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Whenever in a fraction, there is $0$ in the denominator, the fraction becomes $\infty$ or indeterminate. But why do we consider those fractions valid that have some negative numbers in the denominator since those numbers are even smaller than $0$ ?

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It depends on what you mean by smaller. If you mean "smaller in magnitude," then negative numbers are not smaller than zero. –  Joel Reyes Noche Apr 2 '13 at 6:59
As below, denominators are normalized to be positive simply as a matter of convenience $$\rm x = \dfrac{a}{-b} \iff (-b)\,x =a \iff b\, x = -a \iff x = \dfrac{-a}{b} = -\dfrac{a}b$$ More generally, one often normalizes the leading coefficient of a polynomial to be positive (or, further, to be $1$ if the coefficient ring is a field). –  Math Gems Apr 2 '13 at 17:12

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Firstly, $\frac{p}{0}$ never becomes $\infty$, it is simply undefined.

Now, a fraction $\frac{p}{q}$, with $p,q$ integers and $q\ne 0$, is introduced in order to solve a problem. The problem is to solve the equation $q\cdot x = p$. This solution, even though described with integers alone, does not always have a solution in the integers. So, we extend the integers by introducing solutions to all such equations as long as it makes sense. An example where it does not make sense is the equation $0\cdot x = p$, where $p\ne 0$, since we wish to retain the fact that $0\cdot x=0$. There are some subtleties here, solutions to different equation may lead to essentially that same fraction. This is solved by the familiar equations of the form $\frac{1}{2}=\frac{2}{4}$ and so on.

Now, fractions with negative denominators simply are solutions for equations of the form $-7\cdot x = 2$ or $-5x=17$. There is no reason to exclude such solutions. In fact, by adjoining all solutions to all equations of the form $qx=p$, where $p,q$ are integers and $q\ne 0$ (and properly identifying fractions that only appear to be different in form) one obtains the field of the rational numbers.

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Fractions, like all the rest of mathematics, consist of certain symbols written on a page. As the creators of these symbols, we can decide how to manipulate them however we want, and what meaning to give them. The usual meaning we have decided to give to $$\frac{a}{b}$$ is "this symbol, when multiplied by $b$, produces $a$". Thus, for example, $\frac{1}{2}$ is a symbol satisfying $$\frac{1}{2}\times 2=1.$$ Now, the reason why fractions whose denominator is $0$ are usually taken to be "undefined" is that (unless we are very careful / in a more advanced setting) no matter how hard we try, there is no way of making the symbol $$\frac{a}{0}$$ satisfy all of the rules we want it to; for example, since we want multiplication to be associative, we should have $$\frac{1}{0}\times \mathopen{\bigg(}0\times 2\mathclose{\bigg)}=\left(\frac{1}{0}\times 0\right)\times 2$$ But the left side produces $$\frac{1}{0}\times \mathopen{\bigg(}0\times 2\mathclose{\bigg)}=\frac{1}{0}\times 0=1,$$ while the right side produces $$\left(\frac{1}{0}\times 0\right)\times 2=1\times 2=2.$$ Thus, we give up on defining division by $0$.

But this problem does not exist for fractions with negative denominators; everything works just fine. There's no reason to consider them invalid. In fact, once we start considering complex numbers like $2+3i$, we often like to divide by them too!

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