 20:38
Division is sometimes tricky to think about. It is often easier to "see" things if we get rid of denominators. –  André Nicolas Nov 20 '12 at 20:41
If you prefer, use your $2m+1$ and $2n$, and write $\frac{2m+1}{2n}=k$. Then "flatten" it out, getting $2m+1=2nk$. –  André Nicolas Nov 20 '12 at 20:47

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 $0$.

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$$\frac{2m+1}{2n} = \frac{m+\frac12}{n}$$ 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 $0$.

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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 $\frac{2m+1}{2n}=\frac{m}{n}+\frac{1}{2n}\neq \frac{m}{n}+\frac{2}{n}= \frac{m+2}{n}$.

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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.

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