Topological Properties of Subsets of the Reals
Absolute Value Inequalities
In mathematics one of the most used functions is the absolute value function. First we can define the absolute value and some of its properties.
Absolute Value Function
The absolute value of a real number x, |x|, is
|x|= { x if x ≥ 0
{ -x if x ≤ 0
We will now go over some properties that are associated with the absolute value function.
Property 1:
For any two real numbers x and y, we have |xy|= |x| |y|
This equality can then be check by considering some case. One case to check is as follows:
Suppose x < 0 and y ≥ 0. Then xy is ≤ 0 and we have
|xy| = – (xy) = (-x) y= |x| |y|
Propery 2:
For any real number x, and any nonnegative number a, we have
|x| ≤ a
If and only if –a ≤ x ≤ a.
This is verified by other cases x ≥ 0 and x < 0. We will consider the case of x ≥ 0. Suppose that
x ≥ 0 and |x| ≤ a. Now we have –a ≤ 0 ≤ x = |x| ≤ a
Since |x| ≤ a then –a ≤ x ≤ a. Conversely we can suppose x ≥ 0 and –a ≤ x ≤ a , then we have
|x| = x ≤ a
Property 3:
For any real number x, and any nonnegative number a, we have
|x| ≥ a
If and only if
x ≥ a or x ≤ -a
There are two cases to consider:
X is either positive or negative,
Suppose x ≥ 0. If |x| ≥ a, then we will have
x = |x| ≥ a
But if we have x ≥ a, then we have |x| = x ≥ a. Conversely we suppose x < 0 and |x| ≥ a, then we get
-x = |x| ≥ a
x ≤ -a
In the case where x < 0, we then have
|x| = -x ≥ -(-a)= a
Property 4:
For any two real numbers x and y we have
|x+y| ≤ |x| + |y|
There are four different cases but the simplest one to do is when both x and y are nonnegative, which is
|x + y| = x + y= |x| + |y|
Example:
Prove, using the triangle inequality and properties of absolute value, that for any x, y we have
|x – y| ≤ |x| + |y|
We can see that x – y = x + (-y) so the triangle inequality (*) can be applied as follows:
|x – y| = | x + (-y) |
≤ |x| + |-y|
= |x| + |-1| |y|
= |x| + |y|
Another example is to explain briefly why this inequality may not give the best (least) upper bound for | x – y|.
First we can assume for instance that 99 < x < y < 100. Then |x| < 100 and |y| < 100 so that inequality above would then tell us that
| x – y | ≤ |x| + |y| = 100 + 100 = 200
The distance between x and y is no greater than 200. Since x and y both reside in an interval whose diameter is 1 unit, it will then give a far better upper bound to notice that we have
| x – y | > 1
Because we know the absolute value function properties we can use the triangle equality to prove that, for all x contained in the real numbers, we have
|x| < |x – 1| + |x + 1|
= |x – 1 + 1 + x|
= | 2x|
< |2| |x|
|x| < 2|x|
By one of the properties from the absolute value function we are able to break up the |2x| giving us 2|x| which therefore is greater than |x| which is what we want to prove in the beginning.
Distance Function:
For two real numbers x, y define the distance between them by the function d(x, y) =|x – y|
By using the distance function we can find the upper bound between the distance of n and m. The example below is a great way to apply the distance function with the inequalities.
Let L , m , n be three real numbers, and assume that there exists an epsilon >0 such that both
|n – L| < 1/2E and |m – L| < 1/2E(epsilon)
Determine the best upper bound on the distance between n and m, and write a proof below.
Solution:
By adding and subtracting L, we see that
d(n, m)=|n – m| = |n – L + L – m|
< |n – L| + |L – m|
= |n – L| + |m – L|
< 1/2E + 1/2E = E(epsilon)
Thus |n- m|< E(epsilon) is our best upper bound on the distance between n and m.
http://www.math.tamu.edu/~stecher/171/absoluteValueFunction.pdf
http://www.clayton.edu/portals/164/Math-3005-Alt-Triangle-Inequality.pdf (proof using the fact that abs val x^2 = abs val x)
http://www.math.cmu.edu/~wgunther/127m12/notes/day2.pdf (proof by cases)
https://youtu.be/iQ8_ol_tHHg (youtube video with examples
-This is a great you tube video going into detail about the triangle inequality even showing a proof to prove that it is always right.