Are the members of (x,y) if and only if (y = x + 3) & (y = x – 3) an empty set?

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SUMMARY

The set of all integers where ∈ R if and only if (y = x + 3) & (y = x – 3) is indeed an empty set. The contradiction arises when setting the equations equal to each other, leading to the false statement 3 = -3. This confirms that no elements satisfy the given relationship, thus proving the set is empty. Additionally, the identity holds in the context of Z_3, where 3 = -3.

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Is the set of all integers where <x,y> \in R if and only if (y = x + 3) & (y = x – 3) an empty set?
 
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Hey Bob4040 and welcome to the forums.

Your definition is a little vague. You mention integers but then you say x and y are real numbers which is a bit of a contradiction.

Are you trying to ask whether x and y are real numbers if the relation holds?
 
I do not mean all real numbers. R is the set. x and y are members of the set on the condition that (y = x + 3) & (y = x – 3). These two lines never intersect, so I think that means it is an empty set?
 
Ohh I see what you mean now.

Well the easiest way to check is to set them equal to one another and see if you get a solution or a contradiction.

y = x + 3, y = x - 3 implies x + 3 = x - 3 which implies 3 = -3 which is a contradiction, so you have proven that no element exists satisfying the relationship so you have the empty set.
 
Thank you for the help!
 
I know this is not what you meant/intended , Bob4040, but, strictly speaking your

identity would hold if x were in the set called Z_3, where 3=-3 :

http://www.wolframalpha.com/input/?i=Z+mod+3

Then,for all x, x+3=x-3 .
 

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