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^{+} which is assumed to satisfy the following axioms.
Axiom 1.4.3 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.
Definition 1.4.4 x < y exactly when y +
Theorem 1.4.5 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 1.4.3 −x > 0 and −y > 0. Therefore, by 1.4.2 and what was proved about −x =

Is

Therefore, 1 =
Next consider 4. If x > 0 why is x^{−1} > 0? By 1.4.3 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 1.4.2 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. This proves the theorem. ■
Note that trichotomy could be stated by saying x ≤ y or y ≤ x.
Definition 1.4.6
Note that
Theorem 1.4.7
Proof: You can verify this by checking all available cases. Do so. You need consider both x,y nonnegative, both negative, and one negative and the other positive. ■
Theorem 1.4.8 The following inequalities hold.

Either of these inequalities may be called the triangle inequality.
Proof: First note that if a,b ∈ ℝ^{+} ∪

because b^{2} − ab = b

By the above theorem on order,

Thus b ≥ a.
Now
To verify the other form of the triangle inequality,

so

and so

Now repeat the argument replacing the roles of x and y to conclude

Therefore,

This proves the triangle inequality. ■
Example 1.4.9 Solve the inequality 2x + 4 ≤ x − 8
Subtract 2x from both sides to yield 4 ≤−x − 8. Next add 8 to both sides to get 12 ≤−x. Then multiply both sides by
Example 1.4.10 Solve the inequality
If this is to hold, either both of the factors, x + 1 and 2x− 3 are nonnegative or they are both nonpositive. The first case yields x + 1 ≥ 0 and 2x − 3 ≥ 0 so x ≥−1 and x ≥
Example 1.4.11 Solve the inequality
Here the problem is to find x such that x^{2} + 2x + 4 ≥ 0. However, x^{2} + 2x + 4 =
Example 1.4.12 Solve the inequality 2x + 4 ≤ x − 8
This is written as (−∞,−12].
Example 1.4.13 Solve the inequality
This was worked earlier and x ≤−1 or x ≥
Example 1.4.14 Solve the equation
This will be true when x − 1 = 2 or when x − 1 = −2. Therefore, there are two solutions to this problem, x = 3 or x = −1.
Example 1.4.15 Solve the inequality
From the number line, it is necessary to have 2x − 1 between −2 and 2 because the inequality says that the distance from 2x − 1 to 0 is less than 2. Therefore, −2 < 2x − 1 < 2 and so −1∕2 < x < 3∕2. In other words, −1∕2 < x and x < 3∕2.
Example 1.4.16 Solve the inequality
This happens if 2x − 1 > 2 or if 2x − 1 < −2. Thus the solution is x > 3∕2 or x < −1∕2. Written in terms of intervals this is
Example 1.4.17 Solve
There are two ways this can happen. It could be the case that x + 1 = 2x− 2 in which case x = 3 or alternatively, x + 1 = 2 − 2x in which case x = 1∕3.
Example 1.4.18 Solve
In order to keep track of what is happening, it is a very good idea to graph the two relations, y =
Equality holds exactly when x = 3 or x =
Example 1.4.19 Suppose ε > 0 is a given positive number. Obtain a number, δ > 0, such that if
First of all, note

Now let δ = min
