Answer to Question #275103 in Algebra for Upichu

Question

What can you say about the upper bound of E(x) for numbers of the form 3^(2n) + 1?

Answer 1

Consider the set

$\begin{aligned} &E(x)=\left\{3^{3 n}+1: n \in N 1\right\} . \\ &E(x) \neq \phi . \text { since } 3^{2}+1 \in E(x) . \\ &E(x) \subseteq \mathbb{R} . \end{aligned}$

Now if c is the least upper of E(x).

$c=\ln _{b}((-(x)).$

$\forall \epsilon>0, \quad c-\epsilon<3^{2 n+1}, \quad[\because c-\epsilon, is not * n \in N. lub \ of \ E(x)]$

Take

$c-2 \cdot 3^{2 n}<3^{2 n}+1 \quad \forall n \in N \text {. }$

$\begin{aligned} &c<3^{2 n}+8 \cdot 3^{2 n}+1, \quad \forall n \in N . \\ &c<3^{2 n}(1+8)+1 . \\ &c<3^{2 n} \cdot 3^{2}+1 \\ &c<3^{2(n+1)}+1 . \end{aligned}$

c is not lub of E(x)

Thus E(x) has no upper bound.