Set Relations Question

Question

I understand these laws when applied to certain situations but can't seem to understand how to apply it to these problems. I know that if Jon is Mike's cousin, then Mike is Jon's cousin and that is a Symmetric relation. I also know that if Jon is taller than Mike and Mike is taller than Maddie, then Jon is taller than Maddie and that is a Transitive relation. How do I make the step to these type problems using the same thought system?

For each the following relations on the set of integers list all that apply
(Reflexive, Symmetric, Antisymmetric, or Transitive):

R1 = {(a, b) | a * b <1}

Answer 1

Quick hints: for $\,a,b\in\Bbb Z\;$ :

$$ab<1\iff ba<1\;,\;\;\\0^2<1\;,\;\;\text{but if}\;\; a\neq0\;\;\text{then}\;\;\;a^2\ge 1\\2\cdot(-1)<1\;\;\wedge\;\;(-1)\cdot 1<1\;,\;\;\text{but}\;\;2\cdot 1>1$$

Answer 2

You have the definition that $\langle a,b\rangle\in R_1$, or $a\mathbin{R_1}b$ for short, if and only if $ab<1$. To decide whether $R_1$ is symmetric, you should ask yourself:

If $a\mathbin{R_1}b$, is it always the case that $b\mathbin{R_1}a$?

In the case of cousinship, you’re asking:

If Jon is cousin to Mike, is it always the case that Mike is cousin to Jon?

For cousinship the answer is clearly yes. For $R_1$ you want to use the definition of $R_1$ to translate $a\mathbin{R_1}b$ into something more fundamental. By definition $a\mathbin{R_1}b$ means that $ab<1$, and $b\mathbin{R_1}a$ means that $ba<1$. Thus, the first question becomes:

If $ab<1$, is it always the case that $ba<1$?

And in this form it’s clear that the answer is yes, since we know that $ab=ba$, so $R_1$ is symmetric.

To decide whether $R_1$ is transitive, you should ask yourself:

If $a\mathbin{R_1}b$ and $b\mathbin{R_1}c$, is it always the case that $a\mathbin{R_1}c$?

In the case of the relation taller than, you’re asking:

If Jon is taller than Mike, and Mike is taller than Maddie, is it always the case that Jon is taller than Maddie?

Of course the answer is yes, and we conclude that the relation is transitive. To answer the question for $R_1$, we should again use the definition to translate it into something more basic. The statements $a\mathbin{R_1}b$, $b\mathbin{R_1}c$, and $a\mathbin{R_1}c$ mean $ab<1$, $bc<1$, and $ac<1$, respectively, so the question can be rewritten as follows:

If $ab<1$ and $bc<1$, is it always the case that $ac<1$?

When is a product of two integers less than $1$? This happens when at least one of integer is $0$, and when one is positive and the other negative, and that’s it. So if $a$ and $c$ are positive, and $b$ is negative, the products $ab$ and $bc$ will both be negative, but the product $ac$ will be positive. To make this concrete, let $a=c=1$ and $b=-1$: then $ab=-1<1$, and $bc=-1<1$, but $ac=1\not<1$. This counterexample shows that $R_1$ is not transitive.

The question for reflexivity:

Is it always the case that $a\mathbin{R_1}a$?

It translates to:

Is it always the case that $a^2<1$?

Of course the answer is no: it’s almost never the case that $a^2<1$ when $a$ is an integer — it’s the case only when $a=0$. $R_1$ is not reflexive.

For antisymmetry the question is:

If $a\mathbin{R_1}b$ and $b\mathbin{R_1}a$, can I always conclude that $a=b$?

And the translation:

If $ab<1$ and $ba<1$, can I always conclude that $a=b$?

Again it’s easy to find counterexamples showing that the answer is no and hence that $R_1$ is not antisymmetric: for example, take $a=1$ and $b=-1$.