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This is a question from book "Discrete Mathematics and Its Applications".


Determine whether the relation R on the set of all integers is reflexive, symmetric, antisymmetric, and/or transitive, where(x, y) ∈ R if and only if

b) $xy \ge 1.$

The answer provided by the book is: $R$ is symmetric and transitive.

Why isn't $R$ reflexive?

I think $R$ is reflexive because $x$ and $y$ are integers, since $xy \ge 1$, they are positive integers or negative integers, that $xx \ge 1$ should be true, that $R$ should be reflexive.

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Note that we do not have $xy \geq 1$ if one or both of $x, y$ is zero, and zero is an integer. – Old John Jan 6 '13 at 16:58
If x or y is zero, that xy≥1 will be false, so (x,y)∉R. – Freewind Jan 6 '13 at 17:01
Yes, exactly that. – Old John Jan 6 '13 at 17:02
@Freewind: You forgot about $0$, easy to do, it is so small. The relation is not reflexive because $(0,0)\not\in R$. For all $x\ne 0$, $(x,x)\in R$. So reflexivity almost holds. – André Nicolas Jan 6 '13 at 17:04
I understand now! I misunderstood the definition of reflexive: A relation R on a set A is called reflexive if (a, a) ∈ R for every element a ∈ A.. There is every element a ∈ R, so we must consider 0. – Freewind Jan 6 '13 at 17:04
up vote 5 down vote accepted

Recall that $xRy$ if and only if $(x,y)\in R$, and $R$ is defined to be reflexive if and only if $(x,x) \in R$ for all $x$.

In this case, $0$ is an integer, and the statement $xy \geq 1$ is false if one or both of $x, y$ is zero, so your relation fails to be reflexive.

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A relation $\,R\,$ on a set $A$ is reflexive if and only if, for every element $\,a\in A,\,$ it is true that $\;(a, a) \in R\;$.

In this problem, suppose $x = 0$. Since $\;0 \in \mathbb{Z}$, but $xx = 0\cdot 0 \ngeq 1$, thus $(0, 0) \notin R$, and so the relation fails to be reflexive on the set of integers.

It only takes one element as a counterexample to prove the relation on the set non-reflexive.

If your relation were defined on the set of non-zero integers, then it would be reflexive.

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