Binomial inequalities and characteristics of integers

Question

I had the following question on a test, and I'm really confused about how to obtain the answer. The correct answer is F) Less than 1, but I just can't grasp my head around why.

Given a > b, and (a - b) > (a^2 - b^2), then (a + b) must be: 
F) Less than 1
G) Greater than 1
H) Greater than a
J) Greater than (a - b)
K) Equal to (a - b)

I think the method to solve this question involves expanding the binomial (a^2 - b^2), which results in (a + b) * (a - b), but if you take (a - b) to the left side of the original inequality, wouldn't the result just be 0 > (a + b)?

Answer 1

Let's go with your idea about factoring the difference of squares, and thus see that:

$$a-b > a^2 - b^2 \implies a-b > (a-b)(a+b)$$

Since $a>b$ is given, then $a-b \ne 0$ (and in particular, $a-b > 0$) and we can divide both sides of the right-hand inequality by $a-b$. Doing so, we see a cancellation and obtain the following:

$$a-b > (a-b)(a+b) \implies \frac{a-b}{a-b} > \frac{(a-b)(a+b)}{a-b} \implies 1 > a+b$$

I think the main issue is just a slight lapse in thought since you seem to have been thinking along the lines of subtracting $a-b$ (thus the $0$ in your post) instead of dividing (which would give you the $1$).

Answer 2

Okay, I figured out my problem. My brain somehow thought of subtracting (a - b) instead of dividing (a - b). After dividing, the resulting inequality is 1 > (a + b)

Careless mistake!

Edit: Since (a - b) is not negative here, the inequality sign should be kept the same. However, if (a - b) was negative, it would be necessary to reverse the inequality sign.