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

• gwsinger
In summary, Rudin 1.20(b) proves that between any two real numbers x and y, there exists a rational number p. The key strategy used is to use the archimedian property to find two integers m1 and m2 that are greater than nx and have a difference of at least 1. Then, by considering all integers between -m2 and m1, Rudin is able to find an integer m that satisfies m-1 ≤ nx < m and use this to prove the existence of a rational number between x and y. The proof relies on the fact that there are only a finite number of integers between -m2 and m1, and that if none of them satisfy the inequality, it would contradict the
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.

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

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

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.

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 $m-1\leq nx<m$

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?

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.

Last edited:

## 1. How can there be rationals between any two real numbers?

This concept is based on the density property of the rational numbers. This means that between any two rational numbers, there is always another rational number. Therefore, between any two real numbers, there will always be a rational number.

## 2. Is this statement true for all real numbers?

Yes, this statement is true for all real numbers. It is a fundamental property of the real number system.

## 3. How is this statement related to the completeness property of the real numbers?

The completeness property of the real numbers states that every non-empty set of real numbers that is bounded above has a least upper bound. In other words, there are no "gaps" in the real number line. This property is closely related to the statement that there are rationals between any two reals, as it ensures that there will always be a rational number between any two real numbers.

## 4. Can you provide an example to illustrate this concept?

For example, let's say we have the real numbers 1.5 and 2.5. Between these two numbers, we can find the rational number 2, which is exactly in the middle. We can also find other rational numbers such as 1.75, 2.25, and so on. This shows that there are infinitely many rationals between any two reals.

## 5. Why is this statement important in mathematics?

This statement is important because it helps us understand the relationship between rational and real numbers. It also highlights the difference between the two number systems and how they behave. This concept is also fundamental in calculus, where we use it to prove the existence of limits and continuity.

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