0

Given a homogeneous binary relation $\mathcal{R}$ over a set $A$,and is defied as:

$$\mathcal{R}:=\left\{\left(a,b\right)\mid a\le b\right\}$$

Prove $\mathcal R$ is not symmetric.

I don't understand why $\mathcal R$ is not symmetric,indeed it's symmetric if :

$$\forall a,b \in A:a\le b \implies b\le a $$ Or in other words:

$$\forall a,b \in A$$ one of these are true:

$$a < b \implies b< a $$ $$a < b \implies b= a $$ $$a = b \implies b< a $$ $$a = b \implies b= a $$

As it's seen,the last one is true,and that's why I think the relation is symmetric.

2 Answers2

2

$$\forall a,b \in A:a\le b \implies b\le a $$ Or in other words: $$\forall a,b \in A$$ one of these are true: $$a < b \implies b< a $$ $$a < b \implies b= a $$ $$a = b \implies b< a $$ $$a = b \implies b= a $$

You are wrong.

Your second statement is not equivalent to the first. The first statemtent says that for every pair $a,b\in $A, the statement $a\leq b\implies b\leq a$ is true. The statement $a\leq b\implies b\leq a$ is not the same as

$$(a < b \implies b< a) \lor (a < b \implies b= a)\lor (a = b \implies b< a) \lor(a = b \implies b= a) $$

and you can quite clearly show this by taking $a=1$, $b=2$.

More generally, your argument is based on assuming that, for any collection of logical statements $A, B, C$, the statement $$(A\lor C)\implies(B\lor C)$$ is equivalent to the statement

$$(A\implies B)\lor(A\implies C)\lor(C\implies B)\lor (C\implies C)$$

but this is not the case.

(for clarity, in your argument, $A$ was the statement $a<b$, $B$ was the statement $b<a$, and $C$ was the statement $a=b$)

In fact, if you wanted to write out the logical implication (not that I recommend it), then what you would need to consider is that $$(P\lor Q)\implies (R)$$ is equivalent to:

$$\begin{align}(P\lor Q)\implies R&\sim \neg (P\lor Q)\lor R\\&\sim (\neg P \land \neg Q)\lor R \equiv (\neg P\lor R)\land(\neg q \lor R)\\&\sim (P\implies R)\color{red}{\land}(Q\implies R)\end{align}$$

Note the highlighted $\land$ appearing in the statement which is nowhere in your argument.

5xum
  • 123,496
  • 6
  • 128
  • 204
  • @45465 Exactly. Your error was a purely logical one. If you consider my answer useful, also consider accepting it. – 5xum Aug 03 '20 at 05:53
  • @45465 Oh, right! That was juist a typo on my part. – 5xum Aug 03 '20 at 06:01
  • @45465 It's a little sloppy to say the statement is always false. It is actually true if $a=b$ (since the first statement of the conjunction becomes $F\implies T$, which is true, and the second becomes $T\implies T$, which is also true. But yeah, the idea is correct. The statement is not true for every pair $a,b$, and in particular, it is not true if $a<b$ is true. And even more particular, it is not true for $a=0, b=1$ (and this is enough to prove the original relation is not symmetric). – 5xum Aug 03 '20 at 06:10
  • @45465 But then you need to also prove why the left side is not true in general. Remember, a statement $\forall x: P(x)$ can only ever be disproven by showing that the statement $\exists x: \neg P(x)$ is true. – 5xum Aug 03 '20 at 06:23
  • @45465 I know. But you have to do that, otherwise the proof is incomplete. It's best to take a particular $a$ and $b$ just to hammer home the point. – 5xum Aug 03 '20 at 06:25
0

Symmetry works like this: if $(a,b) \in \mathcal R$, then we always have $(b,a) \in \mathcal R$; is this always true in this case?

For a particular example, consider $(1,2)$, which is in $\mathcal R$ as $1 \le 2$. Then if $\mathcal R$ is symmetric, $(2,1) \in \mathcal R$, but does it meet the condition to be in $\mathcal R$?

PrincessEev
  • 43,815
  • Does $a < b$ ever imply $b = a$? Does $a=b$ ever imply $b < a$? This isn't something you should be handling by cases because all you do is confuse yourself: stick to the $\le$ relation and it'll be a lot easier. – PrincessEev Aug 03 '20 at 05:20