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:

|x - 2| = 5.

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:

	\|6 - (-1)\|	= 7,
	\|5 - (-2)\|	= 7,
	\|4 - (-3)\|	= 7,
	\|3 - (-4)\|	= 7,
	\|2 - (-5)\|	= 7.

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

|x - 2| < |x - 4|.

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

|x - 2| = 5.

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

|x - 2| < |x - 4|.

Solution

In terms of distances, the problem is trivial: Which are the points that are closer to 2 than to 4? The midway between 2 and 4 is the point 3. Anything to the left of 3 is closer to 2 than to 4. Solution: x < 3. Formally, we have to consider 4 cases:

x - 2 < 0, x - 4 < 0,
x - 2 ≥ 0, x - 4 < 0,
x - 2 ≥ 0, x - 4 ≥ 0.
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:

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.
x - 2 < 4 - x.

From here 2x < 6, or x < 3.
x - 2 < x - 4.

This is also useless. So we are left with just one inequality from case 2: x < 3, which we know to be true.

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):

|u| ≥ 0 and |u| = 0 only of u = 0.
|u| = |-u|.
|u + v| ≤ |u| + |v|, and the equality is only achieved if u and v are of the same sign.
|uv| = |u||v|.
If v ≠ 0, |u/v| = |u|/|v|.
|u - v| ≤ |u - w| + |w - v|.
|u| < v is equivalent to -v ≤ u ≤ v, for positive v.
||u| - |v|| ≤ |u - v|.
√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.

Absolute Value

40619810