I'm having a little bit of a problem with this; any help would be handy!
Find, with proof the least upper bound of this set of real numbers!
E = {6 - (2/(3 n + 1)) : n is a element of N}
Where N is the natural Numbers
I'm having a little bit of a problem with this; any help would be handy!
Find, with proof the least upper bound of this set of real numbers!
E = {6 - (2/(3 n + 1)) : n is a element of N}
Where N is the natural Numbers
Claim Sup(E) = 6
To prove this, first 6 must be found to be a valid upper bound for E. Then we must ascertain that their is no smaller upper bound then 6. Where y < 6, where y might be a smaller upper bound for E.
Proof:
Let x ∈ E with x>y
$\displaystyle \left(6-\frac{2}{3n+1}\right)$>y
$\displaystyle \left(-\frac{2}{3n+1}\right)$>y-6
$\displaystyle \left(-\frac{2}{3n+1}\right)$>y-6
-2>(y-6)(3n+1)
n>$\displaystyle \frac{1}{3}\left(\frac{2}{y-6}+1\right)$
Then pick $\displaystyle n \in N$
such that n>$\displaystyle \frac{1}{3}\left(\frac{2}{y-6}+1\right)$
This means that $\displaystyle \left(6-\frac{2}{3n+1}\right)$>y
Hence y can not be an upper bound for E!!!
Finally got it after pouring through a book for a very long while!!!
Thanks for the help!
this is incorrect. you assumed x > y. that is, y is not an upper bound. this is what you want to show, not assume. your choosing of n seems to be inconsequential. moreover, you ended up with exactly what you assumed, which means you've shown nothing, just went in a circle.
I would proceed this way:
Clearly 6 is an upper bound for E, since (i leave it to you to show this).
To show that $\displaystyle 6 = \sup E$, we need to show that 6 is the least upper bound. Assume, to the contrary, that it isn't. Rather, let $\displaystyle y = \sup E$ where $\displaystyle y \in \mathbb R,~y < 6$. Then, for all $\displaystyle x \in E$, we have $\displaystyle x \le y < 6$.
Now, let $\displaystyle y = 6 - \epsilon$ for some $\displaystyle \epsilon > 0$. Then we have, for all $\displaystyle n \in \mathbb N$, that
$\displaystyle 6 - \frac 2{3n + 1} \le 6 - \epsilon < 6$
$\displaystyle \Leftrightarrow 0 < \epsilon \le \frac 2{3n + 1}$ for all $\displaystyle n$.
Which is absurd (why?). Hence our original assumption must have been in error and, in fact, $\displaystyle \sup E = 6$