Define a relation $M$ on $\mathbb{Z} \times \mathbb{Z}$...

Question

Update #2 (7.21.15):

Here is a screenshot of the corrected question, in case anyone was interested.

No need to look at the first update or original post to anyone viewing this for the first time.

Mia

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Update #1 (07.19.15):

Sorry about the confusion. This was exactly as the problem was presented to me-- I reached out to stackexchange because I was seriously confused on what she was asking me to do. Seems as though the confusion was warranted... so... yay math?

Here is the problem I am working on:

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Define a relation, $M$, on $\mathbb{Z} \times \mathbb{Z}$ by $(x_1,y_1) M (x_2,y_2)$ if and only if $x_1 \mid x_2$ and $y_1 \mid y_2$ is even.

Prove or disprove that $M$ is each reflexive, symmetric, transitive, and/or antisymmetric.

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This is the problem, as it has been presented to me. I am curious about how to best approach this. Here is what I am thinking so far:

$M= \left\{(x,y) \in \mathbb{Z} \times \mathbb{Z}\mid x-y \in \mathbb{Z} \right\}$

Does this make sense as the start towards my relation? I don't feel like I've addressed the even portion of my problem, which is where I find myself stuck. If $x_1 \mid x_2$ and $y_1 \mid y_2$ is even, then I can state that:

$x_1-x_2 = 2k, k \ \in \mathbb{Z}$ and, $y_1-y_2 = 2j, j \ \in \mathbb{Z}$

But I'm not really sure where to work this into my relation. Suggestions or hints would be appreciated!

Once I get the relation worked out, I think I can figure out the reflexive, symmetric (etc.) portion of the problem.

Update: here is a screen shot of the problem.

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Answer 1

Thanks for uploading the screen shot! Now it's clear that the author of the problem made a mistake somewhere. You could try to guess what they meant, but if possible, it would be better to notify them of the mistake and ask them to fix it.

Answer 2

Assuming the "is even" is a typo, and M is defined by

$$(x_1,y_1)M(x_2,y_2) \Leftrightarrow x_1|x_2 \ \text{and}\ y_1|y_2$$

• You can see that $M$ is reflexive :

$x|x$ and $y|y$ so $(x,y) M(x,y)$

• You can see that $M$ is transitive :

if $x_1|x_2$ and $x_2|x_3$ , then $x_1|x_3$

• You can see that $M$ is not symetric :

You have $(1,1)M(2,3)$ but not $(2,3)M(1,1)$ as $2\nmid 1$

• You can see that $M$ is not antisymetric :

$(-1,-1)M(1,1)$ and $(1,1)M(-1,-1)$ but $(-1,-1) \neq (1,1)$