Question 659859
Because the square of a negative number is a positive number, we know x >= 0.
We are looking for all x such that {{{x^2 < x}}}
Dividing both sides by x (with the requirement that x<>0) gives
{{{(x^2)/x < x/x}}}, {{{x<>0}}}
{{{x < 1}}}, {{{x<>0}}}
So we have narrowed it down to 0 <= x < 1.
And since {{{0^2 = 0}}}, we can eliminate 0.
So we can say {{{x^2 < x}}} for all 0 < x < 1.