Checking reflexive, symmetric and transitive properties of $\neq$ on $\mathbb{N}$

Question

QS: Indicate if the relation on the given set are reflexive on a given set, which are symmetric, and which are transitive.

$\not = \text{on } \Bbb N$

So for this problem I am trying to comprehend why this question is not transitive.

$(i)$ Reflexive: No because $\forall x \in \Bbb N$ thus $x \not = x $.

$(ii)$ Symmetric : Yes because if $ x \not = y \; \Rightarrow y\not = x.$ Thus $xRy \rightarrow yRx.$

$(iii)$ Transitive: $x \not = y , y\not=z, x\not = z.$

At least this is what I think, but this is wrong and I want to understand my mistake. One last question are $(x,y) \in \Bbb N$?

Answer 1

A simple counter example will suffice. Take $x=z=1$, $y=2$. Then $x \neq y$ and $y\neq z$, but $x = z$.

As to your last question, we use $x,y \in \mathbb{N}$ to mean that both $x$ and $y$ are natural numbers. This differs from $(x, y) \in \mathbb{N^2}$ where $(x,y)$ is an ordered pair.

Answer 2

The two things you wanted to know:

1. THE MISTAKE:

$x \not= y \land y \not= z \not \implies x \not =z$

Take any counter-example, as in Parth Kohli's comment.

2. $(x,y)$ is an ordered pair, so essentially $(x,y) \in \mathbb{N} \times \mathbb{N}$.