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Here's the problem

Fix b>1. If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$ where $t$ is rational and $t\leq x$. Prove that $$b^r=\sup B(r)$$ when $r$ is rational.

So I basically know exactly what I need to do. We know that $b^r\in B(r)$ so all that's left is to show that for any $t\leq r$, $b^t\leq b^r$. Seems simple enough but I just can't figure out how to prove this without begging all kinds of questions. All I need is a lemma that says for rationals $r,r'$, and a real number $b>1$, $b^r<b^{r'}$ implies $r<r'$. This is so obviously true and yet so hard for me to find a way to show!

By best attempt so far has been this: suppose $b^r<b^{r'}$. We can write $b^{r'}$ as $b^{r'-r+r}=b^{r'-r}b^r$ (I proved this in a previous problem). Now dividing both sides by $b^r$ gives $b^{r-r}<b^{r'-r}$. Now I really would like to take that and say that it implies that $r-r<r'-r$ and then, clearly, I'm done. But doing that makes me uncomfortable. I feel like I'm missing a logical step there—all I really have to work with is a theorem which says that positive real numbers have unique $n$-th roots, and I don't see how my move there would follow from that.

Any ideas? Maybe a different approach I could take?

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It should say $b^t$ and not $b'$. –  Alex Sep 20 '12 at 0:46

3 Answers 3

up vote 3 down vote accepted

Hint: Bring your two rationals to a common denominator.

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So long as you have the existence of $n$th roots, you should be able to show that $x^{1/n} > 1$ iff $x > 1$. Then you should have no problem showing that when $r$ is rational, $b^r > 1$ implies $r > 0$.

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I struggled with this problem for a bit too, so I'm going to give you a little bit more than a homework hint. First, $B(x) \subset \Bbb R$, so $B(x)$ has the least upper bound property and the supremum exists.

Let $r \in \Bbb Q$. Then let $b^t \in B(r)$. By the previous part of this problem, (part b), we have that $b^r/b^t = b^{r-t}$. Since $r - t > 0$, then $b^r \ge b^t$. (You can show this using Erick Wong's hint, or by showing that $1$ is the supremum of $B(0)$.) This shows that $b^r$ is an upper bound of $B(r)$.

This isn't enough. You need to show that it is the least upper bound. This is a simple contradiction proof -- assume that there is a $\gamma < b^r$ that is an upper bound, and then notice the contradiction this creates with the definition of $B(r)$.

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