Is a negative number proper fractions? I was asked this question by a kid, is -$\frac{4}{7}$ a proper fraction or not? As per my knowledge $\frac{4}{7}$ is a proper fraction. If it has a -ve number does it make any difference? Definition says A number whose numerator is smaller than denominator is called a proper fraction. Can we consider -$\frac{4}{7}$as a proper fraction? If not why not please explain. This is my first question I don't have much idea about tags of mathematics if it is tagged wrongly please edit it.
EDIT: I have got these links from comments wiki link and a link form math world and both contradicting each other.
Thank you
Dibya
 A: From Wikipedia:

Common fractions can be classified as either proper or improper. When the numerator and the denominator are both positive, the fraction is called proper if the numerator is less than the denominator, and improper otherwise. In general, a common fraction is said to be a proper fraction if the absolute value of the fraction is strictly less than one—that is, if the fraction is between $-1$ and $1$.* [Italics mine]

So $\;-1 < \left(-\dfrac 47\right) < 1$ is considered a proper fraction; (alternatively $\;\;0 < \Big|-\dfrac 47 \Big| = \dfrac 47 < 1$.

I think in terms of mathworld's definition: When using the division algorithm, for example, one requires an integer quotient $\times$ an integer divisor, plus a non-negative integer remainder less than the value of the divisor. So if dividing $-4$ by $7$:
$$-4 = -1\cdot 7 + 3, \;\;q = -1;\;\;r = 3, i.e., -\dfrac 47 = -1 + \dfrac 37$$ which would be a mixed fraction.
So the fractional part would be the proper fraction $\dfrac 37$, the integer part, $-1$. Consistent with this, mathworld may require that a proper fraction occurs only when the quotient is $0$, and the remainder a positive integer less than the divisor, hence the fractional part = $\dfrac rd\; d: $divisor,$\;r: $ the remainder when dividing number by $d$.
