Is R Reflexive on N and Z

Question

In Vellemans book "How to prove it" (4th edition, section 4.3 exercise 24) the following question is posed: "Let R = {(m, n) ∈ N × N: |m − n| ≤ 1}, which is a relation on N. Note that R ⊆ Z × Z, so R is also a relation on Z." Note: N is the set of natural numbers {0, 1, 2, ...} and Z the set of integers {..., -3, -2, -1, 0, 1, 2, 3, ...}.

He then asks two questions:

1. Is R reflexive on N?
2. Is R reflexive on Z?

I have the following proof for the first question: Suppose we take some x from N, we want to show that xRx. This means x element of N x N, which is obviously true and also that |x − x| ≤ 1. Because x is positive x - x = 0 and so 0 ≤ 1. But also |0| ≤ 1, because |0| = 0, this shows that |x − x| ≤ 1.

However I'm not 100% sure on how to approach the second question. I assume I have to show that if x element of Z then xRx. I assume the same logic applies for |x − x| ≤ 1 because -x - (-x) = -x + x = 0. However x is not necessarily an element of N, which if I understand correctly is a condition for x to be part of R. As a counterexample (-1, -1) is given but it is not explained why this is a counterexample.

Do I understand correctly it's this first condition, namely (x, x) element of N x N that makes it so that R is not reflexive on Z or am I making a mistake somewhere else?

Answer 1

$(-1,-1)$ is a counterexample to statement 2 ($R$ reflexive in $\mathbb Z$) as

• $-1 \in \mathbb Z$ and so $(-1,-1) \in \mathbb Z \times \mathbb Z$
• $-1=-1$
• $(-1,-1) \not \in R$ because $-1 \not \in \mathbb N$ and $(-1,-1) \not \in \mathbb N \times \mathbb N$

so the answer to 2 is "no, $R$ is not reflexive in $\mathbb Z$".