Question

Is this a linear order?
V where mVn iff
m is odd and n is even, or
m and n are even and \( m \leq n\)
m and n are odd and \( m \leq n\)

Answer 1

Integers or naturals?

Answer 2

Naturals

Answer 3

Like I am gonnnnaaa cry i cant solve a thing

Answer 4

all \( \leq \) in N are lineared ordered

Answer 5

First up, we need to show antisymmetry.

Suppose \(mVn\) and \(nVm\). Note that this means they are both either odd or both even. WLOG, suppose they are both even. So \(m\le n\) and \(n\le m\). Clearly, \(m=n\). So this is antisymmetric.

Answer 6

okkkkk

Answer 7

Now, onto transitivity.

Suppose \(rVs\) and \(sVt\). We have a few more cases to consider here. If they are all even/all odd we can see it's transitive because \(\le\) is transitive in the natural numbers.

Now let's start working through the cases. Suppose \(r\) is even. Then \(s\) must be even, and so must \(t\) (do you see why?), so we've reduced the problem to what I wrote above.

Suppose \(r\) is odd. Suppose \(s\) is then even, forcing \(t\) to be even. Since \(r\) is odd, and \(t\) is even, this is also transitive.

Now suppose \(r\) is odd, \(s\) is odd, and \(t\) is either odd or even. If \(t\) is even, it's transitive by the first condition. If \(t\) is odd, it's transitive by the third condition and transitivity of \(\le\).

Therefore, this is transitive.

Answer 8

AWESOME :)))))))))

Answer 9

Now, we finally need to show totality of the ordering.

Let \(a,b\in\mathbb{N}\) be arbitrary. We have 4 cases here, although it can be simplified WLOG to 2 cases.

Case 1: \(a,b\) are both odd/even. Since \(\le\) is a linear order on \(\mathbb{N}\), we know that either \(a\le b\) or \(b\le a\).

Case 2: One of \(a,b\) is odd, and the other is even. WLOG, suppose \(a\) is even, and \(b\) is odd. Then \(bVa\).

Hence, this relation has totality, and is a linear order.

Answer 10

hahahahha still trying to follow the def of totality lol

Answer 11

How do u see totality with the second case?

Answer 12

Totality is merely: Given \(a,b\), then \(aVb\) or \(bVa\) (or both).

In the second case, there's really two cases there. Let me break it up a bit further.

Answer 13

Case 2a: \(a\) is even, \(b\) is odd. Then \(bVa\), so we're done.
Case 2b: \(b\) is even, \(a\) is odd. Then \(aVb\), so we're done.

No matter what the case is, you'll always find that either \(aVb\) or \(bVa\), and so it has totality.

Answer 14

sorry just gonna ask one more question

Answer 15

but the first condition is symmetric isnt it?

Answer 16

antisymmetric.

Answer 17

lol if a m is odd and n is even or m is odd or n is even

Answer 18

did u get what i was saying?

Answer 19

No, sorry. Could you try and explain again? :/

Answer 20

hahahah okkkk

Answer 21

Like the first condition of this relation is m is odd and n is even

Answer 22

Thats mVn right

Answer 23

now like what wld nVm be?

Answer 24

I am just referring to the first condition

Answer 25

If m is odd, and n is even, then \(n\cancel{V}m\). \((n,m)\) is not actually in V.

Answer 26

okkk gotchhhaaaaaa

Answer 27

Thanks u r awesomeeee

Answer 28

U r seriously saving me here

Answer 29

No problem :P

Answer 30

TTTHHANNNKKSSSSSS :D

Answer 31

You're welcome. :)