Solution for the inequality $x \lt \frac 1 x$

Question

I am unable to find correct solution for the inequality:

$x \lt \frac 1 x, x \in \Bbb R, x \ne 0$.

My solution first takes $x > 0$, then we have:

$x < \frac 1x$ (multiply both sides by $x$)

$\iff x^2 < 1$ (taking square root)

$\iff x < 1 \lor x < -1$

$\iff 0 < x < 1$ since $x > 0$.

Then similarly, for $x < 0$,

$x < \frac 1x$ (multiply both sides by $x< 0$)

$\iff x^2 > 1$ (taking square root)

$\iff x > 1 \lor x > -1$

$\iff -1 < x < 0 $ since $x < 0$

But this does not agree with the solution $(x < -1) \lor (0 < x < 1)$

Which part of my solution is incorrect above?

Answer 1

$$x<\frac { 1 }{ x } \\ x-\frac { 1 }{ x } <0\\ \frac { { x }^{ 2 }-1 }{ x } <0\\ \frac { x\left( x-1 \right) \left( x+1 \right) }{ { x }^{ 2 } } <0\\ x\left( x-1 \right) \left( x+1 \right) <0\\ x\in \left( -\infty ;-1 \right) \cup \left( 0;1 \right) $$

Answer 2

We have $x^2>1 \iff x>1$ or $x<-1$.

Answer 3

We have $x^2<1\iff-1<x<1$.

Answer 4

Your mistake is when taking the negative square root:

$$x^2<a^2\implies -a<x\land x<a$$ and not $$x^2<a^2\implies x<a\lor x<-a.$$

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The correct solution:

If $x>0$, $x<\dfrac1x\implies x^2<1\implies x<1$.

If $x<0$, $x<\dfrac1x\implies x^2>1\implies x<-1$.

So

$$x<-1\lor0<x<1.$$

Answer 5

First, think about the graphs of $y=x$ and $y=\frac{1}{x}$ and you'll see what the answer has to be right away.

To be analytic, the "split-point" method works well here. Find all the points which are solutions to the corresponding equation $x=\frac{1}{x}$ and all the points where there is a discontinuity. That is, $x=-1, 0, $ and $1$. Plot the points on the number line. This divides the line into 4 regions. Each region is either entirely in the solution set or entirely not in the solution set. So pick a point in the interior of each region and test it in the inequality. If it make the inequality true, then so does the whole region. If not, throw the whole region away.

Take $-2$ from the region $(-\infty, -1)$ and test it: $-2<\frac{1}{-2}$ is true.

Take $-1/2$ from the region $(-1,0)$ and test it: $\frac{-1}{2} < \frac{1}{-1/2}$ is false.

Take $1/2$ from the region $(0,1)$ and test it: $\frac{1}{2} < \frac{1}{1/2}$ is true.

Take $2$ from the region $(1,\infty)$ and test it: $2<\frac{1}{2}$ is false.

The solution is the collection of all "true" regions: $(-\infty,1)\cup (0,1).$

Answer 6

You could also just multiply with $x^2$ and you get $x^3-x<0$, so $x(x-1)(x+1)<0$

and thus $x\in (-\infty,-1)\cup (0,1)$

Answer 7

I think the best way here is using the intervals method. Without some cases and without multiplication by some number:

We need to solve $$x-\frac{1}{x}<0$$ or $$\frac{(x-1)(x+1)}{x}<0,$$ which gives the answer: $$(-\infty,-1)\cup(0,1).$$ Done!

I'll try to explain.

Firstly, we need to check, what is the sign of $\frac{(x-1)(x+1)}{x}$ on the right interval $(1,+\infty)$.

The sign is $+$ and since degrees of points $1$, $0$ and $-1$ are odds, we see that the sign of $\frac{(x-1)(x+1)}{x}$ changes and we get the following series of signs: $$-,+,-,+$$ and since we need the sign $-$, we can write the answer.

All these things we can make immediately.