# [SOLVED]Baby Rudin Exercise 6 Part C

#### conscipost

##### Member
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
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
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
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
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
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
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
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.