Phenom wrote:is x<x^2 ?' to 'is 0<x?' i.e, 'Is 'x' positive?' ??

Is x<x²?

0 < x²-x.
0 < x(x-1).

The critical points are x=0 and x=1.
These are the only values of x where x(x-1) = 0.
When x is any other value, x(x-1)<>0.
To determine the range of x, try one value to the left and right of each critical point.

x<0:
Plugging x=-1 into x<x², we get:
-1<(-1)²
-1<1.
This works.
x<0 is part of the range.

0<x<1:
Plugging x=1/2 into x<x², we get:
1/2<(1/2)²
1/2<1/4.
Doesn't work.
0<x<1 is NOT part of the range.

x>1:
Plugging x=2 into x<x², we get:
2<(2)²
2<4
This works.
x>1 is part of the range.

Since the only values not included in the range of x are 0≤x≤1, the question rephrased:

Is 0≤x≤1?

The critical points are x=0 and x=1.
These are the only values of x where x(x-1) = 0.

Pardon my ignorance but, doesn't 0< x(x-1) read as 0<x and 1<x?

How did you conclude that x(x-1)=0? It's still a little foggy for me.

I think it should be rephrased to 1≤x≤0. Below quoted value ( 0≤x≤1 ) would be in the case where x^2<x.

Correct me if i m wrong...

Since the only values not included in the range of x are 0≤x≤1, the question rephrased:

Is 0≤x≤1?

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