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I need to solve for x: $$x - \frac{1}{x} < 0.$$ I realize that $x\neq 0$.

The following is always true: $$x-\frac{1}{x} < 0 \iff x < \frac{1}{x}.$$ If $x > 0$: $$x < \frac{1}{x} \iff x^2 < 1 \iff x < \pm 1$$ $$0<x<1$$ If $x<0$: $$x < \frac{1}{x} \iff x^2 > 1 \iff x > \pm 1$$ $$-1<x<0.$$

So my answer is that $-1<x<1$ with $x \neq 0$. But this is not true. The real answer is $x<1$ and $0<x<1$. Why? Where is my logic wrong?

I_snort_TNT
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6 Answers6

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The solution to $x^2<1$ is not $x<1$ or $x<-1$, but rather $x<1$ and $x>-1$. Choosing the negative square root requires reversing the direction of the inequality.

ziggurism
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  • Thank you, exactly the answer I was looking for. I'll skip using $x > \pm 1$ in the future and use $|x| > 1$ instead. Makes more sense. – I_snort_TNT Dec 17 '16 at 22:40
  • Actually it would be x < 1 and x > -1 in this case. (or if x^2 > 1, however.) The absolute value is the only practical way. – The Vee Dec 17 '16 at 22:53
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A step like $x^2>1\implies x>\pm1$ is terribly dangerous (and you got trapped).

$$x^2>1$$ is actually $$x^2-1=(x-1)(x+1)>0$$ solved by

$$(x>1\land x>-1)\lor(x<1\land x<-1)$$

which simplifies to

$$x>1\lor x<-1.$$

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Your mistake is in the second case: In fact, $x^2 > 1 \Leftrightarrow (x<-1 \,\text{or}\,x>1)$.

Max Freiburghaus
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Note that $x^2 > 1\iff \lvert x\rvert >1$. Hence if $x<0$ and $x^2 > 1$ then $x<-1$.

Sri-Amirthan Theivendran
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We have:

$$x - \frac{1}{x} < 0$$ $$\frac{x^2 - 1}{x} < 0 \text{ (common denominator)}$$ $$\frac{(x-1)(x+1)}{x} < 0 \text{ (common denominator)}$$

Now you can make a chart:

enter image description here

And we get what we are looking for

K Split X
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$\frac{x^2-1}{x}$ may change sign iff it is zero, i.e., $x=\pm1$ or it is undefined, here $ x=0$.

Michael Hoppe
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