doughishere
Junior Member
- Joined
- Dec 18, 2015
- Messages
- 59
I found the following: http://www.math.tamu.edu/~stecher/171/absoluteValueFunction.pdf (attached if the site goes dead.) What ive done is re-work the same thing with the gaps filled in. The goal behind this was to learn absolute values a bit more and writing it down helped I felt like doing it one day. https://www.youtube.com/watch?v=2zKl2wgrlKw
This is my understanding of it and is heavily "borrowed" form the above work. I can remove if necessary or not cool. I also suspect its not 100pct correct. Lemma 3 was the hardest.
Absolute Value Function and its Properties
One of the most used functions in mathematics is the absolute value function. Its definition and some of its properties are given below.
The absolute value of a real number \(\displaystyle x, |x|\), is
The absolute value function is used to measure the distance between two numbers. Thus, the distance between \(\displaystyle x\) and \(\displaystyle 0\) is \(\displaystyle |x-0| = x\). In general, the distance between \(\displaystyle x\) and \(\displaystyle y\) is \(\displaystyle |x-y|\).
Lemma 1.
For any two real numbers \(\displaystyle x\) and \(\displaystyle y\), we have
This equality can be verified by considering cases:
Lemma 2.
For any real number \(\displaystyle x\), and any nonnegative number \(\displaystyle a\), we have
This can be verified by considering the cases, \(\displaystyle x \ge 0\) and \(\displaystyle x < 0\).
Lemma 3.
For any real number \(\displaystyle x\), and any nonnegative number \(\displaystyle a\), we have
This can be verified by considering the cases, \(\displaystyle x \ge 0\) and \(\displaystyle x < 0\).
Conversely, suppose \(\displaystyle x < 0\) and \(\displaystyle |x| \ge a\), then we have
To be done.
This is my understanding of it and is heavily "borrowed" form the above work. I can remove if necessary or not cool. I also suspect its not 100pct correct. Lemma 3 was the hardest.
Absolute Value Function and its Properties
One of the most used functions in mathematics is the absolute value function. Its definition and some of its properties are given below.
The absolute value of a real number \(\displaystyle x, |x|\), is
\(\displaystyle |x|=\left\{\begin{array}{rl}x; &\mbox{ if $x\ge0$}\\ -x; &\mbox{ if $x<0$}\end{array}\right\}\)
Example 1: Solve for a)\(\displaystyle |2|\) and b)\(\displaystyle |-2|\).
a) \(\displaystyle |2| = 2\)
b) \(\displaystyle |-2| = -(-2) = 2\)
The absolute value function is used to measure the distance between two numbers. Thus, the distance between \(\displaystyle x\) and \(\displaystyle 0\) is \(\displaystyle |x-0| = x\). In general, the distance between \(\displaystyle x\) and \(\displaystyle y\) is \(\displaystyle |x-y|\).
Example 2: Solve for the distance from a) \(\displaystyle -2\) to \(\displaystyle -4\) and b) \(\displaystyle -2\) to \(\displaystyle 5\)
a) \(\displaystyle |x-y| = |-2-(-4)|=2\)
b) \(\displaystyle |x-y| = |-2-(5)|=|-7|=7\)
b) \(\displaystyle |x-y| = |-2-(5)|=|-7|=7\)
Lemma 1.
For any two real numbers \(\displaystyle x\) and \(\displaystyle y\), we have
\(\displaystyle |xy| = |x| |y|\).
This equality can be verified by considering cases:
1)Suppose \(\displaystyle x < 0\) and \(\displaystyle y \ge 0\). Then \(\displaystyle xy\) is \(\displaystyle \le 0\) and we have
\(\displaystyle |xy| = -(xy) = (-x)y = |x||y|\).
2)Suppose \(\displaystyle x < 0\) and \(\displaystyle y < 0\). Then \(\displaystyle xy\) is \(\displaystyle > 0\) and we have
\(\displaystyle |xy| = (xy) =(-x)(-y) =|x||y|\).
3)Suppose \(\displaystyle x \ge 0\) and \(\displaystyle y > 0\). Then \(\displaystyle xy\) is \(\displaystyle \ge 0\) and we have
\(\displaystyle |xy| = (xy) = (x)(y) = |x||y|\).
4)Suppose \(\displaystyle x \ge 0\) and \(\displaystyle y < 0\). Then \(\displaystyle xy\) is \(\displaystyle < 0\) and we have
\(\displaystyle |xy| = -(xy) =(x)(-y) = |x||y|\).
Lemma 2.
For any real number \(\displaystyle x\), and any nonnegative number \(\displaystyle a\), we have
\(\displaystyle |x|\le\begin{array}{rl}a; &\mbox{ if and only if $ -a \le x \le a$}\end{array}\).
This can be verified by considering the cases, \(\displaystyle x \ge 0\) and \(\displaystyle x < 0\).
1) Suppose \(\displaystyle x \ge 0\) and \(\displaystyle |x| \le a\). This implies that \(\displaystyle -a \le 0 \le x = |x| \le a\). In other words, \(\displaystyle |x| \le a\) implies that \(\displaystyle -a \le x \le a = |x| \le a\).
Conversely, suppose \(\displaystyle x \ge 0\) and \(\displaystyle -a \le 0 \le x = |x| \le a\). This implies \(\displaystyle |x| = x \le a\).
2) Suppose \(\displaystyle x < 0\) and \(\displaystyle |x| \le a\). This implies that \(\displaystyle -a \le x \le 0 = |x| \le a\). In other words, \(\displaystyle |x| \le a\) implies that \(\displaystyle -a \le x \le a = |x| \le a\).
Conversely, suppose \(\displaystyle x < 0\) and \(\displaystyle -a \le x \le 0 = |x| \le a\). This implies \(\displaystyle |x| = x \le a\).
Lemma 3.
For any real number \(\displaystyle x\), and any nonnegative number \(\displaystyle a\), we have
\(\displaystyle |x|\ge\begin{array}{rl}a; &\mbox{ if and only if $ x \ge a $ or $ x \le -a$}\end{array}\).
This can be verified by considering the cases, \(\displaystyle x \ge 0\) and \(\displaystyle x < 0\).
1) Suppose \(\displaystyle x \ge 0\) and \(\displaystyle |x| \ge a\). This implies that \(\displaystyle |x| = x \ge a\). In other words, if \(\displaystyle x \ge a\) implies that \(\displaystyle |x| = x \ge a\).
Conversely, suppose \(\displaystyle x < 0\) and \(\displaystyle |x| \ge a\), then we have
\(\displaystyle -x = |x| \ge a\)
\(\displaystyle x \le -a\)
\(\displaystyle x \le -a\)
Also, suppose \(\displaystyle x < 0\) and \(\displaystyle x \le -a\). This implies that
\(\displaystyle -x = |x| \ge a\)
2) Suppose \(\displaystyle x < 0\) and \(\displaystyle |x| \ge a\). We can not have this case since we assume that \(\displaystyle x \ge 0\).
Lemma 4.
To be done.
