Thread: [SOLVED] Showing Transitivity of a Relation

1. [SOLVED] Showing Transitivity of a Relation

I'm a bit lost in class so I am not sure if my solution is okay. Will you let me know if I did this okay?
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This is the relation: xRy ⇔ |x| ≤ |y+1|, ∀x,y∈Z.

I have this for Transitivity for this relation:

Let xRy, yRz ⇒ |x| ≤ |y+1|, |y| ≤ |z+1|.
By simple algebra on |y| ≤ |z+1| we get |y+1-1| ≤ |z+1|.
By the triangle inequality we get |y+1| + |-1| ≤ |z| + |1|,
then we have |y+1| + 1 ≤ |z| + 1 ⇒ |y+1| ≤ |z|.
By substituting |z| for |y+1| in xRy we get |x| ≤ |z|.
Thus |x| ≤ |z + 1|. It follows that xRz.

2. Hello yvonnehr
Originally Posted by yvonnehr
I'm a bit lost in class so I am not sure if my solution is okay. Will you let me know if I did this okay?
********
This is the relation: xRy ⇔ |x| ≤ |y+1|, ∀x,y∈Z.

I have this for Transitivity for this relation:

Let xRy, yRz ⇒ |x| ≤ |y+1|, |y| ≤ |z+1|.
By simple algebra on |y| ≤ |z+1| we get |y+1-1| ≤ |z+1|.
By the triangle inequality we get |y+1| + |-1| ≤ |z| + |1|,
then we have |y+1| + 1 ≤ |z| + 1 ⇒ |y+1| ≤ |z|.
By substituting |z| for |y+1| in xRy we get |x| ≤ |z|.
Thus |x| ≤ |z + 1|. It follows that xRz.
I think your proof goes a bit wrong where I've shown you. But the relation isn't transitive, I'm afraid. What about x = 3, y = 2 and z = 1? This gives |x| ≤ |y+1| and |y| ≤ |z+1|, but |x| > |z+1|