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I'm working on a homework problem which states, "Prove the $\sup \{\frac{n}{2n+1} : n \in \mathbb{N} \} = \frac{1}{2}$. In particular, explain where the Archimedean Property of the real numbers is used."

My professor went over this in class today. I'm just not getting the points he made for part 2 of the proof he did. I hope my notes make it clear. So, without further ado.

Part 2
Show that if $b<\frac{1}{2}$, then b is not an upper bound. Find $n\in\mathbb{N}$ so that $b<\frac{n}{2n+1}$. So, look at $\frac{1}{2} - (\frac{n}{2n+1}) \rightarrow \frac{2n+1}{2(2n+1)}-\frac{2n}{2(2n-1)} = \frac{1}{2(2n+1)}$. Find an $n$ so that $\frac{1}{2(2n+1)} < \frac{1}{2}-b$. Given any $\epsilon > 0$ there is $n$ such that $\frac{1}{n} < \epsilon$. We can find $n$ with $\frac{1}{n}<\frac{1}{2}-b \rightarrow \frac{1}{2(2n+1)} < \frac{1}{2}-b \rightarrow b < \frac{n}{2n+1}$.

Now my notes further state that it's in that last sentence that the Archimedean Property is used. I'm just not seeing it. I think I do but I don't get why it helps to be honest. As stated in my text, "Given any real number $y>0$, there exists an $n \in \mathbb{N}$ satisfying $\frac{1}{n}is the $\frac{1}{n} < y$ and the rest follows. Do I have it correct? (By the way, if what I have doesn't make sense, I apologize. I feverishly write my notes while he talks.)

Another question I have is, why is the Archimedean Property applicable when real numbers weren't stated explicitly in the problem?

1 Answers 1

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Yes. The usual formulation of the Archimedean Property states that for any reals $0 < r < s$, there is an integer $m$ so that $mr > s$. Another way of saying that is that if $0 < y < 1$, then we can find $n$ with $ny > 1$; dividing both sides by $n$, $y > \frac{1}{n}$.

In the last sentence, we have a real $\frac{1}{2} - b$, and we need an integer $n$ with $\frac{1}{n} < \frac{1}{2} - b$. Since $\frac{1}{2} - b > 0$ (because, by hypothesis, $b < \frac{1}{2}$) the Archimedean Property guarantees that there is such an $n$.

Real numbers were certainly stated in the problem - $\frac{n}{2n+1}$ is a real number for every $n$. Irrationals weren't stated, but the Archimedean Property is true of the rationals as well. The Archimedean Property is just a true statement: under certain circumstances, a certain thing exists. The circumstances were met - we were looking at a real $> 0$ - so the Property applies.

Technically, I suppose you could nitpick and say (for example) "what if we're working in some number system which has $\frac{n}{2n+1}$ but also has a bunch of non-reals around, so that AP doesn't apply?" Technically, the problem doesn't state that this is not the case; but since all mentioned numbers are real, and an ordering is used (so we're not looking at complex numbers) it's reasonable to assume that the question was intended to ask about the supremum in the reals.

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    Thanks for the reply. So, $\frac{n}{2n+1}$ is a real number because fractions aren't in $\mathbb{N}$. I should have thought of that. I was hung up on the $n\in\mathbb{N}$ part of the statement. – 2017-01-27
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    @AndrewFalanga Well, remember that $\mathbb{N} \subset \mathbb{R}$; even if it were just $n$ it would still be a sequence of reals. In other words, the Archimedean Property still applies to the natural numbers - it's just that the $\frac{1}{n}$ it produces isn't *itself* a natural number. – 2017-01-27
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    One last question if I may, is what the professor did a proof by contradiction? – 2017-01-27
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    @AndrewFalanga Not as you've written it. You can tell because there was no contradiction. The only assumption made about $b$ is that it is less than $\frac{1}{2}$; the conclusion was that $b$ can't be an upper bound on the sequence. This conclusion doesn't contradict the assumption. If he'd started with "suppose $b < \frac{1}{2}$ is an upper bound on the sequence", then it would have been a proof by contradiction. – 2017-01-27