[College Calc 01-01. The Reals] #01

Calculus Level 3

For some positive reals $a,\,b,$ define

$S = \left\{\left.b-\frac{a}{n}\right| n\in\mathbb{Z}_+\right\}.$

Given $\inf S = 2$ and $\sup S = 5,$ find $a+b.$

Notations:

$\mathbb{Z}_+$ denotes the set of positive integers.
$\inf$ and $\sup$ denotes infimum and supremum , respectively.

This is a part of the College Calc problem set. You can find more problems here .

-3

-1

0

3

7

8

1 solution

Boi (보이)
Oct 28, 2020

Note that from trivial arithmetics, we have

$b - a\leq b-\frac{a}{n}< b$

for all $n.$ Furthermore, if $n=1$ then $b-\frac{a}{n} = b-a$ , and we also have

$\lim_{n\to\infty} \left(b-\frac{a}{n}\right) = b.$

The set of lower bounds of $S$ is clearly $\{x \,|\, x \leq b-a\},$ and due to the limit the set of upper bounds of $S$ is indeed $\{x\,|\,x\geq b\}.$

Thus $\inf S = b -a = 2$ and $\sup S = b = 5,$ such that $a = 3.$ We have $a+b = \boxed{8}.$

[College Calc 01-01. The Reals] #01

1 solution

0 pending reports