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How to Prove It, 2nd Ed. Sec. 4.4 #1b

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data
    A = ℝ, R = {(x, y) [itex]\in[/itex] ℝ X ℝ l lxl ≤ lyl }

    Say whether R is a partial order on A. If so, is it total order?


    2. Relevant equations
    Suppose R is a relation on a set A. Then R is called a partial order on A if it is reflexive, transitive and antisymmetric.

    1. R is said to be reflexive on A if [itex]\forall[/itex]x [itex]\in[/itex]A((x,x)[itex]\in[/itex]R.

    2. R is said to be transitive on A if [itex]\forall[/itex]x[itex]\in[/itex]A[itex]\forall[/itex]y[itex]\in[/itex]A[itex]\forall[/itex]z[itex]\in[/itex]A((xRy[itex]\wedge[/itex]yRz)[itex]\rightarrow[/itex]xRz).

    3. R is said to be antisymmetric if [itex]\forall[/itex]x[itex]\in[/itex]A((xRy[itex]\wedge[/itex]
    yRx)→x=y).

    3. The attempt at a solution
    1. For all x element of A, lxl ≤ lxl. Reflexive, Yes.
    2. [itex]\forall[/itex]x[itex]\in[/itex]A[itex]\forall[/itex]y[itex]\in[/itex]A[itex]\forall[/itex]z[itex]\in[/itex]A((xRy[itex]\wedge[/itex]yRz)[itex]\rightarrow[/itex]xRz). Transitive, Yes.
    3. If x = -2 and y = 2, then xRy [itex]\wedge[/itex]yRx, but -2 ≠ 2. So it is not antisymmetric, and thus not a partial order.

    Is this correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 16, 2011 #2
    Looks good, I assume you worked out the transitivity property yourself, but if not you should show your work. xRy ^ yRz => |x|<= |y| and |y|<= |z|, and since <= is transitive (do you need to or have you shown this before?) |x|<=|z|, so xRz.
     
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