8
$\begingroup$

Note: In the post by number I mean real number. What is a definition of the relations "bigger" and "less" ? The definition I saw in my book was "$a>b$ if $a-b$ is a positive number" and "$a<b$ if $a-b$ is a negative number". But the problem with this definition is number $a$ is positive if $a>0$ and similarly with negative number and this definition seems to be kind of circular. So how can we define " less" and "bigger" ?

$\endgroup$
  • $\begingroup$ What the set of numbers you are working with? $\endgroup$ – Ennar Mar 9 '18 at 18:42
  • $\begingroup$ @Ennar Real numbers, I will note that in the post. $\endgroup$ – Юрій Ярош Mar 9 '18 at 18:43
  • $\begingroup$ Are you sure that this was meant as definition? How are the reals constructed? $\endgroup$ – Ennar Mar 9 '18 at 18:44
  • 1
    $\begingroup$ There are various ways around it. The first way is to avoid defining $>$ and $<$ in such a way that it makes any reference to positive or negative numbers (for example using dedekind cuts). Another way is to avoid defining positive and negative with any reference to $>$ or $<$. The set of positives, $P$, can be defined as the set satisfying the properties: i) $x,y\in P\implies x+y\in P, x\cdot y\in P$, ii) $x\in P, y\notin P\implies x\cdot y\notin P$, iii) $x,y\notin P$ and $x\neq 0,y\neq 0\implies x\cdot y\in P$, and iv) $1\in P,0,-1\notin P$. $\endgroup$ – JMoravitz Mar 9 '18 at 18:56
  • 1
    $\begingroup$ @Acccumulation: I don't believe JMoravitz claimed to be able to do that. They merely said they could define Dedekind cuts without using positive and negative numbers, which is much easier since Dedekind doesn't care about zero in particular. $\endgroup$ – Kevin Mar 10 '18 at 0:25
16
$\begingroup$

The axioms for the real numbers provide one way to settle this issue. The axioms are broken into two portions: the field axioms concerning algebraic properties of the binary operations $+$ and $\times$; and the order axioms which require the existence of a total order $x<y$ satisfying three axioms:

  1. if $x < y$ then $x+z < y+z$ for all $z$;
  2. if $0 < x$ and $0 < y$ then $0 < x \times y$.
  3. The completeness axiom.

You can then define the positive numbers to be all numbers $x$ such that $0<x$.

On the other hand you don't have to be satisfied with the axiomatic approach to the real numbers themselves, because the real numbers can be constructed from the rational numbers, which can be constructed from the integers, which can be constructed from the natural numbers, which satisfy Peano's Axioms. This is a long drawn out story, but the salient point for your question is that in Peano's Axioms for the natural numbers there is no need to even mention the concept of positive or negative numbers, but on the other hand the order relation on natural numbers can be defined like this (addition and multiplication having already been defined):

  • Given natural numbers $m,n$ define $m < n$ if there exists a nonzero natural number $k$ such that $m+k=n$.

Once that task is accomplished, one embeds the natural numbers into the integers and defines order similarly:

  • Given integers $m,n$, define $m<n$ if there exists a nonzero natural number $k$ such that $m+k=n$.

Then one embeds the integers into the rationals and defines order on the rationals:

  • Given rationals $r=\frac{a}{b}$, $s=\frac{c}{d}$ where $a,c$ are integers and $b,d$ are nonzero natural numbers, define $r < s$ if $ad<bc$.

Finally, one embeds the rationals into the reals and defines order on the reals:

  • Given reals $x \ne y$, define $x < y$ if there exist Cauchy sequences of rational numbers $(r_n)$, $(s_n)$ converging to $x,y$ respectively, such that $r_n < s_n$ for all $n$.
$\endgroup$
11
$\begingroup$

It all depends on how you define the real numbers. The important thing is that you can define "positive" without any reference to $<$.

In the Dedekind cuts approach, where a real is a pair $(A,B)$ of subsets of $\mathbb{Q}$ such that $A\sqcup B = \mathbb{Q}$ and $\forall a\in A, b \in B, a<b$, then a real $(A,B)$ is said to be positive if and only if there exists $a\in A, a>0$ (note that this $>$ is the ordering on $\mathbb{Q}$, so it's not circular).

In the Cauchy sequences approach, where a real is an equivalence class of Cauchy sequences in $\mathbb{Q}$ (the equivalence relation is induced by the ideal of sequences converging to $0$), then a real $[(q_n)]$ is said to be positive if there is $\epsilon >0$ (again, this is in $\mathbb{Q}$ so no circularity here) and $n$ such that $\forall k \geq n, q_k \geq \epsilon$ (again, in $\mathbb{Q}$). One easily checks that this doesn't depend on the choice of representative of the class $[(q_n)]$, and hence yields a sound definition.

There are of course other approaches to defining the real numbers, and they all come with different ways of defining positivity.

What matters is that the ordering always comes from the ordering of $\mathbb{Q}$, which has been defined before, so it's never circular. Now one might ask how the ordering on $\mathbb{Q}$ is defined. And once again, this depends on the definition of $\mathbb{Q}$.

For instance, defining $\mathbb{Q}$ as $\mathbb{Z}\times \mathbb{N}^*$ modulo the equivalence relation $(a,b) \sim (c,d) \iff ad= bc$, then you can say that the class of $(a,b)$ is positive if and only if $a>0$ (this is in $\mathbb{Z}$, so no circularity here). One again checks that this doesn't depend on the representative, and so this is indeed well defined. Once you have the definition of positivity you can define an ordering as you mentioned in the beginning of your post.

It remains to define an ordering on $\mathbb{Z}$, and similarly this can be done without circularity by using the ordering on $\mathbb{N}$ (of course it also depends on the definition of $\mathbb{Z}$ one is using).

The last step is defining an ordering on $\mathbb{N}$. Obviously this also depends on the definition of $\mathbb{N}$ , but it's most usually defined as the first infinite ordinal (existing because of the axiom of infinity) and so it comes with a "natural" ordering : $\in$. Indeed, in this definition, $0 := \emptyset, 1:= \{0\}$ and more generally, $n+1 := n\cup \{n\}$ and so $n=\{0,1,...,n-1\}$, so $<$ coincides with $\in$. Now $\in$ (membership) is a primitive notion in set theory and so we don't define it.

$\endgroup$
  • $\begingroup$ With the $\mathbb{N} = \omega$ approach, we do define equality via the axiom of extensionality. OTOH, the Peano axioms consider equality primitive and have to define $<$ in terms of successorship or addition instead. It's all a matter of perspective. $\endgroup$ – Kevin Mar 10 '18 at 0:20
  • $\begingroup$ @Kevin : well it depends on the person you're asking. Some take $\in$ and $=$ to be primitives of set theory, and some indeed define $=$ via extensionality. Similarly for Peano, some authors take $<$ to be a primitive; I don't know whether there's an overwhelming consensus on these matters $\endgroup$ – Max Mar 10 '18 at 11:12
6
$\begingroup$

$a\ge b$ if there is a number $c$ with $a=b+c^2$

$\endgroup$
3
$\begingroup$

The positive numbers are those numbers which are the square of some other non-zero real number. The negatives are the rest (except zero).

$\endgroup$
  • $\begingroup$ Then how you prove that you can't express -1 as a square of real number ? $\endgroup$ – Юрій Ярош Mar 9 '18 at 20:36
  • $\begingroup$ @ЮрійЯрош This depends on the axioms you are given and what is your definition of $-1$. I would say $-1$ can be defined as the additive inverse of the multiplicative neutral element (if these terms are defined). If your axioms are strong enough to prove that $1/x$ is never the same as $-x$, then I can prove it. But as I said, I don't know your axioms. $\endgroup$ – M. Winter Mar 9 '18 at 21:01
1
$\begingroup$

You can't.

An order an a set is any relationship with the symbol "$<$" so that 1) at most one the following its true: $a < b; b < a$ or $a=b$ for any $a,b$ in the set. 2) if $a < b$ and $b < c$ then $a < c$.

For example for $\mathbb N$ we can define $a < b$ if $a$ is a non-trivial multiple of $b$. 1) Given $a, b \in \mathbb N$ at most $a = b$ or $a = k*b; k > 1$ (as $a < b$) or $b = k*a; k > 1$ (so $b < a$). Only one, or none, of those can be true. and if $a < b$ so $a = k*b$ and $b< c$ so $b = m*c$ then $a=(mk)*c$ so $a < c$.

....

But, for the one we know and love, we define the rationals $\mathbb Q$ to be an "ordered field" which means:

For any $a,b\in \mathbb Q$ there are two operations: multiplication and addition so that $a+b \in \mathbb Q$ and $a*b\in \mathbb Q$.

A1) $a+b = b+a$.

A2) $(a+b) +c = a+(b+c)$.

A3) There is an element called $0$ so that $a+0 =a$ for all $a\in R$.

A4) For every $a$ there is an element called $-a$ so that $a + (-a) = (-a) + a = 0$.

M1) $ab = ba$

M2) $a(bc) = (ab)c$

M3) There is an element called $1$ so that $a*1 = 1*a = a$ for all $a$.

M4) If $a \ne 0$ then there is an element called $\frac 1a$ so that $\frac 1a*a = a*\frac 1a = 1$.

D) $a(b+c) = (ab) + (ac)$

O1)There is a "total order" so that for $a,b$ exactly one of the following is true: $a < b; b < a$ or $a=b$

O2) If $a < b; b< c$ then $a< c$.

O3) If $a<b$ then for any $c$, $a + c < b+c$.

O4) If $c > 0$ and $a < b$ then $ac < bc$.

With those axioms we can prove all sorts of stuff, the most pertainent being:

Claim: $1 > 0$.

Pf: $1 \ne 0$ so either $1 > 0$ or $1 < 0$. If $1 < 0$ then $0 < 1 +(-1) < 0 + (-1) = -1$. Then $(-1)*(-1) > 0$.

Lemma: $0*a = 0

Pf: $0*a = (0+0)*a = 0*a + 0*a$. So $0 = 0*a + (-(0*a)) = 0*a + 0*a + (-(0*a)) = 0*a + 0 = 0*a$.

Lemma: $(-1)*(-1) = 1$.

Proof $(-1)*(-1) + (-1)=(-1)*(-1) + 1*(-1) = (-1 + 1)(-1) = 0*(-1) = 0$

$ (-1)*(-1) =(-1)*(-1) + 0 = (-1)*(-1) + (-1) + 1 = (-1)*(-1) + 1*(-1) + 1 = (-1)(-1 + 1) + 1 = -1*0 + 1 = 0+1 = 1$.

So we have $(-1)*(-1) = 1 > 1$ which is a contradiction.

So $1 > 0$.

From there you can determine which rational numbers are greater than $0$ and which aren't.

$\endgroup$
1
$\begingroup$

The definition of less-than via positive numbers is only circular if you in turn define the positive numbers in terms of less-than. But that is not the only way to define the positive numbers. Another way to define then is as follows:

We define a positive cone as a set $P\subset\mathbb R$ with the following properties (the exact axioms vary, but apart from the fact that sometimes $0$ is included in the positive cone, all those definitions are equivalent):

  • $P$ is closed under addition and multiplication (that is, if $a\in P$ and $b\in P$, then also $a+b\in P$ and $ab\in P$).

  • If $a\in P$, then $-a\notin P$.

  • If neither $a\in P$ nor $-a\in P$, then $a=0$.

A number $a$ is then called positive if $a\in P$.

Note that this definition does in no way refer to a relation “less than” or “greater than”. Therefore it is now in no way circular to define $a<b$ as “$b-a$ is positive”, that is, $a<b :\iff b-a\in P$.

For the real number, we want also have completeness, which can also be expressed using the positive cone. We call a set $S\subset\mathbb R$ upper-bounded if there exists a number $b\in\mathbb R$ so that $\{x-b|x\in S\}\cap P=\emptyset$. A number $b$ with this property is then called an upper bound. Now $b$ is called a least upper bound if there is no $c\in P$ such that $b-c$ is also an upper bound. Completeness now demands that every upper-bounded set has a least upper bound.

One can indeed easily show that with this definition, you get a strict total order with the additional properties that one demands when defining “$<$” directly.

  • Transitivity: If $a<b$ and $b<c$, then $a<c$.

    Proof: $a<b$ means $a-b\in P$, $b<c$ means $b-c\in P$. But then, $a-c = (a-b)+(b-c)\in P$ because $P$ is closed under addition.

  • Trichotomy: For any $a,b\in R$, exactly one of $a<b$, $a=b$ or $b<a$ is true.

    Proof: Either $a-b\in P$, then by definition $a<b$. Otherwise, either $b-a\in P$, then by definition $b<a$. Otherwise since neither $a-b\in P$ nor $b-a=-(a-b)\in P$, then by definition of $P$, $a-b=0$, that is, $a=b$.

  • If $a<b$ then $a+c

    Proof: $(a+c)-(b+c) = a+c-b-c = a-b \in P$

  • If $0<a$ and $0<b$ then $0<ab$

    Proof: $0<a$ means $a = a-0 \in P$, same for $0<b$. Because $P$ is closed under multiplication, $ab\in P$, therefore $0<ab$.

  • $b$ is an upper bound to $S$ according to the definition above iff it is an upper bound according to the usual order definition.

    Proof: Be $b$ an upper bound as defined above, that is, $\{x-b|x\in S\}\cap P=0$. Then we have: \begin{align} \{x-b|x\in S\}\cap P=0 &\iff \forall x\in S:x-b\notin P\\ &\iff\forall x\in S: x=b \lor b-x\in P\\ &\iff\forall x\in S: x=b \lor x<b \end{align} But that is exactly the definition of an upper bound for an ordered set.

  • $b$ is a least upper bound according to the definition above iff it is a least upper bound according to the usual order definition.

    Proof: Assume $b$ is an upper bound. Then it is a least upper bound according to the above condition iff there exists no $c\in P$ (that is, $0<c$) such that $b-c$ is also an upper bound. Note that the equation $x=b-c$ can always be solved uniquely both for $x$ given $c$ and for $c$ given $x$. Moreover $0<c\iff -c<0 \iff x=b-c<b$. Thus the condition is fulfilled iff there exists no $x<b$ that is also an upper bound. But that is exactly the condition for $b$ being a least upper bound in the usual order definition.

  • Completeness according to the definition above is equivalent to order-completeness

    This is evident from the fact that the definitions of upper bound and least upper bound agree, and the formulation in terms of those concepts is the same.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.