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

Question

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??

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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

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.