Recent content by pendesu

I would start by assuming (P→Q) and (Q→R) and then try to prove the rest. I can get you started in this direction. For example, Suppose P. Then Q. Then R. Thus, P→R. Then show the disjunction with the bijections is true too and then the "backwards" direction of assuming RHS and proving LHS of...

Thought for a moment this morning that the following would be true but I don't think that it could be true for this case since a_{m+1}+b_{m+1} could be greater than b-1. The following is along the lines I was thinking but again I don't think in this case it would work out this way for the...

I realized I typed my question incorrectly. Given the following: c_{0},c_{1},...,c_{m}\in\{0,...,b-1\} and \{a_{k}\},\{b_{k}\}\subseteq\{0,...,b-1\} and \sum\limits _{k=0}^{m+1}c_{k}b^{k}=(\sum\limits _{k=0}^{m+1}(a_{k}+b_{k})b^{k})\text{mod }b^{m+2} and 0\leq c_{m+1}\leq b-1 , show that...

Homework Statement Given that b>1,b\in\mathbb{Z},c_{0},c_{1},...,c_{m}\in\{0,...,b-1\}, 0\leq c_{m+1}\leq b-1, and c_{m+1}b^{m+1}=(\sum\limits _{k=0}^{m+1}c_{k}b^{k})\text{mod }b^{m+2}-c_{0}-c_{1}b-c_{2}b^{2}-...-c_{m}b^{m}, show that c_{m+1}\in\mathbb{Z}. Also, \sum\limits...

That is the thing. This was something I was going to ask my professor that I am going to be doing research with in p-adic analysis. I am starting to think this may just be a definition since if I recall my professor he was saying how a p-adic integer which is a power series may not be a...

Homework Statement Given \overset{\infty}{\underset{n=0}{\sum}}a_{n}(x-a)^{n} and \overset{\infty}{\underset{n=0}{\sum}}b_{n}(x-a)^{n} that are in R. Then, \overset{\infty}{\underset{n=0}{\sum}}a_{n}(x-a)^{n}=\overset{\infty}{\underset{n=0}{\sum}}b_{n}(x-a)^{n} if and only if a_{n}=b_{n}...

The question came up because I was watching these real analysis lectures and when the topic of least upper bound came up, the professor gave the example of the set A = {x in Q | x2 < 2} saying that this had no least upper bound. For some reason, I was thinking that there was a least upper bound...

Homework Statement Prove that there is no upper bound in A, where A = {x in Q | x2 < 2} The Attempt at a Solution My attempt has been to assume that there is an upper bound p in A and then I have been trying to find a way to show that there is a number that is larger than p but still in A...