(adsbygoogle = window.adsbygoogle || []).push({}); 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

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

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