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

Question

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

Answer 1

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

Answer 2

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

Answer 3

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 $\dfrac{x}{x+1}$:

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

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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: $\dfrac{x}{x+1} = 1 - \dfrac{1}{x+1}$