for which real numbers x will the cube of x be less than the square of x?

Question

for which real numbers x will the cube of x be less than the square of x?

Answer 1

Formally, one can derive the result as follows:

 

x^3<x^2

x^3-x^2<0

x^2 (x-1)<0

 

two real numbers multiply to less than 0 if one is positive and the other negative

 

Case I: x^2 is negative and x-1 is positive--->leads to a contradiction b/c x^2 can only be negative if x is non-real

 

Case II: x^2 is positive and x-1 is negative.  x-1<0 if x<1 ; note that x^2 is always positive IF x is not equal to 0

 

So by Case II, we have that ANY x that is less than 1 (except 0) will satisfy the statement.

Answer 2

There are two number sets meeting this criterion:

1) Any negative number. The square of such a number is always positive, the cube always negative.

2) Any positive number < 1. The higher the power (exponent) of such a number, the smaller the result.

 

Negative numbers whose absolute value is < 1 fit both categories.