S = R where a ∼ b if and only if |a| = |b|. Symmetric and transitive?

Question

I'm learning math just for fun and found these problems on google, however I really don't understand this one.

A set S and a relation ∼ on S is given. For each example, check if ∼ is (i) reflexive, (ii) symmetric, and/or (iii) transitive. If ∼ satisfies the property that you are checking, then prove it. If ∼ does not satisfy the property that you are checking, then give an example to show it.

Here R equals the real numbers.

  S = R where a ∼ b if and only if |a| = |b|.
  Solution:

  (i) Yes, ∼ is reflexive. Proof: Let a ∈ R. Then |a| = |a|. So a ∼ a.

  (ii) Yes, ∼ is symmetric. Proof: Let a, b ∈ R and suppose that a ∼ b. Then |a| = |b|. So |b| = |a|. Thus b ∼ a.

  (iii) Yes, ∼ is transitive. Proof: Let a, b, c ∈ R and suppose that a ∼ b and b ∼ c.Then |a| = |b| and |b| = |c|. So |a| = |c|. Thus a ∼ c.

(i) makes sense but (ii) doesn't why are we supposing that "a ∼ b. Then |a| = |b|. So |b| = |a|. Thus b ∼ a"? couldn't b = 3 and a = 4 thus a ∼ b, which means |a| = |b|, is obviously not true? (iii) follows the same trend as I can easily choose a,b, and c values that don't follow this relation at all.

My only assumption for why this might be is that we are just letting a ∼ b mean |a| = |b| and not allowing any values that might not allow that to happen, but then that feels like it just defeats the purpose. And in an earlier question it doesn't do that.

 (a) S = R where a ∼ b if and only if a ≤ b.

 Solution:

 (i) Yes, ∼ is reflexive. Proof: Let a ∈ R. Then a ≤ a. So a ∼ a.

 (ii) No, ∼ is not symmetric. Counterexample: 3 ≤ 4, but 4 6 ≤ 3.That is, 3 ∼ 4 but 4 6 ∼ 3.

 (iii) Yes, ∼ is transitive. Proof: Let a, b, c ∈ R and suppose that a ∼ b and b ∼ c. Then a ≤ b and b ≤ c. So a ≤ c. Thus a ∼ c.

edit: funnily enough, after thinking about it for a few more seconds, I think I get why. However, I'm not so confident in this so as to delete the post.

Answer 1

The symmetry property is a universal implication:

For all $a$ and $b$, if $a\sim b$, then $b\sim a$.

In principle, we want to prove a statement of the form $P\to Q$ ($P$ implies $Q$; or equivalently, "if $P$ then $Q$"). We almost never prove such statements directly, because we can invoke the Deduction (Meta)Theorem. Paraphrasing it, it says:

There is a proof that $P\to Q$ if and only if from the assumption that $P$ is true, there is a proof of $Q$.

So let us say that you want to prove that

For all $a$ and $b$, if $a\sim b$ then $b\sim a$.

To prove this, we are allowed to assume that $a\sim b$ is true. If, from that assumption alone we are able to show that $b\sim a$ follows, then this argument also establishes that the implication is provable.

So here, we can start by saying "Well, assume that $a\sim b$ is true." We then continue: "This means that $|a|=|b|$, because that's what $a\sim b$ means. And this is the same as $|b|=|a|$; and this yields that $b\sim a$. Therefore, from the assumption that $a\sim b$, we may conclude that $b\sim a$."

Then, because our argument doesn't use anything particular about $a$ and $b$ other than that they are real numbers and that $a\sim b$ is assumed to hold, we can then do a Universal Generalization to say that this holds "for all $a$ and all $b$".

Now, by contrast, how would we prove that the symmetric property does not hold for some relation? We want to prove the negation of "For all $a$ and $b$, if $a\sim b$ then $b\sim a$.

The negation of "For all $a$, $P$" is "There exists an $a$ such that not($P$)."

So we would need to show that:

There exists an $a$ and a $b$ for which "if $a\sim b$ then $b\sim a$" is false.

The implication is false exactly when the antecedent, $a\sim b$ is true, and the consequent, $b\sim a$ is false.

So to prove that a relation is not symmetric, what we need to do is:

Exhibit specific (in this case numbers) $a$ and $b$ for which $a\sim b$ is true, but $b\sim a$ is false.

That's why in the second example you quote you are given specific numbers to verify that $a\sim b$ holds for them, but $b\sim a$ does not. That shows that the relation is not symmetric.

And example doesn't prove that something always happens, but an example where it fails does prove that something does not always happen.