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Given

$$\frac1x + \frac1{1-x} > 0 \tag{1}$$

we can do

$$\frac{(1-x) + x}{x(1-x)} = \frac1{x(1-x)} > 0 \tag{2}$$

So, $x(1-x) > 0$ and we arrive at $x>0$ and $x<1$.

However, if we take another path from $(1)$, we arrive at $0<1$. Here are the steps:

$$\begin{align} \frac1x &> -\frac{1}{1-x} \tag{3} \\[4pt] -x &< 1-x \tag{4} \\[4pt] 0 &< 1 \tag{5} \end{align}$$

This result, $0<1$ seems to imply that $-x$ is always smaller than $1-x$, which is true; but is that what it always means, and given where we started from, is this derivation even correct?

Is there a step that is incorrect here? Why does the first approach differ from the second approach?

Blue
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xoux
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  • Your transformation in the second approach is only correct for $0<x<1$ and so both are equivalent – Gono Dec 30 '20 at 06:09
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    I don't have time to write a complete answer but $1/a<1/b$ does not imply that $a>b$. – Taladris Dec 30 '20 at 06:41
  • The first way isn't entirely correct either. $x(1-x) >0$ you will arrive at EITHER $x> 0$ and $x < 1$ *OR* at $x<0$ and $1-x < 0;x>1$. The second is a contradiction and is thus false. But you must consider it and rule it out to correct solve the problem – fleablood Oct 31 '22 at 20:50
  • "I don't have time to write a complete answer but 1/a<1/b does not imply that a>b" But we do have time to point out why. You can assume what we are multiplying are positive. We must consider what happens if one or both are negative. – fleablood Oct 31 '22 at 20:52

1 Answers1

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The error stems from the fact that you took the reciprocal of both sides of the inequality (and changed the direction of the inequality) when the two sides do not necessarily have the same sign. if $x > x'$, it is not in general true that $1/x < 1/x'$. The latter will only hold if the two sides of the inequality have the same sign.

For example, supposed in your example we set $x=0.5$. As you have found, the inequality should hold for this value of $x$. However, subbing in this value into $1/x > -1/(1-x)$ we find $2 > -2$, which is true, but of course we do not have that $\frac{1}{2} < -\frac{1}{2}$, which is what your second to last step implies.

Enforce
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  • So we must consider the signs of the expressions. $1/x > -1/(1-x)$ gives us three cases: 1) x < 0 2) 0 < x < 1 3) x > 1.

    In case 1), 1/x < 0 and -1(1-x) < 0, so both sides have the same sign. We can multiply both sides by x < 0, and that changes the sign of the inequality, so we are left with 1 < -x/(1-x).

    The expression on the right is negative, so we already know this is a dead end, and if we multiply both sides by (1-x), a positive number, we are left with (1-x) < -x and 1 < 0.

     – xoux Dec 30 '20 at 06:56
  • In case 3), we similarly reach 1 < 0.

    For case 2), 1/x is positive and -1/(1-x) is negative. Multiply by x > 0, and get 1 > -x(1-x), where we know the right side is negative. Multiply both sides by the positive (1-x) and we get (1-x) > -x and 0 < 1.

    Is this a common final result, ie 0 < 1. Does this signify that the assumption of 0<x<1 is the correct solution?

     – xoux Dec 30 '20 at 07:00
  • Basically yes. Essentially you have shown here that $x > 1$ and $x < 0$ lead to a contradiction. Although this does not prove that $0<x<1$, but it excludes the other possibilities. – Enforce Dec 30 '20 at 07:11