# Bounded above and bounded below

• May 10th 2010, 09:19 AM
Mathman87
Bounded above and bounded below
See attached document Question 3: Question is almost unwriteable on here. Basically i have these type of questions to answer in a forthcoming exam and i have no idea how to tackle them. It massively throws me knowing how to change my answer depending on whether the set of numbers is in R, Z or N.
Could someone walk me through the thinking behind these type of questions?
Thank you
• May 10th 2010, 12:49 PM
Opalg
Quote:

Originally Posted by Mathman87
See attached document Question 3: Question is almost unwriteable on here. Basically i have these type of questions to answer in a forthcoming exam and i have no idea how to tackle them. It massively throws me knowing how to change my answer depending on whether the set of numbers is in R, Z or N.
Could someone walk me through the thinking behind these type of questions?

Essentially, you have to work out, for each of these sets, the range of values of the elements in the set.

In (i), the set $A = \{3^{-n}:n\in\mathbb{Z}\}$, you have to look at the set of all powers of 3, where n is a (positive or negative) integer. If n is positive, $3^{-n}$ will always be positive, but it can be arbitrarily close to 0. So 0 is a lower bound, but $3^{-n}$ is never actually equal to 0. Therefore the g.l.b. is 0, but it is not in the set. If n is negative, $3^{-n}$ will be a positive power of 3, and it can be arbitrarily large, so there is no upper bound to this set.

For (ii), the set is $\Bigl\{\frac x{3+x^4}:x\in\mathbb{R}\Bigr\}$. So x runs through the real numbers, and you'll need to use calculus to check that this function has a maximum value 1/4 when x=1, and a minimum value –1/4 when x=–1. The function tends to 0 as $x\to\pm\infty$ and it never takes values outside the range from –1/4 to +1/4. So the l.u.b. is 1/4, the g.l.b. is –1/4, and both of those bounds lie in the set (because the function actually attains both of its extreme values).

In (iii), the set is $\Bigl\{\frac{2n}{3n-1}:n=1,2,3,\ldots\Bigr\}$. The best way to tackle this is probably just to write down the first few terms of the sequence and see what they look like. For n=1,2,3,4 you get the values 1, 4/5, 6/8, 8/11. These numbers are decreasing, and you'll need to convince yourself that the sequence continues to decrease, with limiting value 2/3 as $n\to\infty$. The initial value 1 is the l.u.b., and it is in the set. The g.l.b. is 2/3, and that value is never actually attained so it is not in the set.

Finally, in (iv), a real number x satisfies the condition $x^2\leqslant2$ if $-\sqrt2\leqslant x\leqslant\sqrt2$, so $-\sqrt2$ is the g.l.b and $+\sqrt2$ is the l.u.b. But we are told that $x\in\mathbb{Q}$, and the numbers $\pm\sqrt2$ are not rational. So neither the g.l.b. nor the l.u.b. is in the set in this case.

I hope that illustrates the sort of reasoning that you need to use to answer problems of this sort.
• May 10th 2010, 01:19 PM
Mathman87
wow, great explanation, thanks! :-)