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 2.4.3 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 2.4.4 x < y exactly when y +
Theorem 2.4.5 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 2.4.3 −x > 0 and −y > 0. Therefore, by 2.4.2 and what was proved about −x =

Is

Next consider 4. If x > 0 why is x^{−1} > 0? By 2.4.3 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 2.4.2 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 2.4.6
Note that
Theorem 2.4.7
Proof: You can verify this by checking all available cases. Do so. ■
Theorem 2.4.8 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,

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 2.4.9 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 2.4.10 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 2.4.11 Solve the inequality
Here the problem is to find x such that x^{2} +2x+4 ≥ 0. However, x^{2} +2x+4 =
Example 2.4.12 Solve the inequality 2x + 4 ≤ x − 8
This is written as (−∞,−12].
Example 2.4.13 Solve the inequality
This was worked earlier and x ≤−1 or x ≥
Example 2.4.14 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 2.4.15 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 2.4.16 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 2.4.17 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 2.4.18 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 2.4.19 Suppose ε > 0 is a given positive number. Obtain a number, δ > 0, such that if
First of all, note

Now let δ = min
