# Thread: Begining Analysis question

1. ## Begining Analysis question

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 $b\in\mathbb{R}>1$. If $x\in\mathbb{R}$, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t\leq{x}$. Prove that:
$\quad\quad\quad{b^x=\sup{B(x)}}$"

Solution: $x\in\mathbb{R}\implies\exists{t}\in\mathbb{Q}\leq{ x}\quad\therefore{B(x)\ne\emptyset}$. Furthermore if $t\leq{x}\implies{b^t\leq{b^x}}$(Can I assume this, or would I have to prove it?). So we can state that $\forall{t}\in{B(x)}~b^t\leq{b^x}$. So $B(x)$ is bounded above by $b^x$. And if $t $b^t$ would not be an upper bound of $B(x)$. Therefore $\sup(B(x))=b^x$

Is that ok? If it's horrific I apologize, it's my first time doing these kind of questions.

2. Originally Posted by Mathstud28
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 $b\in\mathbb{R}>1$. If $x\in\mathbb{R}$, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t\leq{x}$. Prove that:
$\quad\quad\quad{b^x=\sup{B(x)}}$"

Solution: $x\in\mathbb{R}\implies\exists{t}\in\mathbb{Q}\leq{ x}\quad\therefore{B(x)\ne\emptyset}$. Furthermore if $t\leq{x}\implies{b^t\leq{b^x}}$(Can I assume this, or would I have to prove it?). So we can state that $\forall{t}\in{B(x)}~b^t\leq{b^x}$. So $B(x)$ is bounded above by $b^x$. And if $t $b^t$ would not be an upper bound of $B(x)$. Therefore $\sup(B(x))=b^x$

Is that ok? If it's horrific I apologize, it's my first time doing these kind of questions.
I think every thing is there but the structure of the proof is not clear.

I would present it this way.

1. Show that $b^x$ is an upper bound for $B(x)$ .

2. Now suppose that $b^x$ is not a least upper bound, but that there exists a $x_1$ such that $b^{x_1} and for all $y\in B(x), y. Then derive a contradiction.

CB

3. The big question here is : How is $b^x$ defined, when x is irrational?

There are two possible approaches. The first is to define $b^x$ as the limit of $b^t$ as the rational number t approaches x. The second is to define $b^x = \exp(x\ln b)$, 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 $x\ln b = \sup\{t\ln b: t\in\mathbb{Q}, t, given that b>1. Since the exponential function is increasing and continuous, it follows that $b^x = \sup\{b^t:t\in\mathbb{Q}, t.

If you are using the first definition then the argument suggested by Mathstud28, as clarified by CaptainBlack, looks fine.

4. Originally Posted by CaptainBlack
I think every thing is there but the structure of the proof is not clear.

I would present it this way.

1. Show that $b^x$ is an upper bound for $B(x)$ .

2. Now suppose that $b^x$ is not a least upper bound, but that there exists a $x_1$ such that $b^{x_1} and for all $y\in B(x), y. Then derive a contradiction.

CB
Thank you very much CaptainBlack, that is exactly what I was looking for. Is it best to number steps, subsequently breaking the proof down into sections, oppossed to having one fluid paragraph like proof?

Originally Posted by Opalg
The big question here is : How is $b^x$ defined, when x is irrational?

There are two possible approaches. The first is to define $b^x$ as the limit of $b^t$ as the rational number t approaches x. The second is to define $b^x = \exp(x\ln b)$, 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 $x\ln b = \sup\{t\ln b: t\in\mathbb{Q}, t, given that b>1. Since the exponential function is increasing and continuous, it follows that $b^x = \sup\{b^t:t\in\mathbb{Q}, t.

If you are using the first definition then the argument suggested by Mathstud28, as clarified by CaptainBlack, looks fine.
Thank you Opalg, the definition you alluded to is how I would define the exponentials, but I thought it prudent to follow the example of my book. Which is "Principles of Mathematical Analysis" (Rudin) by the way. It seems a pretty good book as of yet.

5. Originally Posted by Mathstud28
Thank you very much CaptainBlack, that is exactly what I was looking for. Is it best to number steps, subsequently breaking the proof down into sections, oppossed to having one fluid paragraph like proof?
The aim is for clarity, so if numbering the steps help do so, but it is not usually necessary. The structure of a proof can often be made clear just by having blank lines between the main parts.

CB

6. 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

7. Originally Posted by Moo
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
I would normally give a informal description of the result and its proof before starting the more formal presentation rather than interpolating it in the main discussion.

CB

8. 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.

. 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
I doubt this.

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 $b^x$ 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.

>Opalg

I think defining $e^x$ as a power series is enough to omit "..define b^x as the limit of b^t...". And the continuity of $e^x$ 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)

9. Originally Posted by deinol
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.

Firstly, you will have to deal with what you mean by "if $t", though it is a minor problem, compared to this: even if $t, we still have $sup(B(x))=b^x$. 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 $b^x$ 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 $b^x$ 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.

>Opalg

I think defining $e^x$ as a power series is enough to omit "..define b^x as the limit of b^t...". And the continuity of $e^x$ 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)
Ok, how would you go about proving that it is the LUB?

10. As CaptainBlack as mentioned:

2. Now suppose that is not a least upper bound, but that there exists a such that and for all . Then derive a contradiction.
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.

Suppose $$\exists s$$ s.t. $\forall p \in B(x) , p

We will have to prove that [LaTeX ERROR: Convert failed] s.t. $$s<s'<b^x$$, i.e. $$s$$ 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]
then
[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?

11. Originally Posted by deinol
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.

Suppose $\exists s$ s.t. $\forall p \in B(x) , p

We will have to prove that $\exists s' \in B(x)$ s.t. $s, i.e. $s$ is not an upper bound, for the contradiction. I think this relies on

1. the continuity of $b^t$
2. $b^t$ is strictly increasing
3. the Intermediate Value Theorem
4. between any two distinct numbers there is a rational number

, which, in conjunction, effectively state

$\forall x,y \in \mathbb{R}$,
$b^x s.t. $b^x
then
$\exists q \in (x,p) s.t. q \in \mathbb{Q} and also b^x

and complete the proof. Note that although 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?
Oh, I thought you meant some way of proving it by some delta-epsilon analogue. Yeah, I actually did something similar to that in my actual response (not as in depth). And as for your question $\notin$=\notin