Proof Transitivity of given Relation which is not reflective or symmetric

Question

I have to proove that following relation is transitive:

$\sim \ :=\ \{ (n,0) \ |\ n \in \mathbb{N}\setminus \{0\} \} \subseteq \mathbb{N} \times \mathbb{N}$

For me it is not transitive, because $n\sim0$, but $0$ is not in relation to some number $n$. Can anybody help me to understand this? Am i right with that $(n,0)$ is a well-ordered pair?

I have to show for $x,y,z$ that if $x\sim y \wedge y\sim z \Rightarrow x\sim z$.

score 1 · Answer 1 · edited Nov 08 '21 at 20:40

1

Okay i know the answer now!

$(n,0)$ is ordered.

The proof for the relation being transitive is following: $x$ is in relation to y (e.g. $x=3$ and $y=0$) but $y$ is not in relation to $z$ (e.g. $(0,z)$ because the first component has to be unequal zero per definition.

So because the left side of the material implication x~y $\wedge$ y~z $\Rightarrow$ x~z isn't fullfilled, per definition of the implication (if false then true), the given relation has to be transitive!

I hope it will help someone else :-)

edited Nov 08 '21 at 20:40

FreeMind

2,539
3
20
41

answered Nov 08 '21 at 17:33

swag

21

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Harish Chandra Rajpoot Nov 08 '21 at 17:35
i'm sorry, it was my first problem here and i answered it i think. Should i better delete the problem? – swag Nov 08 '21 at 18:52
1

Hey swag, you can ignore the above comments. It is perfectly acceptable to answer your own question when you figure it out (I've done this before). Welcome to MSE! – Mike Earnest Nov 10 '21 at 20:57

Proof Transitivity of given Relation which is not reflective or symmetric

1 Answers1