Foundation of ordering of real numbers

Question

This might be a silly question, but what is the mathematical foundation for the ordering of the real numbers? How do we know that $1<2$ or $300<1000$... Are the real numbers simply defined as being ordered in this way by construction?

Thanks for any contribution.

Answer 1

If you accept the following:

• The standard ordering among real numbers agrees with the standard ordering among natural numbers, i.e. nonnegative integers. This takes some work to prove rigorously, but the sociological proof is simple: Any sane mathematician would refuse to accept as standard any ordering among real numbers that didn't agree with the standard ordering among natural numbers.
• For two natural numbers $x$ and $y$, we write $x\geq y$ if $x = y+z$ where $z$ is also a natural number. This is just a definition that constrains the above agreement.
• $2 = 1+1$, and $1$ is a natural number.
• $1000 = 300 + 700$, and $700$ is a natural number.

Then $2>1$ and $1000>300$.

One can go on to prove that every decimal (finite string of 0-9s) represents a natural number, and that the conventional right-aligned dictionary ordering among decimals agrees with the ordering among natural numbers. Then it suffices to recall that the digit "2" comes after "1", and to note that "1000" has more digits than "300". This is how most people would reason in practice.

Answer 2

The ordering on the reals comes in the end from that on the integers, via that on the rationals. The ordering on the integers is the unique transitive relation satisfying $0<1$ and $a>0\implies (b>c \iff b+a>c+a)$ and not permitting $a<b$ while $b<a$. Then the ordering on the rationals is determined, though I think you need to separately require totality: e.g. $0<1/2$ because $0\neq 1/2$ and $0<1$. To extend to the real numbers requires knowing a construction of the real numbers, for instance as Dedekind cuts: then the ordering on reals just becomes containment, or equivalently $r<s$ means that every rational in $r$ is less than some rational in $s$.

In short, the facts about the ordering follow from assuming $0<1$ and that the ordering behaves in some reasonable way with respect to arithmetic.

Answer 3

Usually, the field of real numbers is constructed by taking Cauchy sequences of rational numbers. The rational numbers have an ordering derived from the ordering on the integers: $p / q > r / s$ just in case $(p / q) - (r / s)$ has either a positive numerator and denominator, or a negative numerator and denominator. You can then get the ordering on the real numbers by saying that a Cauchy sequence is positive if it contains only finitely many terms $\le 0$. From this you can prove things like: every real number is either positive, negative or zero; the sum of two positive numbers is positive, etc. We say $x > y$ just in case $x - y$ is positive.

However, there's another way, of interest to the model theory of the real numbers. If you have the real numbers as a field, you can define that $x > y$ if and only if there exists a real number $z$ such that $y - x = z^2$. But I am not sure it is easy to prove properties of the $<$ order if you take this approach.

Answer 4

You ask that as a foundational question. The answer is simple. It depends on your foundational approach to the real numbers.

1. We can begin by constructing the natural numbers, then the integers, then the rationals, and then the real numbers by one mechanism or another. In this approach, the order on $\Bbb N$ is extended in a way which is compatible with the extension of the operations, and so the order comes from that.

2. We can begin by writing a few axioms, stating that we have an ordered field, and that it satisfies certain properties (which may or may not be expressible in first-order logic, but let's not be worried by this issue now). Then the real numbers are kinda just there. We have terms for $0$ and $1$, so we can write closed-terms for every natural number, and we can define every rational number and even a few irrational numbers.

   Then the question why is $1<2<3$ and so on is answered by proving this from the axioms that we wrote. For example, if we wrote that "Whenever $x<y$ then for all $z$, $x+z<y+z$" and "Whenever $x<y$ and $0<z$, then $x\cdot z<y\cdot z$", then we can prove that $0<1$ and therefore $1<2$ and $2<3$ and so on.

3. You can be extremely Platonist, and simply declare that this is true in the universe of mathematics. Of course, just declaring something to be true, in this day and age, will probably not be accepted by other members of the mathematical community. But you still see people do it (although usually not in something as rudimentary as $300<1000$).

Answer 5

To my knowledge, if we want to tell a tale about the construction of the real numbers, it would go like this:

Natural numbers are defined using set theory: $0 = \{ \}$, $1 = \{ \{ \} \}$, $2 = \{ \{ \}, \{ \{ \} \} \}$, and so on.

"$+$" is a binary operation defined on $\mathbb N$. Through it, we define the negative integers as the inverses with respect to $+$ of the positive integers.

Next, we construct the rational numbers.

Finally, the irrational numbers are defined as the limit of sequences of rational numbers.

Q: How do we know that, for instance, $1 < 2$?

A: Looking at $1$ and $2$ as sets, we can find an injection from $1$ to $2$, but not an injection from $2$ to $1$.