Michael B.

asked • 02/19/16

Discrete Mathematics

A={ -4 ,-3 ,-2 ,-1 ,0 ,1 ,2 ,3 ,4} R is defined on A as follows: For All (m,n) ∈A,m R n ⇔ 5 | (m^2- n^2 )

1 Expert Answer

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Arnold F. answered • 02/20/16

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Michael B.

It is actually. But I don't really know what is asking for.
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02/20/16

Arnold F.

You are given a set A with relation R such that mRn ⇔ 5 .
 
1. Is this saying mRn = 5 ?
 
2. then it says such that "m2 - n2 " but nothing is following that?
 
 
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02/20/16

Michael B.

I think it's asking for 5/(m2-n2), so the Relation set would be anything that results in an integer, I think. So, for example, 5/(-42-(-42)) = 5/(16-16) = 5/0 = undefined, so (-4,-4) wouldn't be in the Relation set. But 5/(-32-(-22)) = 5/(9-4) = 5/1 = 5. So the ordered pair (-3,-2) would be in the relation set. I have trouble with discrete math so, I don't really even know if I'm on the right track.
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02/20/16

Arnold F.

The notation is throwing me a little bit.
 
If we are looking for a relation set the we need a subset of  AxA.
 
Maybe 5|(m2-n2) actually means 5 divides (m2-n2)
 
so (4,1) works since 5 | 15 and so on.
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02/20/16

Michael B.

That's what I'm thinking, thanks for the help.
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02/20/16

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