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

1. Oct 16, 2011

### IntroAnalysis

1. The problem statement, all variables and given/known data
A = ℝ, R = {(x, y) $\in$ 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 $\forall$x $\in$A((x,x)$\in$R.

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

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

3. The attempt at a solution
1. For all x element of A, lxl ≤ lxl. Reflexive, Yes.
2. $\forall$x$\in$A$\forall$y$\in$A$\forall$z$\in$A((xRy$\wedge$yRz)$\rightarrow$xRz). Transitive, Yes.
3. If x = -2 and y = 2, then xRy $\wedge$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. Oct 16, 2011

### aeroplane

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.