Cartesian product notation and plotting the set {(y,z)|y-z≥3}

[ptex]

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

 

[HallsofIvy]

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}"?

 

[ptex]

You helped me with a question like this before and I got it. Now I am studing for a final.
Ok the whole question;

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

 

[gnome]

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 Y_i is related to Z_j 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)

 

[ptex]

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

 

[gnome]

Looks correct.

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