Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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


share|cite|improve this question

migrated from Jan 9 '13 at 17:07

This question came from our site for users of Mathematica.

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 '13 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 '13 at 17:06
It's not true for theta = 1. – murray Jan 9 '13 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 '13 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 '13 at 2:43
up vote 5 down vote accepted

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

share|cite|improve this answer
... Where $\theta \neq 1$. – NeilRoy Oct 21 '15 at 5:41


share|cite|improve this answer
This "rule" is called "common denominator". Used for addition involving fractions. – GEdgar Jan 10 '13 at 2:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.