# Is there a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$?

I'm asked in an exercise from an algebra textbook if there exists a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$

I'm trying to prove that such pair of numbers does not exist, but I'm not sure my proof and my reasoning are correct. Could anyone please check my proof attempt?

First of all, $\exists\frac{1}{a+b}\in\Bbb{R}\iff a\neq -b$

and $\exists(\frac{1}{a}+\frac{1}{b})\in\Bbb{R}\iff a\neq 0 \text{ and } b \neq 0$

$\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b} \iff \frac{1}{a+b}=\frac{a+b}{ab}$

$\iff \frac{ab}{a+b}=a+b$

$\iff ab=(a+b)^2$

$\iff ab= a^2 + 2ab + b^2$

$\iff 0 = a^2+2ab+b^2-ab$

$\iff 0= a^2+b^2+ab$

Since $a \neq 0$ and $b \neq 0$; $\exists (ab)^{-1}\in\Bbb{R}$. This allows me to continue like this:

$0= a^2+b^2+ab \iff 0(ab)^{-1}=(a^2+b^2+ab)(ab)^{-1}$

$\iff 0 = \frac{a^2+b^2}{ab} + 1$

$\iff -1 = \frac{a^2+b^2}{ab}$

$a^2+b^2 > 0$ because $a \neq 0$ and $b \neq 0$

If a and b have the same sign, then $ab>0$.

So $\frac{a^2+b^2}{ab}$ could be negative only if:

• $a>0 \text{ and } b<0$ or
• $a<0 \text{ and } b>0$

Taking into account that $a \neq -b$, any of these two options would be possible only if either:

• $|a| > |b|$
• $|b| > |a|$

Assuming that $|a| > |b|$ we have the following:

$|a| > |b| \implies |a|*|a| > |b|*|a|$

$\implies |a|^2 > |ab|$

$\implies \frac{|a|^2}{|ab|}>1$

$\implies |\frac{a^2}{ab}|>1$

And $b \neq 0$ so we also have $|\frac{b^2}{ab}|>0$. Therefore $|\frac{a^2}{ab}|+|\frac{b^2}{ab}|> 1 + 0$.

$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.

This implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.

Assuming $|b|>|a|$ we arrive at the same conclusion:

$|b| > |a| \implies |b|*|b| > |a|*|b|$

$\implies |b|^2 > |ab|$

$\implies \frac{|b|^2}{|ab|}>1$

$\implies |\frac{b^2}{ab}|>1$

And $a \neq 0$ so we also have $|\frac{a^2}{ab}|>0$. Therefore $|\frac{b^2}{ab}|+|\frac{a^2}{ab}|> 1 + 0$.

$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions.

So this also implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.

$\blacksquare$

Is this correct??

• A simplification, starting from $a^2+ab+b^2=0$: If $ab>0$ then $a^2+ab+b^2>a^2-2ab+b^2=(a-b)^2\ge 0$; and if $ab<0$ then $a^2+ab+b^2>a^2+2ab+b^2=(a+b)^2\ge0$. Aug 22, 2017 at 20:47
• Note that $4(a^2+ab+b^2)=4a^2+4ab+4b^2=(2a+b)^2+3b^2$ Aug 22, 2017 at 20:51
• I find it quite amusing that there are no numbers for which this is true. When someone writes $1/(a+b) = 1/a + 1/b$, he/she is always wrong!
– user370967
Aug 22, 2017 at 21:04

A simple proof for $a^2 + ab + b^2 \neq 0$ for non-zero reals $a$ and $b$ is as follows. $$2(a^2+ab+b^2) = (a+b)^2 + a^2 + b^2=0$$ implies $a=b=0$. Hence, a contradiction.

• Very nice idea @math lover . Aug 22, 2017 at 23:02

It gives $$(a+b)^2=ab$$ or $$a^2+ab+b^2=0$$ or $$\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2=0,$$ which gives $b=0$ and $a+\frac{b}{2}=0$, which gives $a=b=0$, which is impossible.

Thus, we have no these numbers.

$\iff 0= a^2+b^2+ab$

Multiplying by $a-b$ gives $a^3-b^3=0 \iff a^3 = b^3\,$, then taking the (real) cube roots $a=b\,$. But it is readily verified that $\frac{1}{a+a} \ne \frac{1}{a} + \frac{1}{a}\,$ for any $\forall a \in \mathbb{R}\,$, so there are no real solutions.

It is always true that $$0\le (|a|-|b|)^2=a^2+b^2-2|a||b|$$ $$|a||b|\le \frac{a^2+b^2}{2}.$$ $$-ab\le|a||b|\le \frac{a^2+b^2}{2}.$$ As $a\not =0$ and $b\not =0$ in this case, it follows $$-ab\le|a||b|\le \frac{a^2+b^2}{2}<a^2+b^2,$$

and it holds that

$$-ab<a^2+b^2.$$

This result leads to a contradiction considering your development. Therefore, $\not \exists \{a,b\} \subset \Bbb R$ such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$. The existence of this subset $\{a,b\}$ would imply that $-ab=a^2+b^2$, as you've already shown, and that is not possible if $a\not =0$ and $b\not =0$, a necessary condition to prevent division by zero.

If $a$ and $b$ have the same sign, the magnitude of $\frac 1{a+b}$ is less than either $\frac 1a$ or $\frac 1b$ so they must have opposite signs. By symmetry we can demand $a$ be positive. If $a \gt -b, \frac 1{a+b}$ is positive while $\frac 1a +\frac 1b$ is negative. The opposite happens if $a \lt -b$ so there is no solution.