Equivalence Relations on Set S: Description and Number of Classes

Bonafide
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Hello!

I'm a bit lost on these questions pertaining to equivalence relations/classes. If someone could run me through either, or both, of these questions, I'd be very thankful! I'm completely lost as to what to do with the z in terms of set S...

Homework Statement



Show that the given relation R is an equivalence relation on set S. Then describe the equivalence class containing the given element z in S, and determine the number of distinct equivalence clases of R.

16. Let S be the set of all subsets of {1,2,3,4,5}. let z = {1,2,3}, and define xRy to mean that x \bigcap {1,3,5} = y \bigcap {1,3,5}.

18. Let S be the set of ordered pairs or real numbers, let z = (3, -4) and define (x1, x2) R (y1, y2) means that x1 + y2 = y1 + x2.
 
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Then describe the equivalence class containing the given element z in S,

So for number 16, two objects are equivalent if they share the same elements out of {1,3,5}. z contains both 1 and 3, so if you have another subset equivalent to it:

Does it contain 1?
Does it contain 3?
Does it contain 5?

And does it matter whether it contains 2 or 4?
 
You really have to try a little harder than that. First of all what do you need to show a relation is an equivalence relation. It's not that hard to show for either one. As for what to do with the z, you want to find all x such that xRz in each case. Try the second one first. If (x,y)R(3,-4) what does that tell you about (x,y)?
 
In order to show a relation is an equivalence relation, you need to show it has the three properties characteristc of such relations. Once that is done, you need to find what sets must have in common in order to be R-related.

As a hint for this, both S and the set {1, 3, 4} are R-related to z. Do you see why?

EDIT: S is not R-related to z. I overlooked that 5 is in S. However {1, 2, 3, 4} is R-related to z.


A hint for the second questions is to change the relationship equation so that the x-coordinates are on one side and the y-coordinates on the other. What does this seem to indicate about the ordered pairs?

--Elucidus
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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