# B Distance between real numbers

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1. Oct 11, 2016

### Frank Castle

Why is it that the distance between two real numbers $a$ and $b$ in an ordered interval of numbers, for example $a<x_{1}<\ldots <x_{n-1}<b$, is given by $$\lvert a-b\rvert$$ when there are in actual fact $$\lvert a-b\rvert +1$$ numbers within this range?!

Is it simply that, when measuring the distance between two real numbers we are counting the number of unit intervals that separate the two of them, and there will always be one less unit interval between the two numbers than the range of numbers between them?!

For example, suppose I have the ordered set of integers $(0,1,2,3,4,5)$, then the distance between 4 and 1 is of course $\lvert 4-1\rvert = \lvert 1-4\rvert = 3$, which is to say, there are 3 unit intervals between 1 and 4. Equivalently, one could arrive at this result by counting the number elements between 1 and 4, including the endpoint (4) but not the start point (1). However, if one includes both the start point and the endpoint then the number of elements between 1 and 4 is 4. Is the qualitative difference here that in the former case I am determining a relative quantity - the separation between 1 and 4, whereas in the latter case I am determining an absolute quantity- the number of elements ranging from 1 to 4?!

Apologies if this is a really stupid question, but it's something that I've been thinking about recently, and how I would reason the answer.

2. Oct 11, 2016

### TeethWhitener

Is the distance between 1 and 4 greater than, less than, or equal to the distance between 1.5 and 4.5? Why?

3. Oct 11, 2016

### Frank Castle

It is equal to, since there are still 3 unit intervals between 1.5 and 4.5. However, there are more than 4 numbers between 1.5 and 4.5...

4. Oct 11, 2016

### TeethWhitener

How many numbers are you counting between 1.5 and 4.5?

5. Oct 11, 2016

### Frank Castle

Well, if we are counting in halves, i.e. 1.5, 2.0, 2.5,... then there are 6 numbers.

Is the point that we can arbitrarily partition up each unit interval into as many numbers as we please and still end up with the same number of unit intervals between two numbers so the distance between two numbers is something that is intrinsic as opposed to the arbitrary partitioning of the interval between them?!

6. Oct 11, 2016

### TeethWhitener

Yes. Well, kind of. There are three answers to your question. The first is that we simply define the distance between $a$ and $b$ to be $|a-b|$in $\mathbb{R}$. The second, more intuitive, answer is that the number of real numbers between $a$ and $b$, where $a \neq b$, is uncountable, so the notion of counting the numbers between $a$ and $b$ to determine a distance doesn't make any sense in $\mathbb{R}$. The third, most intuitive, answer is this: say we live at mile marker 1 and the grocery store is at mile marker 4. If the Department of Transportation comes along and relabels our mile marker as 1.5 and the grocery store's as 4.5, the simple act of relabeling the mile markers shouldn't change the distance between our house and the grocery store, should it?

7. Oct 11, 2016

### Frank Castle

So, is the intuitive answer simply that the distance between two points along the real number line should not depend on how we separate up the distance between them - it is an intrinsic property. It is not well-defined to determine the distance between two real numbers in terms of counting the numbers between them, since the answer would depend on how we partition up the interval, e.g. in to integers, halves, quarters, etc. Is this why the distance between two numbers is defined as it is, since it only depends on the two numbers themselves and not on what is inbetween them?!

The reason I originally started pondering this was because I was considering an interval of numbers $(-10,10)$ which I partitioned up into intervals of $0.01$, i.e. such that $(-10,10)=(-10,-9.99,-9.98,\ldots ,0,0.01,0.02,\ldots ,9.98,9.99,10)$. Naively, I thought that there would be $2000$ numbers in this interval, but on further analysis I found that there were $2001$...

8. Oct 11, 2016

### Staff: Mentor

Between any two distinct real numbers a and b, there are an infinite number of numbers, so it makes no sense to try to count the real numbers between a and b. This is a consequence of the reals being dense. The same is true for the rational number -- between any two distinct rationals there are an infinite number of rational numbers between them.

9. Oct 11, 2016

### Frank Castle

Is this the reason for defining the distance between two numbers as $\lvert a-b\rvert$?

10. Oct 11, 2016

### Staff: Mentor

Yes - the length of the interval between the two points.

11. Oct 11, 2016

### Frank Castle

I guess what's messed me up is the intuitive picture that I picked up at school that the length of the interval between a number $a$ and $0$, i.e. $\lvert a-0\rvert=\lvert a\rvert$ is the number of units $a$ is from $0$, in this case $a$-units. But perhaps this is too much of a simplification?!

12. Oct 11, 2016

### TeethWhitener

Technical aside: is this true? The rationals have measure zero in the reals.

13. Oct 11, 2016

### Staff: Mentor

No it's not. There's a difference between the length of an interval in some units and the number of points between two points, if I'm understanding you correctly.
The distance between 1 and 3 is 2 "1 units" or 4 "1/2 units" and so on, which is very different from what you said in post 7: "counting the numbers between them". Possibly what you meant differs from what you actually said.

14. Oct 11, 2016

### Staff: Mentor

Yes, but they are still dense in the number line. Between any two rational numbers there are an infinite number of other rational numbers. In contrast, the integers are not dense in the number line.

15. Oct 11, 2016

### TeethWhitener

Right. But the rationals are countably infinite. I don't know that you can use the same notion of distance in the rationals that you can use in the reals.
EDIT: To be clear, it's absolutely true that the rationals are dense in the reals, but specifically, I'm hesitant to talk about the concept of "distance" over a set of measure zero using the same metric that you would use over a set of non-zero measure.

16. Oct 11, 2016

### Frank Castle

So is the point that the distance between two numbers is intrinsic, but its numerical value depends on the unit of distance that we choose as a basis (e.g. "1 units" or "1/2 units")?

Apologies, I didn't word this very well in post #7. What I meant is that I wanted to determine the number of elements in the (ordered) set $\lbrace -10,-9.99,-9.98,\ldots ,0,0.01,0.02,\ldots ,9.98,9.99,10\rbrace$. Naively, I just worked out the distance between 10 and -10, which was of course 20, and then, as my unit of distance (between consecutive points) is 0.01, this gives 20x0.01=2000 points. But it turns out that I didn't take into account the end point (-10), which gave 2001 points in total.

17. Oct 11, 2016

### Staff: Mentor

I don't know if the right word is "intrinsic". There are different metrics that can be used to measure distance. For example, in R2, there's the Euclidean norm, with $d(x, y) = \sqrt{(x_2 - x_1 )^2 + (y_2 - y_1 )^2}$ and there's also the so-called "taxicab" norm, with $d(x, y) = |x_2 - x_1| + |y_2 - y_1|$, just to name two of them. See https://en.wikipedia.org/wiki/Norm_(mathematics).
This is the classic "fence-post" error.

If you have fenceposts every 10 feet, starting at 20 ft up to 110 ft, there are $\frac{110 - 20}{10} + 1 = 9 + 1 = 10$ fenceposts.

18. Oct 11, 2016

### Frank Castle

I guess what I meant by this is that, given a particular metric, the distance between two real numbers $a$ and $b$ is independent of the units that we use. For example, in the case I gave, given the metric $d(a,b)=\lvert a-b\rvert$, the distance between $a=10$ and $b=-10$ is always $d(10,-10)=\lvert 10-(-10)\rvert =20$ units, but the value can change depending on what one chooses one's units to be. In this example I further divided up each unit into intervals of 0.01, in which case the distance between -10 and 10 is 2000 "0.01 units".

Is it correct to intuitively think of a real number $x$ as being a distance of $\lvert x\rvert$ units from $0$?

Is it the case then, that for $\mathbb{R}$ one usually chooses $d(x,y)=\lvert x-y\rvert$ as the metric (which I guess is simply the one dimensional case of the Euclidean metric)?!

What is the intuitive reason for why this error arises?

Last edited: Oct 11, 2016
19. Oct 11, 2016

### Staff: Mentor

Yes
Yes
(Re: fencepost error) The intuitive reason is forgetting to count the starting fencepost. The distance from 0 to 10 is (obviously) 10 units, but if you have fenceposts at 0, 1, 2, 3, ..., 9, 10 feet, there are 10 + 1 = 11 fenceposts.

20. Oct 11, 2016

### Frank Castle

Ah ok, so it is simply the case that one has forgotten to include start point (at which the first unit interval starts).

Is there any intuitive reasoning/ motivation for defining the distance between two real numbers as it is? Does it simply follow from geometrical analysis - thinking of the real numbers as lying along a geometric line (the real number line) and then using Pythagoras's theorem in one dimension?!

Also, does one take a unit interval (from 0 to 1), i.e. 1, as the basis for defining the distance between two real numbers? By this I mean, for example, the number 10 is a distance of 10 units from 0. It is always a distance of 10 units from 0 regardless of whether I split the interval between 0 and 10 up further into fractions of one unit.

Last edited: Oct 11, 2016