Question

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

Answer 1

Can you read it? Its a lil unclear for some reason

Answer 2

I can read it. Give me a minute to think it over.

Answer 3

ohhhh ik y its missing the brackets { (m,5): m element N}

Answer 4

whtvr i give up on latex lol

Answer 5

I think it's a linear order, but I'm not entirely sure yet. Let's work through the necessary steps.

Antisymmmetry: Suppose \((m,n),(n,m)\in S\). If \(m=5\), then \(n=5\) by the second condition. If \(m\neq5\), then \(n\neq 5\) as well (do you see why?). Since neither are equal to 5, they must both fall into the first condition. Hence, \(m\leq n\leq m \implies m=n\) So it is antisymmetric.

Answer 6

i dont get if \( m \neq 5, then, n\neq 5 \)

Answer 7

Remember that we're starting with the assumption that \((m,n),(n,m)\in S\). Now suppose \(m\neq5\) and \(n=5\). Note that \((5, m)\) does not fit either of the conditions to be in S. Hence, \(n\neq 5\).

Answer 8

UUGHHHHHH wait can we translate the relation into words maybe it will be easier to follow

Answer 9

The way I'm reading it is \(m\,S\,n\) iff (\(m\le n\) and \(m\neq5\)) or \(n=5\).

Answer 10

well its (m,5) so m has to be less than or equal to 5 is that correct?

Answer 11

okkkkk i got ittttt yayyyyyyyyy

Answer 12

Sorry, gone for a minute there. So you're clear on why it's antisymmetric now?

Answer 13

yup

Answer 14

Next up, transitivity.

Suppose \((r,s),(s,t)\in S\). If \(r=5\) then \(s=5\), and \(t=5\), so \((r,t)\in S\) as well.

If \(r\neq 5\), and \(s=5\), then \(t=5\), and we can see that \((r,t)\in S\) as well.

If \(r\neq5\), and \(s\neq5\), then \(r\le s\). If \(t\neq5\), then \(r\le s\le t\implies r\le t\) so \((r,t)\in S\). If \(t=5\), then we see that \((r,t)\in S\) as well.

Therefore it's transitive. That make sense?

Answer 15

yuuuppp I got that soo farrrrr

Answer 16

Finally, we just need to show that it has totality.

Let \(a,b\in \mathbb{N}\). It's obvious that if either of them is 5, then either \((a,b)\) or \((b,a)\) will be in S. Therefore, suppose \(a,b\neq5\). Now, since \(\le\) is a linear ordering in \(\mathbb{N}\), it must be that either \(a\le b\) or \(b\le a\). Thus, we fall into the first condition, and it must be linear ordering.

Answer 17

the easiest to follow was totality

Answer 18

idk the rest r hurting my brain

Answer 19

There was a reason I wasn't sure whether it was linear or not at first, and instead had to work this out by hand.

Answer 20

k let me reread this stuff. If i am still having issues ill just call u :DDDDDD

Answer 21

KGGGGGGGGGGGGG I GOOOOTTTTTTTTTTTTTTTTTTT ITTTTTTT FINNAAALLYYYYYY

Answer 22

I wasnt gonna move on until i figured this one out. Last nite I had an event so I had no time to read it over. When i came back like at one I just cldnt get it but like now its clearrrrr.

Answer 23

THHANNKKKSSSS U R SERIOUSLY AWESOMEEEEEEEEEEEE

Answer 24

You're welcome :)