I am just working through some sample problems in a introduction to discrete mathematics.
And i encountered this problem:
Task: Express in the simplest possible form, with positive exponents
Concrete problem: $\frac{u^{-2}}{v^{-3}}$
So normally when i attack this sort of algebra thing, i do: $$ \frac{\frac{1}{u^2}}{\frac{1}{v^3}} $$
And then i would for example multiply the fraction with $v^3$
$$ \frac{v^3\frac{1}{u^2}}{v^3\frac{1}{v^3}} = \frac{v^{3}\frac{1}{u^{2}}}{1} = v^{3}\frac{1}{u^{2}} = \frac{v^{3}}{u^{2}} $$
And then i guess i wouldn't simplify it more than that, but didn't give it more thought beyond that.
But in the book i think they want me to use the rule: $(ab)^{x} = a^{x}b^{x}$, without explicitly saying so, because in their solution they do: $\frac{u^{-2}}{v^{-3}} = (u^{-2})(v^{-3})^{-1} = u^{-2}v^{3}=\frac{v^{3}}{u^{-2}}$
Can anyone explain what the logic is behind their way of doing it? and can their result even be considered a positive exponent? it has a $u^{-2}$ as a denominator.
EDIT, add picture of solution in book:
I agree with the comments. The book has made a mistake in the last line. I believe it should read:
$$\frac{u^{-2}}{v^{-3}}=(u)^{-2}(v^{-3})^{-1}=u^{-2}v^3=\frac{v^3}{u^2}$$