Solving $\frac{1}{x} + \frac{1}{1-x} > 0$. Where's my error?

Question

For which $x$ is the following true? $$\frac{1}{x} + \frac{1}{1-x} > 0$$

I’ve worked through the question a couple ways and got the correct answer, but the initial approach has a flaw and I’m not sure where in which step it is.

I got the correct answer by adding the two numbers to get $\frac{1}{x-x^2} > 0$ and going from there.

However my first attempt went like:

$$\begin{align}\frac{1}{x} &> - \frac{1}{1-x} \tag1\\[4pt] \Rightarrow \qquad \frac{1-x}{x} &> -1 \tag2\\[4pt] \Rightarrow \qquad 1-x &> -x \tag3 \\[4pt] \Rightarrow \qquad x-1 &< x \tag4 \end{align}$$

For which the answer is all $x$. However, the correct answer, which is clear from the first approach is $0 < x < 1$.

I’m new to working with inequalities and I figure there is just a simple property of them I’m being ignorant about which means one of my rearrangements isn’t true.

All help appreciated. Thanks.

Answer 1

"I figure there is just a simple property of them I’m being ignorant about which means one of my rearrangements isn’t true": that's it.

Let us take a few examples :

• $\color{blue}5<\color{green}8$ and $\color{blue}5\times2<\color{green}8\times2$
• $\color{blue}{-5}<\color{green}{\frac83}$ and $\color{blue}{-5}\times2<\color{green}{\frac83}\times2$
• $\color{blue}{-8}<\color{green}{-5.3}$ and $\color{blue}{-8}\times2<\color{green}{-5.3}\times2$

But

• $\color{blue}5<\color{green}8$ and $\color{blue}5\times{(-2)}\color{red}>\color{green}8\times(-2)$
• $\color{blue}{-5}<\color{green}{8}$ and $\color{blue}{-5}\times(-2)\color{red}>\color{green}{8}\times(-2)$
• $\color{blue}{-8}<\color{green}{-5}$ and $\color{blue}{-8}\times(-2)\color{red}>\color{green}{-5}\times(-2)$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{{1 \over x} + {1 \over 1 -x} > 0} \sr{x\ \not\in\ \braces{0,1}}\implies x^{2}\pars{1 - x}^{2}\ \overbrace{{1 \over x\pars{1 - x}}} ^{\ds{= 1/x + 1/\pars{1 - x}}} > 0 \\[5mm] & \implies x\pars{1 - x} >0 \implies x\pars{x - 1} < 0 \implies \bbx{\color{#44f}{0 < x < 1}} \end{align}

Answer 3

you must check the inequality on the three intervals on which the expression is defined. These are $(-\infty,0) , \ (0,1), \ $and $(1,\infty)$. of you take one test point from each interval you will find that the inequality holds only when $0<x<1$

The flaw in your initial approach is when you multiply by $x-1$. If $x-1<0$ the direction of the inequality changes

also see Solving $\frac1x + \frac1{1-x} > 0$ two ways gives different results, one of which is simply $0<1$. What does this mean?

Answer 4

When you multiply by $1 - x$, if $1 - x < 0$ you flip the direction of the inequality. Then inequality $(2)$ would be backwards; it should be $\frac{1-x}{x} < -1$ in that case.

Again, when you multiply by $x$, if $x < 0$ you flip the direction of the inequality.

In no case can either $x$ or $1 - x$ be zero, because then the original inequality would be comparing something with an undefined result.

If $x$ and $1 - x$ are both positive, all your steps are correct and moreover you can make the implications bidirectional, so the true statement at the end implies your initial statement was true; that is, it is sufficient for $x$ and $1 - x$ both to be positive.

If $x$ and $1 - x$ are both negative, your first two steps are wrong; to make them correct you need to reverse the $>$ to $<$ in inequality $(2).$ But with that correction, you can make the implications bidirectional, so the true statement at the end implies your initial statement was true; that is, it is sufficient for $x$ and $1 - x$ both to be negative.

If one of $x$ and $1 - x$ is positive and the other is negative, the direction of the inequality gets flipped in only one step, so in the final step (correcting the directions of the inequalities) you have $x - 1 > x.$ Since this is always false, you can never satisfy the original inequality when $x$ and $1 - x$ have opposite signs.

So the original inequality is true exactly when $x$ and $1 - x$ have the same sign (both positive or both negative), that is, when $$ x (1 - x) > 0. $$

Answer 5

$a < b$ does not imply $ac < bc$. For example $3 < 4$ but $3\times (-1) =-3$ and $4\times (-1) = -4$ and we do not have $-3 < -4$.

That trivial example is to warn you when you attempt something like $\frac ab < k\implies a < bk$ you are tacitly making the assumption that $b > 0$ and if $b \le 0$ that is simply not true. You must, if you are going to try that, verify that $b$ is indeed positive, or, you must do both cases.

So to do the problem your way

$\frac 1x + \frac 1{1-x}> 0 \implies$

$\frac 1x > -\frac 1{1-x}$.

So far so good adding and subtracting things from both sides of an inequality is always okay. It's multiply and dividing that we have to check if positive or not.

$\frac 1x\cdot (1-x) > -\frac 1{1-x}\cdot (1-x)$.

!!WRONG!! we don't know if $1-x$ is positive, negative, or zero.... well we do know that $1-x$ can't be zero because we are not allowed to divide by zero.

So we must do cases

Case 1: $1-x > 0$. Then $\frac {1-x}x > -1$.

Case 2: $1-x < 0$. Then $\frac {1-x}x < -1$.

Case 3: $1-x =0$. This is impossible.

Then we continue. If we multiply both sides by $x$ we must do cases if $x>0$, $x< 0$ or $x=0$.

Case 1a: $1-x > 0$ and $x>0$. Then $1-x > -x$ and $1> 0$

Case 1b: $1-x > 0$ and $x < 0$. Then $1-x < -x$.

Case 1c: $1-x> 0$ and $x =0$. THis is impossible as we are dividing by $0$.

Case 2a: $1-x < 0$ and $x > 0$. Then $1-x < -x$

Case 2b: $1-x < 0$ and $x < 0$. Then $1-x > -x$

Case 2c: $1-x< 0$ and $x =0$. This is impossible as we are dividing by $0$.

Then we check those cases.

Case 1a: $1-x > 0$ and $x>0$. Then $1-x > -x$ and $1> 0$. This is valid.

Case 1b: $1-x > 0$ and $x < 0$. Then $1-x < -x$ or $1< 0$. This is wrong. This is invalid.

Case 1c: $1-x> 0$ and $x =0$. THis is invalid.

Case 2a: $1-x < 0$ and $x > 0$. Then $1-x < -x$ or $1 < 0$. THis is invalid.

Case 2b: $1-x < 0$ and $x < 0$. Then $1-x > -x$ or $1>0$. This is valid.

Case 2c: $1-x< 0$ and $x =0$. This is invalid.

So ...

Case 1a: $1-x > 0$ and $x>0$; $1>x$ and $x>0$; or $0< x < 1$ will be possible solutions.

Case 1b: $1-x> 0$ and $x < 0$; $1> x$ and $x < 0$ or $x < 0$ make the statement false.

Case 1c: $1-x > 0$ and $x =0$; or $1>x$ and $x=0$; or $x=0$ will make the statement false.

Case 2a: $1-x < 0$ and $x>0$; or $1< x$ and $x >0$; or $x > 1$ will make the statement false.

Case 2a: $1-x < 0$ and $x< 0$; or $1< x$ and $x< 0$. That's a contradiction. This can never actually happen and it doesn't matter that it would make the statement true, as it can never happen.

Case 2c: $1-x < 0$ and $x=0$. Well, that is both impossible and will make the statement false.

Conclusion: $0 < x < 1$.

... Now, you will probably note that that was too much work and redundant.

We can make it a lot simpler by using common sense:

$\frac 1x + \frac 1{1-x} >0 \implies$

$\frac 1x > -\frac 1{1-x}$

We want to multiply by $1-x$ so we must do two cases: $1-x > 0$ or $1>x$ or $x < 1$. Or $1-x < 0$ or $1< x$ or $x> 1$.

Let's start with $x> 1$. And go all the way through.

Case 1: $x > 1$.

Then $1-x < 0$ and so $\frac 1x > -\frac 1{1-x}\implies$

$\frac {1-x}x < -1$. We want to multiply both sides by $x$. Is $x > 0$. Well, yes, $x > 1 > 0$ so

$1-x < -x$ and so

$1< 0$. This is impossible so Case 1: $x > 1$ is impossible.

Note: we are breaking the real number line into cases and just moving on down.

Case 2: $x =1$. Don't need to do this point. If $x =1$ then $1-x =0$ and we can't divide by $0$.

Case 3: $x < 1$. Then $1-x > 0$.

So $\frac 1x >-\frac 1{1-x} \implies \frac {1-x}x > -1$.

We want to multiply by $x$ but is $x >0$. We don't know $x <1$ so it may or may not be.

Case 4: $0 < x <1$.

So $\frac {1-x}x > -1\implies 1 -x > -x$ and $1 > 0$. That is always true so $0 < x< 1$ is a solution.

Let's keep moving down the number line.

Case 5: $x=0$. We can rule that out as that is dividing by $0$.

Case 6: $x < 0$. Then $\frac {1-x}x > -1\implies 1-x < -x \implies 1<0$. That is false.

So.... we are done. $0< x < 1$ is the only solution.

=====

BTW "However, the correct answer, which is clear from the first approach"

I suspect it is not as clear as you thought it was and you lucked out.

$\frac 1{x-x^2} > 0$ is valid.

So $x-x^2 > 0$. [We have a rule that $a > 0 \iff \frac 1a > 0$ and $a< 0\iff \frac 1a < 0$. This is because if $\frac 1a =0$ then $1=0\cdot a = 0$ and that's impossible. ANd if $a > 0$ and $\frac 1a < 0$ then we would have $\frac 1a < 0\implies a\cdot \frac 1a < a\cdot 0\implies 1 < 0$. And if $a<0$ and $\frac 1a >0$ we would have $a < 0\implies \frac 1a \cdot a < \frac 1a \cdot 0\implies 1 < 0$.]

Now we can go two ways here:

$x > x^2$ or $x(1-x) > 0$

If we do $x > x^2$ we know $x^2 \ge 0$ so $x > 0$. And we can divide both sides by $x$ WHICH IS POSITIVE to get $1 > x$ and $x > 0$ we have $0 < x < 1$.

If we do $x(1-x) >0$ we know that EITHER $x, 1-x$ are both positive or $x,1-x$ are both negative.

If $x, 1-x$ are both positive we get $0 < x < 1$.

If $x,1-x$ are both negative (I suspect you forgot to check this) we get $x < 0$ and $1-x < 0$ so $1 < x$. So $x < 0 < 1 < x$ which is a contradiction. So that is not the case.