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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

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:

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.

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.

enter image description here

Mr. Meeseeks
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  • What do you mean by $x|y$? – Michael Hoppe Jul 19 '15 at 19:35
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    What does it mean that "$y_1\mid y_2$ is even"? Usually "$y_1\mid y_2$" is a proposition -- it can be true or false, but not even or odd. – hmakholm left over Monica Jul 19 '15 at 19:35
  • @MichaelHoppe I don't have $x \mid y$, so I'm not sure to what you are referring. – Mr. Meeseeks Jul 19 '15 at 19:38
  • @HenningMakholm I don't believe that $y_1 \mid y_2$ has to be a proposition. This is parameter set by how I should come up with a relation. – Mr. Meeseeks Jul 19 '15 at 19:39
  • @mia.avery: You can define the notation to mean something that is not a proposition if you want to. But then you need to tell us what your new meaning for that notation is, if you want us to answer questions about it. – hmakholm left over Monica Jul 19 '15 at 19:41
  • @HenningMakholm It means "divides". Sorry, this is the standard notation used in my text book, I didn't realize this would be perceived as unknown or "new" notation. – Mr. Meeseeks Jul 19 '15 at 19:45
  • @mia.avery: Yes, usually it means "divides", but "$y_1$ divides $y_2$" is a claim that can be true or false, not a number that can be odd or even. So with that meaning of $y_1|y_2$, your definition of $M$ does not make sense. – hmakholm left over Monica Jul 19 '15 at 19:47
  • @HenningMakholm So, I have written the problem exactly as it was presented to me. I do not know how to solve it, I do not know how to properly develop my relation, $M$- this is what I am asking for help with. I do not realize why my question/post is unclear. Sorry...? – Mr. Meeseeks Jul 19 '15 at 19:51
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    @mia.avery: You will need to ask the people who gave you the problem what it means. It is not using standard mathematical terminology -- neither "$y_1$ divides $y_2$ is even" nor "$y_1|y_2$ is even" makes any sense in common mathematical English. Was it translated from some other language? – hmakholm left over Monica Jul 19 '15 at 19:53
  • @mia.avery It's unlikely that that's exactly how the problem was presented, because it doesn't make sense. Try taking a picture of the problem and adding that to your question. – Chris Culter Jul 19 '15 at 19:55
  • You wrote that "$x_1$ divides $x_2$ is even" means that $x_1-x_2=2k, k \in \Bbb Z$, or $x_1\equiv x_2 !\pmod{2}$. Is this what you meant? – coldnumber Jul 19 '15 at 19:58
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    @ChrisCulter I attached the problem. So, unless I suck at transcribing... – Mr. Meeseeks Jul 19 '15 at 20:00
  • @HenningMakholm No... she wrote the question herself, and she has English as a first language. This certainly explains why I'm confused... ha. – Mr. Meeseeks Jul 19 '15 at 20:03

2 Answers2

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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.

Chris Culter
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  • Thanks. This certainly explains why I was seriously confused on how to solve this problem... didn't mean to cause MathStack drama, but I seriously couldn't figure out what to do with this thing. Guess I'll go ask my professor... (she wrote this question, so I dunno what she was thinking). – Mr. Meeseeks Jul 19 '15 at 20:07
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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 $(-1,-1) \neq (1,1)$

hmakholm left over Monica
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