Question

Nita finds that the expression (a + 1) 2 is sometimes, but

Answer 1

I guess the expressions are supposed to be (a + 1)² and (a + 1)³.

Consider the inequality,

(a + 1)² > (a + 1)³

Move everything to one side:

(a + 1)² - (a + 1)³ > 0

Factorize the left side:

(a + 1)² (1 - (a + 1)) > 0

-a (a + 1)² > 0

a (a + 1)² < 0

Notice that the left side is exactly 0 if either a = 0 or a = -1.

Now consider the following 3 cases:

• If a < -1, then a is negative, while (a + 1)² is always non-negative, which means a (a + 1)² will always be negative.

• If -1 < a < 0, then again a is negative, so a (a + 1)² is also negative.

• If a > 0, then a (a + 1)² is always positive.

So the inequality is satisfied for all a in the interval (-∞, -1) ∪ (-1, 0). This is to say that (a + 1)² is always greater than (a + 1)³ if a is chosen from this domain.

Another way to look at this: assume a + 1 falls between -1 and 1. Whenever you scale a number between -1 and 1 by another number in the same range, the product will always be smaller than the original number.

More concretely, let's say a + 1 = 1/2. So for instance, 1/2 * 1/2 = (1/2)² = 1/4, and 1/4 is clearly smaller than 1/2. If we multipy again by 1/2, we get 1/2 * 1/2 * 1/2 = (1/2)³ = 1/8, which is smaller than 1/4.