4
votes
Accepted
Showing that $\frac{x - \frac{x}{y}}{y - 1} = \frac{x}{y}$ by directly manipulating $\frac{x - \frac{x}{y}}{y - 1}$.
First factor out $x$ and then $\frac{1}{y}$ leads us to\begin{align*}
\frac{x - \frac{x}{y}}{y - 1} &= \frac{x(1 - \frac{1}{y})}{y - 1} \\
&= \frac{\frac{x}{y}(y - 1)}{y-1} = \frac{x}{y}\end{...
- 5,298
1
vote
Can I write $(x^p)^{q} = x^{pq}$, where either of $p$ or $q$ are rational.
Let us first recall the conventional definition of rational exponentiation:$$x^{\frac{m}{n}} = \left ( x^{\frac{1}{n}} \right )^m = \left ( \sqrt[n]{x} \right ) ^m,$$where $m, n \in \mathbb{Z}$ and $x ...
- 702
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