2
$\begingroup$

Fix $b>1$. Show the following:

(a) If $m,n,p,q$ are integers, $n>0, q>0$, and $r = m/n = p/q$, prove that $(b^{m})^{1/n} = (b^p)^{1/q}$. So $(b^r) = (b^m)^{1/n}$.

(b) Prove that $b^{r+s} = b^{r}b^{s}$ for $r,s \in \mathbf{Q}$.

(c) If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, there $t$ is rational and $t \leq x$. Prove that $b^r = \sup B(r)$ where $r$ is rational. So $b^x = \sup B(x)$ for every real $x$.

(d) Show that $b^{x+y} = b^{x}b^{y}$ for all real $x$ and $y$.

For (a), we know that $mq = np$. Thus $b^{mq} = b^{np}$. Then $(b^{m})^{q} = (b^{n})^{p}$. Or $(b^{m})^{q} = (b^{p})^{n}$. So the result follows if we note that $m = np/q$ and $n = mq/p$?

For (b) just start with $b^{\frac{mq+pn}{qn}}$?

For (c) and (d), we just need to show that $B(x)$ is non-empty. Then use a proof by contradiction?

  • 0
    For (a), yes, that's one way. For (b), you can use (a) and the properties known for integers, using the expression you do. For (c) you need to show more than just that $B(x)$ is nonempty; also, bounded above. Then you can use the definition of suprema as the theast upper bound. For (d), you use (c) and show you can express $B(x+y)$ in terms of $B(x)$ and $B(y)$.2010-12-20
  • 1
    The plural of "supremum" is "suprema".2010-12-20
  • 1
    It looks like you have just started with Rudin's Principles. I've just started with the book as well, if you'd like to discuss it with me, my email address is sqznegva.92@tznvy.pbz (it's been rot-13d).2010-12-20
  • 0
    So for (d), show that $\sup B(x+y) = \sup B(x) \sup B(y)$?2010-12-20

1 Answers 1

1

I've been working on this exercise today as well. I managed to prove (I think) (a), but it's a bit complicated. Let's see:

Lemma 1: $(a^x)^{1/x} = a$, for all $a>0,x\in\mathbb{Q}$. This follows from the uniqueness of positive roots.

Lemma 2: $(a^x)^{1/y} = (a^{1/y})^x$, for all $a>0\in\mathbb{Q}$, $x,y\in\mathbb{Z}$. This follows from Theorem 1.2.1 from Rudin's.

Lemma 3: $(a^x)^y = a^{xy}$, for all $a\in\mathbb{Q}$, $x,y\in\mathbb{Z}$, since $\prod_1^y\prod_1^x a = \prod_1^{xy} a$

To prove (a), we start from $b = (b^{1/q})^q$ because of (1).

Then we raise both sides to the $m$th power to get $b^m = ((b^{1/q})^q)^m= (b^{1/q})^{qm}$, because of (3).

Since $m/n = p/q$, then $qm = pn$. Then $b^m = (b^{1/q})^{pn}=((b^{1/q})^p)^n$, once again because of (3).

Then $b^m = ((b^p)^{1/q})^n$ because of (2). Taking the $n$th root to both sides we get $(b^m)^{1/n} = (b^p)^{1/q}$, which is what we wanted to prove.

To prove (b), consider $b^{mq+np}=b^{mq}b^{np}$ (which holds because $m,n,p,q\in\mathbb{Z}$). This is equivalent to $(b^{m/n})^{nq}(b^{p/q})^{qn}$ because of the previously mentioned lemmas.

Taking the $nq$th root to both sides, we end up with $b^{\frac{mq+np}{nq}} = b^{m/n}b^{p/q}$, which is what we wanted to prove.

(c) is proved easily since $b^r \geq x\in B(r)$ and $b^r\in B(r)$, therefore $b^r=\sup(B(r))$

I haven't proved (d) yet but will edit this if I do.

Anyway, I'm not sure about this so feedback is really appreciated.

  • 0
    For part a), is it ok to "raise both sides" to whole or fractional powers? Rudin does start off the book with the assumption that we are all aware of the rational numbers, but the way he introduces exponents in terms of the secondary field operation makes me wonder if we can take the basic algebra rules for granted.2012-11-12