x^2 - x < 0

x^2 < x

x < sqrt(x)

x must be greater than 0 since sqrt(x) is imaginary for negative x, and 0 < 0 is not true.

Visualizing the graphs of x and sqrt x, one can imediately see that x < sqrt(x) in the range 0 < x < 1. To see this, consider that 1 = sqrt(1), therefore the range will be either greater or less than 1.

0.5 < sqrt(0.5) = .707

while

4 > sqrt(4) = 2

Therefore the range is (0,1)

x^2 -x < 0

Let us factor x:

x(x-1) < 0

In order for the inequality to hold, we have two options:

x < 0 and x-1 >0

x<0 and x > 1

This is impossible becayse x can not be <0 and >1 at the same time.

The other option is:

x>0 and x-1 <0

x>0 and x<1

Then, 0<x<1

**Then x belongs to the interval (0,1)**

First of all, let's solve the expression like an equation. For this reason, we'll put:

x^2 -x = 0

We'll factorize and we'll get:

x(x-1) = 0

We'll put each factor equal to zero.

x = 0

x-1 = 0

x = 1

Now, we'll follow the rule: between the solution of the equation, the expression will have the opposite sign of the coefficient of x^2, that means that x(x-1)<0, outside the solutions, the expression will be positive.

**So, the conclusion is that the expression is negative for values of x which are in the interval (0,1).**

To solve x^2-x < 0

Solution:

We factorise the left and then the inequation is:

x(x-1) < 0.

So to have the negative values the values of should belong to the interval (0,1) where 0 and 1 are the rooots of x(x-1) = 0.

So the solution set of x belongs to (0,1) for which x^2-x < 0.