Question

Is this a linear order on N?
\(S=\{(m,n):m,n \in N, m\leq n \text { and } m \neq 5\} \cup \{(5,m):m \in N \} \)

Answer 1

\[S=\{(m,n):m,n∈N;m≤n \text{ and }m≠5\}∪\{(5,m):m∈N\}\]

Answer 2

?

Answer 3

yupppppp

Answer 4

just so you know what the trick is in latex:
you need to use the \text{...} command to type words clearly
you also need to put a forward slash before brackets if you want them to be visible \{ \}
so:

S=\{(m,n):m,n\in N;m\le n \text{ and }m\neq5\}\cup\{(5,m):m\in N\}

\[S=\{(m,n):m,n\in N;m\le n \text{ and }m\neq5\}\cup\{(5,m):m\in N\}\]that said I have no idea how to answer your question lol

Answer 5

hahahhaha thannkkkksss ill edit my question

Answer 6

okkk Thats better

Answer 7

yep :)

Answer 8

now who could answer this I wonder...
@nbouscal @myininaya @across @KingGeorge maybe...

Answer 9

I think its quite similar to the question I asked yesterday I bet I cld just flip around KG's answer

Answer 10

what is this class called?

Answer 11

Transition to higher mathematics

Answer 12

an appropriate name :)

Answer 13

Like i think the book sucks. I can barely follow what the book is saying. And like there are no examples in the book so like its getting a lil tough. I took a course similar to this but the questions were simpler and there were loads of examples in the book to follow and it was just easy to read the text tooo. idkkkk I am getting myself a tutor

Answer 14

@eliassaab help?

Answer 15

@mukushla help?

Answer 16

sorry ...any information on this topic...

Answer 17

In order for it to be a linear order on N we must prove antisymmetry, transitivity and totality

Answer 18

Anti-Symmetry:
Suppose (a,b) and (b,a), both are in S. Then, for anti-symmetry we need to show that a = b.

Answer 19

Transitivity:
Suppose that (a,b) and (b,c) are in S. Then to prove transitivity of S, we must show that (a,c) is also in S.

Answer 20

Totality:
Either (a,b) is in S, or (b,a) is in S, or both

Answer 21

This is not a linear order.

Take for example, \(m=4\), \(n=5\). Here, we see that \((4,5)\in S\) by the first condition, and \((5,4)\in S\) by the second condition. This is a counterexample to antisymmetry, so it is not a linear order.

Answer 22

Thanksssssssss. u r one brilliant guy :)