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No real exceeds every whole number, so $1/n$ can be made below any $\varepsilon$; and a rational lies in every gap.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to prove the Archimedean property from the completeness axiom, use it to turn a threshold into a particular whole number, and run the density argument that finds a rational strictly between two reals. You will also be able to say which of the three principles of this unit supplies what, and why the order of the two choices in the density proof cannot be swapped.
The completeness axiom, and the habit of using a supremum by producing an element close to it. Both of this lesson's results are proved from that axiom and nothing else, so this is also the first sight of what the axiom is actually for.
An ordered field is Archimedean when no element of it exceeds every natural number — equivalently, when $1/n$ can be made smaller than any given positive amount. A set is dense in the line when some member of it lies strictly between any two reals.
The Archimedean property. For every real $x$ there is a natural number $n$ with $n > x$. Equivalently, for every $\varepsilon > 0$ there is $n$ with $1/n < \varepsilon$.
Proof. If not, $\mathbb{N}$ is bounded above; being non-empty, it has a supremum $M$ by completeness. Then $M - 1$ is not an upper bound, so some $n \in \mathbb{N}$ has $n > M - 1$, giving $n + 1 > M$ — and $n + 1$ is a natural number. Contradiction.
The second form is the one that gets used. Every $\varepsilon$-$N$ proof in this course ends by choosing $N$ with $1/N < \varepsilon$, and this is the theorem that says such an $N$ is there.
Density of $\mathbb{Q}$. Between any two reals $a < b$ there is a rational.
Proof. Choose $n$ with $1/n < b - a$ — Archimedes. Let $m$ be the least integer with $m/n > a$. Then $(m-1)/n \le a$, so $m/n \le a + 1/n < b$. The order of the two choices is not negotiable: the step size is fixed so that one step cannot jump the gap, and only then is the first step past $a$ taken.
The same argument with $a\sqrt{2}$ and $b\sqrt{2}$ shows the irrationals are dense too.
Another way: picture
Lay a ruler marked in steps of $1/n$ along the line. Archimedes says the marks are eventually as fine as you like. Density says that once the marks are finer than the gap between $a$ and $b$, a mark has to fall inside the gap: a step that small cannot straddle it.
Another way: steps
To meet a threshold $\varepsilon$:
To find a rational in a gap, do step 1 to 3 for the gap width, then take the first multiple of $1/n$ past the left-hand end.
It feels like nothing. Of course some whole number is bigger than any given number — what else could happen?
Quite a lot else. There are ordered fields, perfectly consistent ones, containing an element larger than $1, 2, 3, \ldots$ all at once. The rational functions ordered by eventual size are such a field: the function $x$ is above every constant. In that field $1/x$ is positive and smaller than every $1/n$, and any sequence argument that assumes otherwise is simply false there.
So the property is not a fact about numbers in general; it is a fact about $\mathbb{R}$, and it comes from completeness. That is the pattern of the whole unit: the things that feel too obvious to prove are exactly the things the axiom is buying.
| Need | Choose | Because |
|---|---|---|
| $1/N$ below $\varepsilon$ | $N > 1/\varepsilon$ | Archimedean property |
| $n/N$ below $\varepsilon$ for fixed $n$ | $N > n/\varepsilon$ | the same, rescaled |
| a rational in $(a,b)$ | $n > 1/(b-a)$, then the least $m$ with $m/n > a$ | density |
| a rational within $\varepsilon$ of $x$ | a rational in $(x - \varepsilon, x)$ | density |
The last row is worth keeping. It says every real can be approximated by rationals to any accuracy, which is why decimal expansions exist and why a computer can work with real numbers at all.
Taking $N = 1/\varepsilon$. That is a real number, not a whole one, and the definition asks for a whole one. Take the first whole number past it.
Thinking density implies completeness. $\mathbb{Q}$ is dense in itself and is not complete. Density fills gaps between numbers that are there; completeness is about the gaps where no number is.
Using the property without naming it. An $\varepsilon$-$N$ proof that says 'choose $N$ large enough' has skipped the only step that needed a theorem. Name it once and the proof is honest.
Show that $\inf\{1/n : n \in \mathbb{N}\} = 0$. Certainly $0$ is a lower bound.
The easy half first.
Let $\varepsilon > 0$. By the Archimedean property there is $n$ with $n > 1/\varepsilon$, so $1/n < \varepsilon$.
The property, named where it is used.
So no positive number is a lower bound, and the infimum is $0$ — unattained, since no $1/n$ equals it.
Greatest lower bound, not smallest element.
Find a rational between $2.7$ and $2.75$. The gap is $0.05$, so choose $n = 100$, since $1/100 < 0.05$.
Denominator first, from the gap.
The least integer $m$ with $m/100 > 2.7$ is $m = 271$.
Numerator second, from the denominator.
And $271/100 = 2.71$, which is below $2.75$. A coarser ruler — steps of $1/10$ — would have jumped from $2.7$ straight to $2.8$ and missed.
The order of the choices is the proof.
Given $a < b$, apply density to the pair $a/\sqrt{2}$ and $b/\sqrt{2}$ to get a rational $r$ strictly between them.
Rescale so the known theorem applies.
Then $r\sqrt{2}$ lies strictly between $a$ and $b$.
Rescale back.
It is irrational provided $r \ne 0$, and if the rational density hands back $0$ the interval can be shrunk to one side of the origin first. So both the rationals and the irrationals are dense, which is why neither can be recognised by looking at a small piece of the line.
Put the steps of the proof that some rational lies strictly between two reals $a < b$ into the order they must be made in.
Number the steps in order (write the number in the box):
The Archimedean property gives a whole number $N$ with $\dfrac{1}{N} < \dfrac{1}{7}$. What is the smallest such $N$?
Answer:
For each threshold, give the smallest whole number $N$ with $1/N$ strictly below it.
| smallest whole number N | |
|---|---|
| threshold $1/6$ | |
| threshold $1/9$ | |
| threshold $1/12$ |
Match each thing you might need to the principle that supplies it.
| the completeness axiom | the Archimedean property | the density of the rationals | |
|---|---|---|---|
| a whole number larger than a given real number | |||
| a rational number strictly between two given reals | |||
| the least upper bound of a non-empty set bounded above | |||
| a value of $1/n$ below the threshold $1/7$ |
The Archimedean property is proved, not assumed. Suppose no whole number exceeded the real number $9/3$. What does the proof do with that supposition?
Find a rational strictly between $\dfrac{5}{6}$ and $\dfrac{7}{6}$ whose denominator is $6$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For each threshold, give the smallest whole number $N$ with $1/N$ strictly below it.
| smallest whole number N | |
|---|---|
| threshold $1/2$ | |
| threshold $1/5$ | |
| threshold $1/8$ |
You can produce the whole number a threshold needs, prove the Archimedean property from completeness, and find a rational in any gap. Say in your own words why an ordered field can fail to be Archimedean. Next: the inequality every estimate in the rest of the course is built from.
10. Your turn: show that between any two reals there is an irrational, step 3