0<x<1

[adthedaddy]

Hi,

I came across following explanation in MGMAT Book-3. Would like to understand how it is derived.

If x^2-x<0 then 0<x<1.

Please help me understand the above statement.

[Ian Stewart]

Hi,

I came across following explanation in MGMAT Book-3. Would like to understand how it is derived.

If x^2-x<0 then 0<x<1.

Please help me understand the above statement.

There are quite a few ways to analyze this type of inequality. Two algebraic solutions:

We can rewrite the inequality: x^2 < x. Now, this inequality can't be true if x is negative, since if x were negative, then x^2 would be positive and would be greater than x. So we know x > 0, and that means we can divide by x on both sides of the inequality to find x < 1. So 0 < x < 1.

Or we can factor. If x^2 - x < 0, then (x)(x-1) < 0. Now we have a product of two terms which is negative, so one of the terms is negative, the other positive. But x-1 is certainly smaller than x, so it must be that x-1 is the negative term and x the positive term. So x - 1 < 0 and thus x < 1, and x > 0, so 0 < x < 1.

Or there's a general 'number picking' approach you can use to analyze any inequality with a simple power of x on either side (so you can use this to analyze inequalities like x^3 > x^2, or 1/x^3 < 1/x, or x^98 < x^97). Test four different values of x: one value less than -1, one value between -1 and 0, one value between 0 and 1, and one greater than 1. If the inequality is true for that one value, it will be true for the entire 'zone' from which you took that value. So if you plug in, say x = -2, x = -1/2, x = 1/2 and x = 2 into the original inequality x^2 < x, we find the inequality only works for x = 1/2. So it only works for the 'zone' of values between 0 and 1, and 0 < x < 1.

That number picking strategy should only be used though in cases where you have simple powers of x on either side of an inequality. It won't typically give you the right answer in more complicated situations.