There are rationals between any two reals? - Rudin 1.20

[gwsinger]

There are rationals between any two reals? -- Rudin 1.20

In Rudin 1.20(b), Rudin proves that between any two real numbers x and y, there exists a rational number p. His strategy is to "blow up" x and y by some integer n (generated by the archimedian property) such that ny - nx > 1. Then he attempts to locate another integer m that is between nx and ny, such that m - 1 < nx < m < ny. From here we have

nx < m < ny

and when we multiply each term by (1/n) we can say that

x < m/n < y where p = m/n. Hence the proof is done.

Now my question is as follows: how can Rudin just pick an m such that m - 1 < nx < m < ny?

I certainly understand that we can use the archimedian property to find an m greater than nx, but how can we be SURE that this m is less than ny and even less than 1 unit distance greater than nx?

Here is the proof in full from the text in case it helps:

Pg 9

If x in R, y in R, and x<y, then there exists a p in Q such that x<p<y.

Proof:
Since x<y, we have y-x>0, and the archimedean property furnishes a positive integer n such that
n(y-x)>1.

Apply the archimedean property again, to obtain positive integers m1 and m2 such that m1>nx, m2>-nx. Then

-m2<nx<m1.

Hence there is an integer m ( with -m2<= m <= m1) such that

m-1 <= nx < m

If we combine these inequalities, we obtain

nx<m<= 1 + nx < ny.

Since n>0, it follows that

x < m/n < y. This proves (b) with p = m/n.

 

[Bacle2]

If 0< y-x<1; y-x± 0.5 , then you can always find n with 2< n(y-x)<1. The least integer N larger than 1/(y-x) will do, i.e.:



2>N(y-x)>1

 

[gwsinger]

Bacle I appreciate your reply but I'm afraid I don't follow. Can you elaborate a little more?

 

[AlephZero]

Informally, you can "see" this must be true by drawing a number line and marking the integers between -m2 and m1.

Formally, there are only a finite number of integers m between -m2 and m1 so you can "check" the inequality for each m individually. If none of them satisfy m-1 <= nx < m, you have contradicted the fact that -m2<nx<m1.

 

[micromass]

Just define

$$m=\inf\{a\in \mathbb{N}~\vert~nx<a\}$$

Some things you need to check:

1) Why is $$\{a\in \mathbb{N}~\vert~nx<a\}$$ nonempty and bounded below? (we need that in order for the infimum to exist.
2) Why is m natural? (or equivalently, why is the inf a minimum)
3) Why is $m-1\leq nx<m$

 

[gwsinger]

Micromass and AlphaZero thank you for your replies. The ability to show the existence of elements within certain boundaries seems to be a pretty fundamental concept in Principles of Mathematical Analysis. I tried to answer both of your questions below to help my understanding.

Micromass:

1) "Why is {a∈N | nx<a} nonempty and bounded below? (we need that in order for the infimum to exist."

Let M = {a ∈ N | nx < a}. M is non-empty because we know by the Archimedian property that for any real number there exists an integer greater than it. Since nx is a real number, we know that there is going to be an integer greater than it. Hence M is non-empty.

Furthermore, we know that M is bounded below since nx is less than any of its members (by definition).


2) "Why is m natural? (or equivalently, why is the inf a minimum)"


Intuitively, M is comprised of naturals that start at some element and expand infinitely rightward on the number line. Visually, we could see that there would be a least member of M that is less than every other member of the set (with no other lower bound greater than it) and hence the infimum. Still, how would I state this formally? I've thought for a while and can't think of how.

3) "Why is m−1 ≤ nx < m?"

We know that nx < m since m ∈ M which is the (non-empty) set of all natural numbers which are greater than nx. To show that m -1 ≤ nx we will suppose the opposite and derive a contradiction. Negating m - 1 ≤ nx is to assert that m -1 > nx (by trichotomy). First note that if m is natural (which as I stated above I haven't really formally shown to be true), m - 1 is natural provided m is greater than or equal to 1. This is true since nx is positive (two positives multiplied returns a positive by ordered field axiom #2), and m is an integer greater than nx. Specifically, consider that 0 < nx < 1 or 1 < nx. If the former than m = 1, and if the latter than m > 1. Hence m - 1 is definitely a natural. Since m - 1 is a natural greater than nx, m -1 ∈ M. Hence we have these facts

m ∈ M
m - 1 ∈ M
m -1 < m

But m is the infimum of M, meaning it is a lower bound of M. This would be an impossible if m -1 ∈ M and hence we have reached a contradiction. Thus m - 1 < nx. Hence m−1 ≤ nx < m.

AlephZero:

(1) "[Observe that] there are only a finite number of integers m between -m_2 and m_1":

First note that -m_2 and m_1 are integers (grabbed from the Archimedean property). We can say that

-m_2 < m_1 since -m_2 < nx < m_1

Hence it is true there is a finite number of integers between -m_2 and m_1 (at least 2 inclusive and at most an infinite amount).

(2) "If none of them satisfy m - 1 ≤ nx < m, you have contradicted the fact that -m_2 < nx < m_1".

This I'm unclear on. Can you elaborate this point further?



Overall, I'm really confused by Rudin chose to find the m between nx and ny such that

-m_2 ≤ m ≤ m_1

This is because the -m_2 lower bound seems completely useless since he could have just said choose the m such that

nx < m ≤ m_1

Or he could have used Micromass' method of finding the infimum of {a ∈ N: a > nx}. What's the motivation of the -m_2 at all from Rudin's perspective?

 

[Bacle2]

Sorry for taking so long to reply:

As others said, using the LUB L of the (non-empty ) set of natural numbers larger than nx with the least integer k greater than 1/(x-y) will do the job:

You get : nx≤ L≤ ny

You can then use the glb of all naturals that are less than nx, which will be one less than L.