Given the rules of basic arithmetic, and defining $<$ as a binary relation with:

  • For all $a\in\mathbb{R}$, either $a<0$, $a=0$, or $0<a$
  • If $0<a$ and $0<b$ then $0<a+b$ and $0<ab$
  • $a<b$ iff $a-b<0$

Prove that $a<b$ and $b<c$ implies $a<c$.

The difficulty seems to be that I don't have the 2nd law also for negative a and b.


  • Addition is commutative
  • Addition is associative
  • For all a, a+0=a
  • -a+a=0
and similar for multiplication

  • 1
    $\begingroup$ What 'rules of basic arithmetic' do you have? And saying 'the obvious' ones doesn't work, because I could also say that it is 'obvious' that $a < b$ and $b < c$ implies $a < c$. Once you start doing this kind of 'axiomatic' approach, proving things from 'first primitives'. In other words, we need to know what 'first primitives' you are given. In other words, what are the exact rules that you have for involving addition, subtraction, multiplication, etc.? Can you please add those to your post? $\endgroup$
    – Bram28
    Oct 23 '19 at 20:49
  • $\begingroup$ And what also needs to go into your Post: your own initial attempts, or at least intuitive/informal thoughts. $\endgroup$
    – Bram28
    Oct 23 '19 at 20:56
  • 1
    $\begingroup$ Can you assume that the 'either' is an exclusive disjunction? $\endgroup$
    – Bram28
    Oct 23 '19 at 20:58
  • $\begingroup$ There seem to be a lot of deleted answers that boil down to adding the positive numbers $b-a,\,c-b$ to verify $c-a>0$. $\endgroup$
    – J.G.
    Oct 23 '19 at 20:58
  • 1
    $\begingroup$ You dont have an axiom that $a< b$ and $b < a$ are mutually exclusive. I don't think you can prove this without it. If we define $a < 0$ iff $a \ne 0$ then all your axioms hold. But then $0\ne a; a\ne b; b\ne 0 \implies$ both $a< b$ and $b < a$ $\endgroup$
    – fleablood
    Oct 23 '19 at 21:45


Suppose we define $k < 0$ if $k\ne 0$ for all $k$.

This satisfies all axioms!

Axiom 1) either $0 < a$ (never happens) or $0 = a$ or $a< 0$ (always happens if $a\ne 0$.

Axiom 2) If $0 < a$ and $0< b$ then.... vacuously anything follows.

Asiom 3) $a < b\iff a-b < 0$.

Well, $a-b =0 \iff a=b$ so $a< b\iff a\ne b \iff b< a$.

There's nothing that says if $a < b$ and $b < a$ are mutually exclusive.

And so we could have $a < b$ and $b < a$ but that doesn't implies $a < a$.


So we can not prove $<$ is transitive.

But if we make $a<b$ and $b<a$ mutually exclusive then we are good.

If $a < b$ then we can't have $b<a$ and so we do have $a-b <0$ and so we can't have $a-b=0$ and so we can't have either $b-a = 0$ or $b-a <0$ so we must have $0 < b-a$.

So $a<b \iff a-b < 0 \iff 0 < b-a$ and $a=b \iff a-b = b-a =0$ and $b< a \iff b-a < 0 \iff 0 < a-b$ and $a<b, a=b, b< a$ are mutually exclusive and exhaustive.

And then transitivity follows:

$b-a > 0$ and $c-b > 0\implies$

$(c-b) + (b-a) > 0$ but $(c-b)+(b-a) = c-a\implies$

$c-a > 0 \implies$

$a < c$.

  • $\begingroup$ Oh.... ooops. hold on.... I'll fix it. $\endgroup$
    – fleablood
    Oct 23 '19 at 20:54

Important note: you don't say it in the question, but I am assuming here that when you write "either $a< 0, a = 0, $ or $a>0$ you are intending that "either... or" to be mutually exclusive. That is, precisely one of the three properties holds. (Without that, as fleablood points out, the claim is false.)

Assuming you intend your "either... or" to be mutually exclusive, the definitions you are working with say that $$a<b \iff a-b<0$$

It's probably easier to solve your problem if you first prove an equivalent form of the above: $$a<b \iff 0 < b-a$$

So let's do that first. Since this is an "if and only if" proof, there are two directions to prove:

  1. Proving $a<b \implies 0 < b-a$.
    Suppose $a<b$. Our goal is to prove that $0 < b-a$. If not, there are two other possibilities: either $0 = b-a$, or $b-a < 0$. The first of these implies that $a=b$, which contradicts the assumption that $a<b$. On the other hand, if $b-a < 0$ then by definition this implies that $b<a$, which also contradicts our assumption. By trichotomy (your first bullet), the only option left is that $ 0 < b-a$.
  2. Proving $0 < b-a \implies a<b$. I'm going to omit this proof, as it's essentially identical to the previous paragraph, written backwards.

Okay, now that this is out of the way, the rest of the argument is simple. If $a < b $ then $ 0 < b-a$; similarly if $b< c$ then $0 < c - b$. Using your second bullet, we conclude that $0< (b-a) + (c-b) = c-a$; this implies (via my lemma above) that $a < c$. Done.

  • $\begingroup$ The $a< 0; a=0; 0< 0$ can be mutually exclusive (and exhaustive) but nothing is said about whether for any $b\ne 0$ whether any of $a<b; b<a;a=b$ need be true or exclusive. "then by definition this implies that $b<a$, which also contradicts our assumption" How is $b < a$ a contradiction of $a < b$, even with $0<a;a<0;a=0$ mutual exclusivity? $\endgroup$
    – fleablood
    Oct 23 '19 at 22:09
  • $\begingroup$ "Without that, as fleablood points out, the claim is false." Even with it the claim could be false. But I think if you have an axiom that there is at least one element $k>0$ I think we can prove $\mathbb kQ^+=\matbbh Q^+$ are all greater than $0$ and $\mathbb Q^-$ are all less.... I think. $\endgroup$
    – fleablood
    Oct 23 '19 at 22:27

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