8

I am referring to two Peano axioms:

If $x = y$ then $y = x$ (symmetry)

If $x = y$ and $y = z$ then $x = z$ (transitivity)

What I don't understand is why we need the transitive axiom. Isn't it already implied by the symmetry axiom? Is it more of a convenience? Can we prove that transitivity is a required axiom? What happens if we were to remove it?

Or should I think of this more like "symmetry just says we can physically flip the order of the equalities, and transitivity lets us swap things around as long as they're all equal to each other"?

Asaf Karagila
  • 393,674
user537069
  • 699

5 Answers5

51

Consider the following relation: $x$ opposes $y$ if and only if $x = -y$.

Can you see that the "opposes" relation is symmetrical but not transitive?

Pedro M.
  • 3,061
  • 18
  • 19
  • No, I don't see this. What do you mean? If $x = -y$ and $-y = z$ then $x = z$ seems correct to me – user537069 Mar 04 '18 at 23:12
  • 16
    @user537069 But $x$ does not oppose $z$. We are considering the relation oppose here. – bames Mar 04 '18 at 23:14
  • So $x$ opposes $y$ and $y$ opposes $z$, but $x$ does not oppose $z$, so transitivity fails here and we can only speak in terms of symmetry: $x$ opposes $y$ and $y$ opposes $x$. – user537069 Mar 04 '18 at 23:17
  • 18
    @user537069 And most importantly, symmetry does not imply transitivity. Therefore you need both axioms for a meaningful definition of equality. – Pedro M. Mar 04 '18 at 23:18
  • 1
    Does transitivity imply symmetry? – user537069 Mar 04 '18 at 23:19
  • 18
    @user537069 Not at all. Consider the "less than" relation. – Pedro M. Mar 04 '18 at 23:19
  • 8
    You got an answer to that question from @bames. – Lee Mosher Mar 04 '18 at 23:19
  • Is there a way to prove the necessity of these axioms logically without the use of things like "opposing" and "less-than/greater-than" counterexamples? Or is it just that these are the axioms and they either apply/don't apply to various kinds of comparisons? – user537069 Mar 04 '18 at 23:20
  • 19
    Why wouldn't you want a counterexample? There are many more, but one counterexample is already sufficient for a proof. – Chris Mar 04 '18 at 23:22
  • @ChrisRandall I am fine with counterexamples, but I am curious if there is a way I could have shown the necessity of these things from logic alone rather than reliance on specific cases. – user537069 Mar 04 '18 at 23:23
  • 7
    @user537069 A proof would be necessary if the transitivity axiom were to imply the symmetry axiom. The counterexamples are actually a proof that no such proof can be produced. – Pedro M. Mar 04 '18 at 23:29
  • Interesting... thanks everyone! This has been very enlightening – user537069 Mar 04 '18 at 23:30
  • 7
    @user537069 Keep in mind that a counter example is a logical proof: $\because A \to B,\quad \because \neg B,\quad \therefore \neg A$. – Vaelus Mar 05 '18 at 01:05
  • In this specific example, $A$ is "Symmetry implies transitivity", and $B$ is "All symmetric relations are transitive" – Vaelus Mar 05 '18 at 01:10
22

Some relations are symmetric without being transitive.

For example if John and Jill are friends and Jill and Jeff are friends, it does not imply that John and Jeff are friends.

So friendship is a relation which is symmetric but not transitive.

Mathematically speaking, in geometry perpendicularity is symmetric but not transitive.

We like the relation $x=y$ be both symmetric and transitive.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005
Mohammad Riazi-Kermani
  • 68,728
  • Are there any examples of relationships that are neither transitive nor symmetric? – user537069 Mar 04 '18 at 23:27
  • 16
    X is a parent of Y? – Air Conditioner Mar 04 '18 at 23:29
  • 9
    Parenthood is neither symmetric nor transitive. If x is the father of y, then y is not the father of x. And if x is the father of y and y is the father of z, then x is not the father of z. – Mohammad Riazi-Kermani Mar 04 '18 at 23:30
  • 1
    "Ancestor" is the transitive relation, not "Parent". If X is an ancestor of Y and Y is an ancestor of Z, then X is an ancestor of Z. (At least grandparent, obviously) – MSalters Mar 05 '18 at 11:12
  • Ancestor is an example of a relationship that's transitive but not symmetric. Sibling would be an example of a relationship that's both symmetric and transitive. So we have examples of all the combinations in families. – Barmar Mar 05 '18 at 16:11
  • 1
    @Barmar I have a brother named James. So, James is my sibling, and I am also James's sibling. Does that mean that I am my own sibling? – mathmandan Mar 05 '18 at 16:40
  • 1
    @mathmandan Good point. https://www.youtube.com/watch?v=eYlJH81dSiw – Barmar Mar 05 '18 at 16:53
  • @user537069 in the finite case, once you get beyond sets with just a few elements, the vast majority of relations are neither transitive nor symmetric – John Coleman Mar 05 '18 at 19:49
7

It's easy to write down an example of a symmetric relation which fails to satisfy transitivity: on the set $\{a,b,c\}$ we say that $a \sim b$ and $b \sim a$ and $b \sim c$ and $c \sim b$ (I'm using a different symbol $\sim$ so as not to confuse the issue). So no, the symmetry axiom does not imply the transitivity axiom.

Perhaps you might want to think of the transitive law as it is stated in some translations of Euclid: thing which are equal to the same thing are equal to each other.

Lee Mosher
  • 120,280
1

Another example of a relationship which is symmetric but not transitive: computer floating point arithmetic is approximate, so for example (1./3.)*3. != 1.0 (some rounding occurs when the division is done, and this destroys the equality). It is thus important when comparing floating point results for equality to compare with an appropriate margin of error. Deciding what this margin should be is a hard problem, but let us use a margin of 0.1:

Then 0.61 is near 0.7 and 0.7 is near 0.61 (symmetric), but whereas 0.7 is near 0.79, 0.61 is not near 0.79 (not transitive).

Martin Bonner supports Monica
  • 111
1

Other answers cover very well why we need a transitivity axiom in the case of axiomatised Peano arithmetic.

But it's worth considering: why does this not feel necessary for the equality relation? The following idea may underly your intuition.

In many logics, we have an additional rule of 'substitution' or 'substitutivity'. Sometimes it's considered an axiom, sometimes a rule part of the logic, and sometimes an axiom schema (for example sometimes in First Order Logic, in which the second order axiom given below can not be written). Read about it here on Wikipedia.

It looks something like this:

For all variables $x$ and $y$ and for all formulae $P$, if we know $x = y$ and we know $P(x)$ then we know $P(y)$.

One way to write it as a second order statement is this:

$\forall P. \forall x. \forall y. x = y \implies P(x) \implies P(y)$.

Interestingly, once we have this axiom, logical rule, or axiom schema in place, all that is required to derive the other properties of equality is Reflexivity.

The derivation of Symmetry relies on the observation that an application of the substitution rule with $P$ as $w \mapsto w = x$ gives us

$\forall x. \forall y. x = y \implies x = x \implies y = x$.

The derivation of Transitivity relies on observation that an application of the substitution rule with $P$ as $w \mapsto w = z$ gives us

$\forall x. \forall y. x = y \implies x = z \implies y = z$.

Oly
  • 303