Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $(m+2)/n$.

share|improve this question
    
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

5 Answers 5

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.

share|improve this answer
    
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 $0$.

That's a contradiction.

share|improve this answer

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

share|improve this answer

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.