Absolute Value

For a real number u, the absolute value |u| is the distance from u to the origin, 0. There are other, seemingly more formal, definitions but this one is extremely handy in solving a great variety of problems. For two real numbers u and v, |u - v| is the distance between u and v on the real line. Indeed, imagine dragging the whole of the real line so as to make v coincide with 0. This is just a shift by v that does not change the distances between the points. Point u is dragged into u - v and v, as we agreed, to 0. It now follows that the distance between u - v and 0, which is |u - v|, is exactly the distance between u and v.

As an example of the utility of this definition, let's solve this equation:

Without getting any algebra involved, we are looking for a point or points whose distance from 2 is 5. Look at the number line:

There are just two points at distance 5 from 2. One to the left of 2 is -3. One to the right is 7.

So, for any two real numbers u and v, |u - v| is the distance between u and v on the number line. Well, what is |2 + 5|. Now 2 + 5 = 7, of course, and |7| is the distance from 7 to 0, which is 7. You may guess this is not exactly what I had in mind. But what was it? Let's see. 0 is 7 units from 7. Looking leftward, what is 7 units from 6? -1, right? This means that |6 - (-1)| = 7. This is true simply because 6 - (-1) = 6 + 1 = 7. Continue in this way:

The latter says that 7 = |2 - (-5)| = |2 + 5| and shows that |2 + 5| = 7 can be said to broadcast the fact that 2 and -5 are at distance 7 on the number line. And so (from the second line above) are 5 and -2. Inside the absolute value marks we are allowed to change signs. This is because the distance is a symmetric function: it is as far from one point to another as it is from the second point to the first. This implies that |u - v| = |v - u| or, which is the same, |w| = |-w|.

Let's solve an inequality

Here we are looking for points x that are closer to 2 than to 4. Again, if you check on the number line, the answer is immediate: x < 3. 3 is the midpoint between 2 and 4 and so separates the points closer to 2 from the points that are closer to 4.

Formally, |u| = u, for u ≥ 0, and |u| = -u, for u < 0. If you marvel at the last part (|u| = -u for negative u), remember that prepending a minus sign to a number means multiplying the number by -1 which changes its sign. So, for a negative u, -u is positive and, of course is at the same distance from 0 as u itself. So again, |u| = |-u|.

Let's now try to solve our examples using the formal definition.

Example 1

Solve

Solution

The (formal) definition of the absolute value consists of two parts: one for positive numbers and zero, the other for negative numbers. So first assume x - 2 ≥ 0. In this case, 5 = |x - 2| = x - 2. Adding 2 to both sides gives x = 7. Now, we have to check the second part of the definition. So assume x - 2 < 0. In this case, 5 = |x - 2| = -(x - 2), or 5 = 2 - x. Adding x to both sides and subtracting 5 gives x = 2 - 5 = -3. As we already knew, -3 and 7 are at distance 5 from 2.

Example 2

Solution

Strictly speaking, we have to consider 4 cases:

1. x - 2 < 0, x - 4 < 0,
2. x - 2 ≥ 0, x - 4 < 0,
3. x - 2 ≥ 0, x - 4 ≥ 0.
4. x - 2 < 0, x - 4 ≥ 0,

However, the last one is obviously impossible because always x - 2 < x - 4. So, we need to consider the first three cases. The difference between the cases is in which part of the definition of the absolute value applies in which case:

1. x - 2 < 0, x - 4 < 0

   2 - x < 4 - x. Which is true but sterile. It provides no useful information that would help in determining x.

2. x - 2 < 4 - x.

   From here 2x < 6, or x < 3.

3. x - 2 < x - 4.

   This is also useless. So we are left with just one inequality from case 2: x < 3.

Absolute value has several properties, some of which we have encountered earlier (I'll omit the repetitive "for all" expression that should precede each of the properties):

1. |u| ≥ 0 and |u| = 0 only of u = 0.

2. |u| = |-u|.

3. |u + v| ≤ |u| + |v|, and the equality is only achieved if u and v are of the same sign.

4. |uv| = |u||v|.

5. If v ≠ 0, |u/v| = |u|/|v|.

6. |u - v| ≤ |u - w| + |w - v|.

7. |u| < v is equivalent to -v ≤ u ≤ v, for positive v.

8. ||u| - |v|| ≤ |u - v|.

9. √u² = |u|.

To prove #3 one needs to consider four cases where u and v are independently positive and negative but, otherwise, is straightforward. #7 just tells us where the points lie whose distance to 0 is less than v: they lie between -v and v.

#6 follows from #3. Indeed, to avoid confusion, rewrite the latter as, say, |x + y| ≤ |x| + |y|, and then replace x and y with u - w and w - v, respectively. The equality holds when (u - w) and (w - v) are of the same sign. This happens only when w is between u and v.

#8 follows from #6 and #7. Similarly to the change of variables we just did, |u| ≤ |v| + |u - v| so that |u| - |v| ≤ |u - v|. Swapping u and v, |v| - |u| ≤ |v - u| = |u - v|. Changing the signs gives |u| - |v| ≥ - |u - v|. Now we only need to apply #7 to get #8.

#8 tells us that the function f(x) = |x| is continuous and indeed is uniformly continuous.