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Math Help - [SOLVED] Showing Transitivity of a Relation

  1. #1
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    Question [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?
    ********
    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.
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  2. #2
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    Hello yvonnehr
    Quote Originally Posted by yvonnehr View Post
    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|

    Grandad
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