Are X,Y,Z Equal? Examining {(y,z)|y-z> or = 3}

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Homework Help Overview

The discussion revolves around the relationships defined by sets X, Y, and Z, specifically examining the relations R1 and R2, which involve matrix representations of these relations. The participants are exploring the implications of the conditions y-z ≥ 3 and x+y ≤ 7 within the context of these sets.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to clarify the definitions and relationships between the sets X, Y, and Z, as well as the matrix representations of the relations R1 and R2. There is a focus on understanding how the orientation of the matrices affects the results.

Discussion Status

Some participants have provided feedback on the matrices presented, indicating that they appear correct. However, there is still a need for further clarification on the relationships and how they are represented in the matrices.

Contextual Notes

There is a mention of needing more detail about the original problem setup, particularly regarding the definitions of X, Y, and Z and how they relate to the matrices. The original poster's reference to a previous discussion suggests a continuity in the topic, but specifics are lacking.

ptex
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X=Y=Z={1,2,3,4,5,6}
Is this the same as
X={1,2,3,4,5,6}
Y={1,2,3,4,5,6}
Z={1,2,3,4,5,6}

If so {(y,z)|y-z> or = 3}
is the y the vertical (or side)and z the Horizontal (or top)?

Like this;
? 1 2 3 4 5 6
1 0 0 0 0 0 0
2 0 0 0 0 0 0
3 0 0 0 0 0 0
4 1 0 0 0 0 0
5 1 1 0 0 0 0
6 1 1 1 0 0 0
 
Last edited:
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I'm afraid you will have to give more detail as to exactly what your question is.

Matrices in general do NOT have X, Y, Z parts- AND do not have 3 dimensions.

Perhaps the problem you are working on specifies in some way how X, Y, Z affects the matrix but you will have to tell us. Where did you get "X=Y=Z={1,2,3,4,5,6}"?
 
You :rolleyes: helped me with a question like this before and I got it. Now I am studing for a final.
Ok the whole question;
Code:
X=Y=Z={1,2,3,4,5,6} in that order and,
R[sub]1[/sub] = {(x,y)| x+y < or = to 7}
R[sub]2[/sub] = {(y,z)| y-z  > or = to 3}

a) The matrix A[sub]1[/sub] of the relation R[sub]1[/sub] (relative to the given orderings)

b) The matrix A[sub]2[/sub] of the relation R[sub]1[/sub] (relative to the given orderings)

c) The matrix product A[sub]1[/sub] A[sub]1[/sub]

d) Use the result of part (c) to find the matrix of the relation R[sub]2[/sub]0 R[sub]1[/sub].

e) Use the result of part (d) to find the relation R[sub]2[/sub]0 R[sub]1[/sub]. (as a set of ordered pairs).

For R[sub]1[/sub] I have;
? 1 2 3 4 5 6
1 1 1 1 1 1 1
2 1 1 1 1 1 0
3 1 1 1 1 0 0
4 1 1 1 0 0 0
5 1 1 0 0 0 0 
6 1 0 0 0 0 0
 
Last edited:
Your A1 looks fine to me. Of course, R1 would come out the same no matter which way you look at it. For R2, orientation matters.

You want each A(i,j) (where i is the row and j is the column) to have a 1 when Yi is related to Zj according to the relation R2. So, for example, A(6,1) = 1 because the value in Y6 minus the value in Z1 ≥ 3 (6-1=5)
 
So are these correct?

For R1 or A1 I have;
? 1 2 3 4 5 6
1 1 1 1 1 1 1
2 1 1 1 1 1 0
3 1 1 1 1 0 0
4 1 1 1 0 0 0
5 1 1 0 0 0 0
6 1 0 0 0 0 0

For R2 or A2 I have;
? 1 2 3 4 5 6
1 0 0 0 0 0 0
2 0 0 0 0 0 0
3 0 0 0 0 0 0
4 1 0 0 0 0 0
5 1 1 0 0 0 0
6 1 1 1 0 0 0
 
Last edited:
Looks correct.

PS: those code windows are annoying -- too much scrolling.
 

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