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[SOLVED] Baby Rudin Exercise 6 Part C

conscipost

Member
Jan 26, 2012
39
I'm having trouble with this exercise:
If x is an element of real numbers, define B(x) to be the set of all numbers b^t,
where t is rational and t<=x
Prove that b^r=sup(B(r)) when r is rational.
I had a proof by contradiction in mind, but I am having trouble furnishing it.
Thanks,
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I'm having trouble with this exercise:
If x is an element of real numbers, define B(x) to be the set of all numbers b^t,
where t is rational and t<=x
Prove that b^r=sup(B(r)) when r is rational.
I had a proof by contradiction in mind, but I am having trouble furnishing it.
Thanks,
Hi conscipost, :)

Is this the original problem exactly as it appears on the book? It seems to me that this problem is incorrect. Take for example,
\(b=-1\) and \(x=r=3\). Then,

\[B(3)=\{(-1)^{t}~:~t\leq 3\mbox{ and }t\in\mathbb{Q}\}\]

It is clear that, \(\mbox{Sup}[B(3)]=1\) and \((-1)^{3}=-1\) and hence,

\[\mbox{Sup}[B(3)]\neq (-1)^{3}\]

Kind Regards,
Sudharaka.
 

conscipost

Member
Jan 26, 2012
39
Hi conscipost, :)

Is this the original problem exactly as it appears on the book? It seems to me that this problem is incorrect. Take for example,
\(b=-1\) and \(x=r=3\). Then,

\[B(3)=\{(-1)^{t}~:~t\leq 3\mbox{ and }t\in\mathbb{Q}\}\]

It is clear that, \(\mbox{Sup}[B(3)]=1\) and \((-1)^{3}=-1\) and hence,

\[\mbox{Sup}[B(3)]\neq (-1)^{3}\]

Kind Regards,
Sudharaka.
Thanks Sudharaka, but no, I'm sorry. Earlier in the exercise Rudin says to fix b>1.
So I can prove that b^r is an upper bound , that is that r>=t implies that b^r>=b^t, but how should I prove that if y<b^r then y is not an upper bound?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Thanks Sudharaka, but no, I'm sorry. Earlier in the exercise Rudin says to fix b>1.
So I can prove that b^r is an upper bound , that is that r>=t implies that b^r>=b^t, but how should I prove that if y<b^r then y is not an upper bound?
In that case the proof is obvious. For \(r\in\mathbb{Q}\)

\[B(r)=\{b^t~:~t\leq r\mbox{ and }t\in\mathbb{Q}\}\]

and we know that for \(b>1\)

\[b^t\leq b^r\mbox{ where }t\leq r\]

So \(b^r\) is the maximum of \(B(r)\).

Kind Regards,
Sudharaka.
 

conscipost

Member
Jan 26, 2012
39
Hi conscipost, :)

Is this the original problem exactly as it appears on the book? It seems to me that this problem is incorrect. Take for example,
\(b=-1\) and \(x=r=3\). Then,

\[B(3)=\{(-1)^{t}~:~t\leq 3\mbox{ and }t\in\mathbb{Q}\}\]

It is clear that, \(\mbox{Sup}[B(3)]=1\) and \((-1)^{3}=-1\) and hence,

\[\mbox{Sup}[B(3)]\neq (-1)^{3}\]

Kind Regards,
Sudharaka.
Would this be true as the complex field is not an ordered field? (-1)^(1/2) is an element of "B(3)" and how would one prove that 1>i? If we consider the sup is in C, then it dne, right? But if R is on our mind can we ignore the undefined things like (-1)^(1/2) and claim that 1 is the sup?

The definition of the least upper bound is given as such.
for a non empty subset E of S that is bounded above the least upper bound, a holds two properties:
1) for all x in E, x<=a (a is an upper bound)
2) if y<a then y is not an upper bound of E

So what I am asking is what is a good way to prove that if y is an element of B(r) and is not b^r then y is not an upper bound.

thanks again,
 
Last edited:

conscipost

Member
Jan 26, 2012
39
I'm sorry, it is rather obvious. It just was not coming until now.
So if y<b^r assume that y is an upper bound of B(r) then y>=b^t for all rational t<=r. Thus y>=b^r but y<b^r thus y is not an upper bound. Thanks to all.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
I'm having trouble with this exercise:
If x is an element of real numbers, define B(x) to be the set of all numbers b^t,
where t is rational and t<=x
Prove that b^r=sup(B(r)) when r is rational.
I had a proof by contradiction in mind, but I am having trouble furnishing it.
I don't have access to Rudin's book, but I wonder whether you have mis-stated the problem. I think that the definition of $B(x)$ should be $B(x) = \{b^t: t<x\}$ (with a strict inequality). The motivation for this exercise is that when $x$ is irrational $b^x$ is defined to be $\sup B(x)$, and the point of the exercise is that this agrees with the existing definition of $b^x$ in the case where $x$ is rational.

If my version of the question is correct then the problem reduces to this: Given $\varepsilon>0$, find a rational number $t<x$ such that $b^t > b^x - \varepsilon.$ To see how this might be done, consider the very simple special case where $b=2$, $x=3$ and $\varepsilon = 0.01$. The problem then is to find a rational number $t<3$ such that $2^t > 7.99.$

Suppose that $t = p/q$, where $p,q$ are positive integers with $p/q<3$. We need to choose $p$ and $q$ so that $2^{p/q} >7.99$, or equivalently $2^p > 7.99^q.$ Since $p/q$ must be less than 3, the largest admissible value of $p$ is $p=3q-1$. The inequality then becomes $2^{3q-1} > 7.99^q$. Take logs: $(3q-1)\ln2 > q\ln7.99.$ The solution to that inequality is $$q>\frac{\ln2}{\ln8-\ln7.99}.$$ So choose $q$ to be any integer large enough to satisfy that inequality, and take $t=(3q-1)/q.$ Then $t<3$ and $2^t>7.99.$

That should give you a strategy for proving the result in general. There will be some extra complications, because in my simple numerical example I chose $x$ to be an integer, $x=3$. In the case where $x$ is a general rational number the proof may need some extra ingredients, but the overall method ought to work.
 

conscipost

Member
Jan 26, 2012
39
I don't have access to Rudin's book, but I wonder whether you have mis-stated the problem. I think that the definition of $B(x)$ should be $B(x) = \{b^t: t<x\}$ (with a strict inequality). The motivation for this exercise is that when $x$ is irrational $b^x$ is defined to be $\sup B(x)$, and the point of the exercise is that this agrees with the existing definition of $b^x$ in the case where $x$ is rational.

If my version of the question is correct then the problem reduces to this: Given $\varepsilon>0$, find a rational number $t<x$ such that $b^t > b^x - \varepsilon.$ To see how this might be done, consider the very simple special case where $b=2$, $x=3$ and $\varepsilon = 0.01$. The problem then is to find a rational number $t<3$ such that $2^t > 7.99.$

Suppose that $t = p/q$, where $p,q$ are positive integers with $p/q<3$. We need to choose $p$ and $q$ so that $2^{p/q} >7.99$, or equivalently $2^p > 7.99^q.$ Since $p/q$ must be less than 3, the largest admissible value of $p$ is $p=3q-1$. The inequality then becomes $2^{3q-1} > 7.99^q$. Take logs: $(3q-1)\ln2 > q\ln7.99.$ The solution to that inequality is $$q>\frac{\ln2}{\ln8-\ln7.99}.$$ So choose $q$ to be any integer large enough to satisfy that inequality, and take $t=(3q-1)/q.$ Then $t<3$ and $2^t>7.99.$

That should give you a strategy for proving the result in general. There will be some extra complications, because in my simple numerical example I chose $x$ to be an integer, $x=3$. In the case where $x$ is a general rational number the proof may need some extra ingredients, but the overall method ought to work.
The book does say t<=x, and I can see the other definition being more suitable for reals but maybe this less than or equal has to do with the fact that this b^r may be rational. So in the case b=4 and r=(1/2) then there is no problem with 2 being part of the set and encourages you to view the sup(B(1/2)) as being in Q rather than R. I'm not sure...

I'll quote:
Fix b>1
If x is real, define B(x) to be the set of all numbers b^t, where r is rational and t<=x. Prove that b^r=sup(B(r))
when r is rational. Hence it makes sense to define
b^x=sup(B(x))
for every real x.

I considered a general "epsilon" before but as you can see I'm not quick on my feet yet and I think I'll come back to it later. Thanks for the insights.