How to pick values for checking if relations are reflexive, symmetric and/or transitive?

Question

I am trying to follow a video (listed below if you want to watch the part I am questioning), where he looks at whether $\le$ is reflexive, symmetric and/or transitive.

When he is determining if it is reflexive he questions "is 1 $\le$ 1" which would satisfy the reflexive condition.

Then for determining if it is symmetric he uses the example "does 3 $\le$ 4 imply that 4 $\le$ 3", which of course it doesn't, so it's not symmetric.

And finallly for checking whether it is transitive he states that "1 $\le$ 2 and 2 $\le$ 3 implies that 1 $\le$ 3" which makes it transitive.

He ends it by stating that the above 3 examples should be true for arbritary values,

My question is, how are these arbitrary values chosen?

As shown above, he used 1 and 1 for checking if it is reflexive, but couldn't we have instead used 9 and 3 which would mean "9 $\le$ 3" therefore it is not reflexive. Or, for the transitive example, instead of using 1, 2 and 3, we instead use 9,8,7 which would give "9 $\le$ 8 and 8 $\le$ 7 implies 9 $\le$ 7" which isn't true.

Sorry if I am not clear, my English is not good. I just want to know why in the video he can only pick values that satisfy the equations, instead of the many values that do not satisfy the equations?

The video is https://youtu.be/FI6j5QZNVx0?&t=426 (important part is from current time-stamp to 8:59).

Answer 1

I just want to know why in the video he can only pick values that satisfy the equations, instead of the many values that do not satisfy the equations?

The short answer is that because   $\le$   meets the definition of reflexive and transitive over numbers, there are no counter examples that could show it isn't. So, any values will show that the definitions of these properties hold for   $\le$   over numbers.

Looking more closely at the implication ($antecedent \Rightarrow consequent$) in the definition of transitive property:

$ \forall \space a, b, c \in A : (a \space X \space b) \wedge (b \space X \space c) \Rightarrow (b \space X \space c) $

With $X$ as the   $\le$   relationship, and $a=9, b=8, c=7$   then we can see that the antecedent is false and the consequent is false, making the implication hold true according to truth table $(false \Rightarrow false) \equiv true$, so this is not a counter example of   $\le$   being transitive.

With $X$ as the   $\le$   relationship, and $a=7, b=6, c=8$   then we can see that the antecedent is false and the consequent is true, making the implication also hold true according to truth table $(false \Rightarrow true) \equiv true$, so this is also not a counter example of   $\le$   being transitive.

To find a counter example, we need to see that the implication as a whole is false, which can only be achieved by $ (true \Rightarrow false) \equiv false $, meaning we need the antecedent to be true at the same time the consequent to be false. And there aren't any such counter examples.

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For symmetric, we can find both examples and counter examples. Because at least one counter example exists, then   $\le$   is not symmetric. All we have to do to find a counter example is use two different values.

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but why is this not the case for the example we were discussing for transitivity: i.e. because 9≤89≤8 and 8≤78≤7 are false, the whole statement is true?

The implication part of the statement is true for that one case of specifically selected $a, b, c$, but because of the   $\forall$   we need to consider all possible cases at the same time, and further, all the possible cases have to be true, since that's what the universal quantifier is saying (for all). Thus, one or two positive example does not prove the larger statement, though one counter example disproves it.

That   $\le$   is transitive and reflexive over the numbers cannot be proven by examples, and so must be proved using other reasoning that involves knowledge of logical reasoning for numbers.

A system for doing that over integers is Peano Arithmetic, which expresses numbers and their properties as logical statements that can be logically reasoned over (instead of just calculated).

Answer 2

A relation $\sim$ on a set $A$ is defined as $\sim=(A,\mathscr{R})$ where $\mathscr{R} \subseteq A \times A$.

We say $\sim$ is reflexive if for all $a \in A$, we have $a \sim a$. So you can't wheck reflexivity by choosing $a,b \in A$ such that $a \ne b$.

We say $\sim$ is symmetric, for all $a,b \in A$, if $a \sim b$ then $b \sim a$.

And we say $\sim$ is transitive, for all $a,b,c \in A$, if $a \sim b$ and $b \sim c$ then $a \sim c$. So you can't check transitivity with an example like $a \nsim b$ and $b \nsim c$. It is contrary to the definition.

In order to be more simple, you can consider $\sim$ as a "true/false" statement. So in your example, is $9 \le 8$ true or false? Is $8 \le 7$ true or false? You need to find two true statements first in order to check transitivity because of its definition. And you can't also choose these statements arbitrarily like $1 \le 2$ and $3 \le 4$ by the definition as well.

Here, I generalize the relation but you can just put $\le$ instead of $\sim$, it won't really change anything since the definitions are general. In addition to that, we call $\le$ a partial ordering since it is reflexive, anti-symmetric and transitive.