To do linear algebra, you need a field which is something satisfying the axioms listed in Theorem 1.5.1. This is generally all that is needed to do linear algebra but for the sake of completeness, the concept of an ordered field is considered here. The real numbers also have an order defined on them. This order may be defined by reference to the positive real numbers, those to the right of 0 on the number line, denoted by ℝ^{+}. More generally, for a field, one could consider an order if there is such a “positive cone” called the positive numbers such that the following axioms hold.
Axiom 1.11.3 For a given real number x one and only one of the following alternatives holds. Either x is positive, x = 0, or −x is positive.
An example of this is the field of rational numbers.
Definition 1.11.4 x < y exactly when y +
Theorem 1.11.5 The following hold for the order defined as above.
Proof: First consider 1, the transitive law. Suppose x < y and y < z. Why is x < z? In other words, why is z − x ∈ ℝ^{+}? It is because z − x =
Next consider 2, addition to an inequality. If x < y why is x + z < y + z? it is because
Next consider 3. If x ≤ 0 and y ≤ 0, why is xy ≥ 0? First note there is nothing to show if either x or y equal 0 so assume this is not the case. By 1.11.3 −x > 0 and −y > 0. Therefore, by 1.11.2 and what was proved about −x =

Is

Next consider 4. If x > 0 why is x^{−1} > 0? By 1.11.3 either x^{−1} = 0 or −x^{−1} ∈ ℝ^{+}. It can’t happen that x^{−1} = 0 because then you would have to have 1 = 0x and as was shown earlier, 0x = 0. Therefore, consider the possibility that −x^{−1} ∈ ℝ^{+}. This can’t work either because then you would have

and it would follow from 1.11.2 that −1 ∈ ℝ^{+}. But this is impossible because if x ∈ ℝ^{+}, then
Next consider 5. If x < 0, why is x^{−1} < 0? As before, x^{−1}≠0. If x^{−1} > 0, then as before,

which was just shown not to occur.
Next consider 6. If x < y why is xz < yz if z > 0? This follows because

since both z and y − x ∈ ℝ^{+}.
Next consider 7. If x < y and z < 0, why is xz > zy? This follows because

by what was proved in 3.
The last two claims are obvious and left for you. ■