I was flipping through the Gmat Math Bible and got to the

Pmenon, I saw that many people make mistakes in inequalities. I guess these advices can help:

1. pick very large and vary small numbers. It helps recognize mistakes faster.
2. use drawing. It is really helpful and makes a solution obvious. For example, x^2<1 ==> bottom part of the parabola y=x^2 limited by the horizontal line: y=1

yeah but verbally what is the rule cause I cannot see it in your illustration. It seems confusing and I've done these problems a millions different ways but just cant spot the rules.

I get confused because I'm not sure when its the upper limit or the lower limit or if its both. For example :
x^2<x
x<0 x<1
X>0 X>1
Answer: 0<x<1 Why though?

x^2>x
x>0 x>1
x<0 x<1
Answer: x<0 and x>1 why? If I knew the rule I could figure it out. I try all combo's but nothing seems to work.

The main rule is:

if we divide left and right sides of an inequality by a positive number, the sign of the inequality remains the same one.
if we divide left and right sides of an inequality by a negative number, the sign of the inequality changes.

Therefore, when we divide an inequality by x than can be both positive and negative, we have to consider two possibilities and also be careful with x=0.

Consider an example:

x^2>x

for x>0: x^2/x > x/x --> x>1. Therefore, x e (1,+∞)
for x<0: x^2/x < x/x --> x<1. Therefore, x e (-∞,0)
for x=0: x does not satisfy the inequality.

Join three findings: x e (-∞,0)&(1,+∞)

Consider other example:

x^2<x

for x>0: x^2/x < x/x --> x<1. Therefore, x e (0,1)
for x<0: x^2/x > x/x --> x>1. There is no x that satisfies the inequality.
for x=0: x does not satisfy the inequality.

Join three findings: x e (0,1)

I'm missing something here...I don't quite follow '3' above. Can someone elaborate further?

x^3 < x^2
Dividing both sides by x^2,
we get x<1 which means x takes any value from -infinity upto 1.

x^3>x.. i think the range of this is x>-1

x^3<x^2 i think x<-1 but u have a -1<x<0 where this doesnt..

for example, x=0.5 and 0.125>0.5 is not true


for x<0 x^3 is negative and x^2 is positive, so x^3<x^2 is true for x<0
but I found my mistake here: should be x e (-∞,0)&(0,1) rather than x e (-∞,1)