# Is the term true? $\frac{\theta}{\theta - 1 } = \frac{1} {\theta-1} + 1$

Why is the following term true? What is the corresponding rule or how is it transformed?

$\frac{\theta}{\theta - 1 } = \frac{1} {\theta-1} + 1$

Thanks!!

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This is a math question, not a Mathematica question. Has to do with adding fractions (general principle: find a common denominator). Probably this should be closed, or maybe migrated if there is an SE group that fields this sort of query. – Daniel Lichtblau Jan 9 at 16:56
agree with @DanielLichtblau and of course it is true since 1/(theta -1) + 1 is an alternative form for the left side of the equation. – Stefan Jan 9 at 17:06
It's not true for theta = 1. – murray Jan 9 at 20:31
OP has not defined what $\theta$ and $1$ are and the operations used. It depends on which algebraic structure we are dealing with. division is not defined for Natural Numbers and Integers. While posting these sort of questions, you have to be explicit. – 007resu Jan 10 at 2:40
Some one said this is not Mathematica question. TrueQ[x(x-1)==1/(x-1)+1] can be used to check this in mathematica. And it returns FALSE UNLESS YOU SPECIFY X<>1 there. – 007resu Jan 10 at 2:43

## migrated from mathematica.stackexchange.comJan 9 at 17:07

$$\frac{\theta}{\theta - 1 } = \frac{1 + \theta - 1}{\theta - 1 } = \frac{1}{\theta - 1 } + \frac{\theta - 1}{\theta - 1 } = \frac{1} {\theta-1} + 1$$

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$$\frac{1}{\theta-1}+1=\frac{1+(\theta-1)\cdot1}{\theta-1}=\frac{1+\theta-1}{\theta-1}=\frac{\theta}{\theta-1}$$

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 This "rule" is called "common denominator". Used for addition involving fractions. – GEdgar Jan 10 at 2:49

A more-or-less general rule is to observe that $$\frac{\theta}{\theta-1}\approx 1$$ so we could try to represent $$\frac{\theta}{\theta-1}=1+A$$ for some $A$. Doing so, we find that $A=1/(\theta-1)$.

In a similar vein, if we're given $$\frac{\theta^2+2}{\theta-1}$$ we could write $$\frac{\theta^2+2}{\theta-1}=\theta+A$$ Solving for $A$ we find that $$\frac{\theta^2+2}{\theta-1}=\theta+\frac{\theta+2}{\theta-1}$$ If we wish, we could do a similar construction on the rightmost term to obtain $$\frac{\theta^2+2}{\theta-1}=\theta+1-\frac{3}{\theta-1}$$ Added. Which, by the way, shows that $y=x+1$ is an asymptote of $y=(x^2+2)/(x-1)$.

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