wolframalpha tells me it's the same but I can not follow how to get from one to another.
$$\frac{-x+1}{-x+2} = \frac{1-x}{2-x} = \>? \dots$$
I don't get any further, always end up where I started.
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@MichaelHardy What's the point in making two dollars ($) ? – user132393 Mar 02 '14 at 00:04
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Try it both ways and you'll see the difference. – Michael Hardy Mar 02 '14 at 00:05
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I see.. it's all about justification ;) – user132393 Mar 02 '14 at 00:06
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It's not just about justification; it also affects sizes and formats: $\frac 2 3$ versus $\dfrac 2 3$, $\sum_{k=0}^n$ versus $\displaystyle\sum_{k=0}^n$, $\max_{x\in S}f(x)$ versus $\displaystyle\max_{x\in S}f(x)$, and numerous other similar things. – Michael Hardy Mar 02 '14 at 00:09
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$$\begin{align} \dfrac{1-x}{2-x} & = \dfrac{-(x - 1)}{-(x-2)} \\ \\ &= \dfrac{x-1}{x-2} \\ \\ &= \dfrac {x-1 \color{blue}{\bf - 1 + 1}}{x-2} \\ \\ & = \dfrac{(x-2) +1}{x - 2}\\ \\ & = \dfrac{x-2}{x-2} + \dfrac 1{x-2}\\ \\ & = 1 + \frac{1}{x - 2}\end{align}$$
amWhy
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Note: When you're stuck on such a rational function, or can't find a "tricky" way to manipulate a rational function, there's always polynomial long division! – amWhy Mar 02 '14 at 00:06
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I just introduced myself to this technique This particular term is actually an used in an integral exercise. – user132393 Mar 02 '14 at 00:13
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You... are a genius. I would have thought the step that you highlighted was pointless until I saw you factorize it. – recursive recursion Mar 02 '14 at 00:15
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@user132393 In the case, because the top and bottom have the same degree and lead coef, the long division algorithm boils down to subtraction, see my answer here, and, more generally, see this answer. – Bill Dubuque Mar 02 '14 at 00:47
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$$\dfrac{-x+1}{-x+2}\, =\, \dfrac{(-x+2)-1}{-x+2}\,=\,1+\dfrac{-1}{-x+2}\,=\, 1+\dfrac{1}{x-2}$$
Bill Dubuque
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