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$a$ and $b$ are two non-zero real numbers that satisfy $ab = a - b$ What is the possible value of: $\frac{a}{b}+\frac{b}{a}-ab$

I found it $2b$

Is it true?

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    $\begingroup$ Please edit your question to show how you got that result. $\endgroup$ Nov 26, 2022 at 13:14
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    $\begingroup$ plug in some numbers (e.g. $a=b=1$) and see if your result is correct in that case. $\endgroup$ Nov 26, 2022 at 13:17
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    $\begingroup$ @Oбжорoв while I agree that checking some cases is a good idea I feel compelled to mention that if $a=b=1$ then $ab = 1 \neq 0 = a-b$ so it's doesn't meet the criteria. You can find some counter-examples pretty quickly though by guess-and-test. $\endgroup$ Nov 26, 2022 at 13:22
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    $\begingroup$ @CyclotomicField you are absolutely right. Although my approach was correct, the example wasn't. $\endgroup$ Nov 26, 2022 at 13:24
  • $\begingroup$ Please, show your work that you did to find the result. $\endgroup$ Nov 26, 2022 at 13:45

1 Answer 1

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$$\frac a b + \frac b a - ab =\\ \frac {a^2 + b^2 - (ab)^2} {ab} =\\ \frac {a^2 + b^2 - a^2 +2ab - b^2} {ab} =\\ \frac {2ab} {ab} =\\ 2$$

EDIT: FWIW, here's another sequence $-$ first worth noting that $a, b \ne 0$ for otherwise the problem statement is not well defined. We have

$$ ab = a - b \iff ab + b = a \iff a + 1 = \frac a b $$

Likewise,

$$ ab = a - b \iff a - ab = b \iff 1 - b = \frac b a $$

So,

$$ \frac a b + \frac b a - ab =\\ a + 1 + 1 - b - ab =\\ 2 + a - b - ab =\\ 2 $$

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