# Thread: Condition for equality to hold

1. ## Condition for equality to hold

Question is

Are this equal? 1/(a+b) =1/a + 1/b

Does this have a condition for equality?

2. ## Re: Condition for equality to hold

You can go ahead and add the right side: $\displaystyle \frac{1}{a}+ \frac{1}{b}= \frac{a+ b}{ab}$.

So you are asking for a and b that satisfy $\displaystyle \frac{1}{a+ b}= \frac{a+ b}{ab}$
That is the same as $\displaystyle (a+ b)^2= a^2+ 2ab+ b^2= ab$.

Solve the quadratic equation $\displaystyle a^2+ ab+ b^2= 0$.

By the quadratic formula $\displaystyle a= \frac{-b\pm\sqrt{b^2- 4b^2}}{2}= \frac{-1\pm\sqrt{-3}}{2}b$
Since that is not a real number, no, there are no real numbers a and b such that $\displaystyle \frac{1}{a}+ \frac{1}{b}= \frac{1}{a+ b}$.

3. ## Re: Condition for equality to hold

$\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{a + b} \implies \dfrac{a + b}{ab} = \dfrac{1}{a + b} \implies a^2 + ab + b^2 = 0 \implies$

$a= \dfrac{- b \pm b\sqrt{-3}}{2} \implies a,\ b \not \in \mathbb R.$

In short, the equality does not apply to any pair of real numbers.

Thank you!