Example 1.2.1.
-
\begin{equation*} |2| = 2 \end{equation*}
-
\begin{equation*} |-2| = -(-2) = 2 \end{equation*}
-
\begin{equation*} |x^2| = x^2 \end{equation*}
Section 1.2 Number Line \(\&\) Intervals
In mathematics, an interval can be defined as a set of real numbers \(a, b \in \mathbb R\) which contains all real numbers lying within any two specific numbers of the set \(\mathbb R\text{.}\) Interval Notation uses inequalities to represent a subset of real numbers by those numbers that bound them.
Consider an interval signified as \(0 \lt x \leq 5\text{,}\) which means a set of numbers lying between 0 and 5 (excluding 0 but including 5) represent the value of x.
Types of Interval Notation: There are different types of notations of intervals that are classified based on the endpoints of intervals. They are: Open intervals, Closed intervals, Half-open intervals, Degenerate intervals, and Bounded and Unbounded intervals.
-
Open Interval: \((a, b)\)
e.g., \(1 \lt x \lt 2\) unfilled circle represents the number is not included. -
Closed Interval: \([a, b]\)
e.g., \(1 \leq x \leq 2\) filled circle represents the number is included. -
Closed - open Interval: \([a, b)\)
-
Open - closed Interval: (a, b]
Subsection 1.2.1 Inequality Rules
Summary of inequality rules in mathematical operation:
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When adding or subtracting \(+ve\) or \(-ve\) number on both sides of inequality there is no sign change of inequality. For example:\begin{equation*} 3x - 5 \lt 10 \, \Rightarrow \, 3x \lt 15 \end{equation*}and\begin{equation*} 2x + 5 \lt 10 \, \Rightarrow \, 2x \lt 5 \end{equation*}
-
When multiplying or dividing \(+ve\) number on both sides of inequality there is no sign change of inequality. For example:\begin{align*} 3x \amp \lt 15 \\ or, \, x \amp \lt \frac{\cancelto{5}{15}}{\cancel{3}} \\ \therefore \, x \amp \lt 5 \end{align*}
-
When multiplying or dividing \(-ve\) number on both sides of inequality there is sign change of inequality. For example:\begin{align*} -2x +4 \amp \gt 12 \\ or, \, -2x \amp \gt 8\\ x \amp \lt \frac{\cancelto{4}{8}}{-\cancel{2}} \\ \therefore \, x \amp \lt -4 \end{align*}Do not forget to flip sign, even if you are cross multiplying both side by \(-ve \) number.\begin{align*} -\frac{2}{x+3} \amp \gt -\frac{5}{x+3}; \, (x \gt -3)\\ -\frac{x+3}{2} \amp \lt -\frac{x+3}{5}\\ \frac{x+3}{2} \amp \gt \frac{x+3}{5} \end{align*}Hence, it is true as \(x+3 \gt 0; \, \forall_x \gt -3\text{.}\)
-
When taking reciprocal of inequality having same \(+ve\) or \(-ve\) number on both sides, there is sign change of inequality. For example:\begin{align*} 10 \amp \gt 2 \\ or, \, \frac{1}{10} \amp \lt \frac{1}{2} \end{align*}or,\begin{align*} -10 \amp \lt -2 \\ or, \, -\frac{1}{10} \amp \gt -\frac{1}{2} \end{align*}However, When taking reciprocal of inequality having one side \(+ve\) and another side \(-ve\) number then, there is no sign change of inequality. For example:\begin{align*} 5 \amp \gt -2 \\ \frac{1}{5} \amp \gt -\frac{1}{2} \end{align*}
Subsection 1.2.2 Modulus Functions
Basics of Modulus functions: The general representation of the modulus function is \(f(x) =|x|,\) where
\begin{equation*}
|x| = \begin{cases} x; & \text{ if } \, x \geq 0 \\ -x; & \text{ if } \, x\lt 0 \end{cases}
\end{equation*}
If \(|x|=0\) we get only one solution as distance of zero from zero is 0.
If \(|x|=-ve \) number, we get no solution as mod of any number is never negative.
If \(|x|=+ve \) number, we get two solution as described in defintion above.
Example 1.2.2.
Solution.
From definition:
\begin{equation*}
|x| = \begin{cases} x; & \text{ if } \, x \geq 0 \\ -x; & \text{ if } \, x\lt 0 \end{cases}
\end{equation*}
We have -
\begin{equation*}
|x|=x; \, x\gt 0 \, \Rightarrow \, x=5.
\end{equation*}
or
\begin{equation*}
|x|=-x; \, x\lt 0 \, \Rightarrow \, x=-5.
\end{equation*}
Example 1.2.3.
Example 1.2.4.
Example 1.2.5.
Find \(x\) if
\begin{equation*}
|-4x+7| =-9
\end{equation*}
No solution as mod never gives any negative output.
Example 1.2.6.
\begin{equation*}
\vert x-1 \vert = \begin{cases} x-1; & \text{ if } \, x \geq 1 \\ -(x-1); & \text{ if } \, x \lt 1 \end{cases}
\end{equation*}
Rememebr: Mod always gives a single value, if there is one.
Example 1.2.7.
Solve the following:
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\begin{equation*} \vert x \vert = 5 \end{equation*}\begin{equation*} then, \, x=+5 \, or \, -5. \end{equation*}
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\begin{equation*} \vert 2x-1 \vert =2 \end{equation*}\begin{equation*} 2x-1=+2 \, or \, -2 \, \Rightarrow \, x = \frac{3}{2} \, or, \, x = -\frac{1}{2} \end{equation*}
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\begin{equation*} \vert 3x-5 \vert =-3 \, \Rightarrow \, x \in \phi \, \text{ (no solution) } \end{equation*}
Observation 1.2.8.
Example 1.2.9.
Solve the following inequalities:
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\begin{equation*} |x| \gt 3 \end{equation*}\begin{equation*} x\gt 3 \, \quad OR, \, x \lt -3 \end{equation*}
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\begin{equation*} |x| \geq \sqrt{2} \end{equation*}\begin{equation*} x \geq \sqrt{2} \, OR \, x \leq -\sqrt{2} \end{equation*}
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\begin{equation*} |x| \leq \pi \end{equation*}\begin{equation*} -\pi \leq x \leq \pi \end{equation*}
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\begin{equation*} |2x+1| \geq 2 \end{equation*}\begin{equation*} 2x+1 \geq 2 \, \quad \Rightarrow \, x \geq \frac{1}{2} \end{equation*}\begin{equation*} OR, \, 2x+1 \leq -2 \, \quad \Rightarrow \, x \leq -\frac{3}{2} \end{equation*}
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\begin{equation*} |x+3| \geq -2 \end{equation*}\begin{equation*} |x+3| \gt -2 \, \end{equation*}\begin{equation*} or, \quad |x+3| = -2\, \text{ (no solution)} \end{equation*}\begin{equation*} x+3 \gt -2 \end{equation*}\begin{equation*} OR, \, x+3 \lt -(-2) =2 \end{equation*}
