# How to get from $\frac{x}{x+1}\;$ to $\;1 - \frac{1}{x+1}$?

Please show me how to manipulate $\dfrac{x}{x+1}\;\;$ to get $\;\;1 - \dfrac{1}{x+1}$

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The key point is $(x+1)=1\cdot (x+1)$, so $\displaystyle\frac{x+1}{x+1}=1$ (unless $x+1=0$). – Berci Dec 15 '12 at 20:42
"How are these two things equal" is a very different question from "given one thing, how/why would I go about finding the other thing that it's equal to?" – Hurkyl Dec 15 '12 at 20:44

Simply $$\frac{x}{x+1}=\frac{x+1-1}{x+1}=\frac{x+1}{x+1}-\frac{1}{x+1}=1-\frac{1}{x+1}$$

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Thank you. My mind was drawing a blank! – PeteUK Dec 15 '12 at 20:52

Sometimes when working with polynomial division, it helps to recall how we handle division of plain old integers:

Consider the fraction $\dfrac{17}{20}$. Note that $\dfrac{17}{20} = \dfrac{20 - 3}{20} = \dfrac{20}{20} - \dfrac{3}{20}$.

We can do the same for $\dfrac{x}{x+1}$:

$$\dfrac{x}{(x+1)}\; = \;\dfrac{(x+1) - 1}{(x+1)} \;=\; \dfrac{(x+1)}{(x+1)} - \dfrac{1}{(x+1)} \;=\; 1 - \dfrac{1}{(x+1)}$$

Also, since you'll likely be moving on to division of more complex polynomials very soon:

You can use "long division", just as you would for dividing, say $17$ by $20$.

In this case you have $x$ as the "dividend" (numerator) and $(x + 1)$ the "divisor" (denominator):

So here, $1$ is your "quotient", and $-1$ is your "remainder", giving us: $\dfrac{x}{x+1} = 1 - \dfrac{1}{x+1}$

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How about start with $1 - \frac{1}{x+1}$ to get $$1 - \frac{1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = \frac{x}{x+1}$$

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Thank you. I can see how starting from the end form of the expression, and converting that 1 into $\frac{x+1}{x+1}$ would've helped now. – PeteUK Dec 15 '12 at 20:55