# Negative Indices

I am not exactly good at evaluating negative indices--can someone please show me how to work out this expression: $$\frac {m^{-3}n^{-2}} {m^{-5}n^6}$$

Both $$m$$'s and the top $$n$$ have negative indices.

Thanks!

• Hint: $-3-(-5)$, $-2-6$. – Yves Daoust Aug 31 '16 at 9:54

Use $a^{-n}=\frac1{a^n}$ $$\frac {m^{-3}n^{-2}} {m^{-5}n^6}= \frac{m^5}{m^3n^6n^2}=\frac{m^2}{n^8}$$

• So basically in a two part fraction, if there is a negative indice the base goes on the other side and the indice becomes positive? – Kiwi Aug 31 '16 at 10:00
• @Kiwi: Yes exactly – Roman83 Aug 31 '16 at 10:02
• Okay thanks so much @Roman83 !! – Kiwi Aug 31 '16 at 10:05

Negative exponents:

$$m^{-3} = \frac{1}{m^3}$$

In the same way for the others and your expressions is nothing but

$$\frac{\frac{1}{m^3}\frac{1}{n^2}}{\frac{1}{m^5}n^6} = \frac{m^5}{m^3 n^2 n^6} = \frac{m^2}{n^8} = \left(\frac{m}{n^4}\right)^2$$

It's all about powers, exponents and their properties.

As indicated in a comment, I consider the best exercise here to use the fact that dividing powers subtracts the exponents, that is: $\frac{a^m}{a^n} = a^{m-n}$. So in this case:

$$\frac {m^{-3}n^{-2}} {m^{-5}n^6} = m^{-3-(-5)}n^{-2-6} = m^{2}n^{-8} = \frac{m^2}{n^8}$$

Assuming that you can subtract negative numbers mentally, then you can usually forego writing the middle steps and just do the whole thing in one step. And if you can't, then that's a sign you need to practice basic operations on negative numbers some more.