I am starting analysis and one of the problems I am encountering is since I do not have a teacher or guide I am not sure if what I write is sufficent or overly verbose etc. I would appreciate if anyone would critique this solution to a problem and maybe give me suggestsions such as "You cannot assume this" or "You needn't state that", things of that nature. Thank you very much in advance.
Question: "Fix . If , define to be the set of all numbers , where is rational and . Prove that:
Solution: . Furthermore if (Can I assume this, or would I have to prove it?). So we can state that . So is bounded above by . And if would not be an upper bound of . Therefore
Is that ok? If it's horrific I apologize, it's my first time doing these kind of questions.
The big question here is : How is defined, when x is irrational?
There are two possible approaches. The first is to define as the limit of as the rational number t approaches x. The second is to define , where the exponential function is defined by its power series, and the logarithm is the inverse of the exponential.
In a rigorous approach to analysis, I think that the second definition is far more satisfactory and easy to deal with. If you are using that definiton then this problem is quite simple: It is clear that , given that b>1. Since the exponential function is increasing and continuous, it follows that .
If you are using the first definition then the argument suggested by Mathstud28, as clarified by CaptainBlack, looks fine.
There's something we're sometimes told to do, it's to announce what you'll prove.
For example, you can make a first paragraph beginning with : let's prove that b^x is an upper bound for B(x)
and a second one beginning with : let's prove that b^x is the smallest of the upper bounds.
When you have a 2-pages long proof, it may be useful
When an assumption is given to you at your progress (up to some page of a beginners' Analysis book, some point of your course etc.)
For instance, was the exponential function properly defined? Was it proven to be strictly increasing and continuous? Was even the continuity of a function defined? When the answers to these are given to us, we would be able provide you with what you've requested.
I doubt this.. So B(x) is bounded above by b^x. And if t<x b^t would not be an upper bound of B(x) Therefore \sup(B(x))=b^x
Firstly, you will have to deal with what you mean by "if [LaTeX ERROR: Convert failed] ", though it is a minor problem, compared to this: even if [LaTeX ERROR: Convert failed] , we still have [LaTeX ERROR: Convert failed] . Think of x being irrational, then the equality part of the inequality sign would mean nothing.
One more note is, even if you were right above, you still haven't actually proven [LaTeX ERROR: Convert failed] is the least upper bound.
You might have thought "since the supremum is a fine line to cross, and proving we have crossed the fine line, by using slightly smaller numbers to generate B(x), we can somehow prove is the supremum". Sometimes this is true (not in this case), but still you may have to elaborate.
This are some general tips for deciding what you should assume:
1. Stick to the proven theorems given to you, unless it is something trivial (e.g. various basic properties of the natural numbers).
2. Check where did you theorems come from - do not try to prove a premise in a theorem using the theorem.
I think defining as a power series is enough to omit "..define b^x as the limit of b^t...". And the continuity of already implies a proper definition of b^x when x is anyway real.
(Edited many times due to my careless and impatient proof-reading... this will be the last 02:18 16/11/2008)
As CaptainBlack as mentioned:
This is a common way of identifying the supremum. In fact, I have used it in almost all of my homework and exams in my first year at university.2. Now suppose that is not a least upper bound, but that there exists a such that and for all . Then derive a contradiction.
Suppose [tex]\exists s[/tex] s.t.
We will have to prove that [LaTeX ERROR: Convert failed] s.t. [tex] s<s'<b^x [/tex], i.e. [tex]s[/tex] is not an upper bound, for the contradiction. I think this relies on
1. the continuity of [LaTeX ERROR: Convert failed]
2. [LaTeX ERROR: Convert failed] is strictly increasing
3. the Intermediate Value Theorem
4. between any two distinct numbers there is a rational number
, which, in conjunction, effectively state
[LaTeX ERROR: Convert failed] ,
[LaTeX ERROR: Convert failed] s.t. [LaTeX ERROR: Convert failed]
[LaTeX ERROR: Convert failed] s.t. [LaTeX ERROR: Convert failed] and also [LaTeX ERROR: Convert failed]
to help completing the proof. I am not sure whether we could use theorems that are more elementary, and whether doing so could maintain a similar degree of simplicity, but here we go.
BTW, is there a "not in" symbol (slashed "in" symbol) in LaTex?