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This is a continuation of Absolute Value.

### Properties of the absolute value:

1. Non-negative: $$\vert x \rvert \geq 0$$ for all real values.

2. $$\lvert x \rvert = \sqrt{ x^2 }$$ and $$\lvert x \rvert^2 = x^2$$ for all real values.

3. Multiplicative: $$\lvert xy \rvert = \lvert x \rvert \cdot \lvert y \rvert$$.

4. Sub-additive: $$\lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert$$.

5. Symmetry: $$\lvert -x \rvert = \lvert x \rvert$$.

6. Distance to origin: $$\lvert x \rvert$$ measures the geometric distance of the real number $$x$$ to the origin $$0$$. Since distance is always positive, the absolute value of a number is always positive. Thinking of absolute value as the distance from zero is also helpful when considering complex numbers $$z = a + bi$$. This distance from $$z$$ to the origin is given by the distance formula: $\text{distance}_{(a,b)\leftrightarrow(0,0)} = \sqrt{(a-0)^2 + (b-0)^2}$ In fact, this is also the definition of the absolute value for a complex number $$z = a+bi$$:

$\left| z \right| = \sqrt{a^2 + b^2}$

## Worked Examples

### 1. Find all real values $$x$$ satisfying $$\lvert x - 3 \rvert = 5$$.

Solution 1: If $$x-3 \geq 0$$, then we need to solve $$x-3 = 5$$, which gives $$x = 8$$. This satisfies the original condition of $$x - 3 \geq 0$$, hence is a valid solution. If $$x-3 < 0$$, then we need to solve $$-(x-3) = 5$$, which gives $$-x+3 = 5$$ or $$x = -2$$. This satisfies the original condition of $$x-3 < 0$$, hence is a valid solution.

Solution 2: Using property 2, we obtain $$(x-3)^2 = \lvert x-3\rvert ^2 = 5^2$$, or $$x^2 - 6x + 9 = 25$$, which reduces to $$0 = x^2 -6x - 16 = (x-8)(x+2)$$. This has solutions $$x=8, -2$$. We can verify that both of these are solutions.

### 2. Find all real values of $$x$$ satisfying $$\lvert x^2 -1 \rvert = \lvert x-2 \rvert$$?

Solution: Let's approach this problem by considering the different regions. We have $$x^2 -1 \geq 0 \Leftrightarrow x\leq -1, x \geq 1$$. Also, $$x -2 \geq 0 \Leftrightarrow x \geq 2$$.

If $$x \leq -1$$ then we have $$x^2 -1 = 2 - x$$, or $$x^2 + x - 3 = 0$$. This has roots $$\frac{-1 \pm \sqrt{ 1^2 - 4 \times 1 \times (-3) }} {2}$$, of which only the choice of $$-$$ satisfies $$x < -1$$. There is one solution in this case.

If $$-1 \leq x \leq 1$$, then we have $$1 - x^2 = 2 - x$$, or $$x^2 -x + 1 = 0$$. This has roots $$\frac { 1 \pm \sqrt{ 1^2 - 4 \times 1 \times 1 } } {2}$$ which are not real. Hence, there are no solutions in this case.

If $$1 \leq x \leq 2$$, then we have $$x^2 -1 = 2 - x$$, or $$x^2 + x - 3 = 0$$. We check that the root $$\frac{ -1 + \sqrt{13} } { 2}$$ is within this domain. Hence, there is one solution in this case.

If $$2 \leq x$$, then we have $$x^2-1 = x - 2$$, or $$x^2 - x + 1 = 0$$. This has no real root.

### 3. Prove the sub-additive property:

$\lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert.$

Solution 1: We apply the triangle inequality to the triangle with vertices $$O = 0$$, $$A = x$$, and $$B = -y$$ on the real number line. Then $$AB \leq OA + OB$$, implying $$\lvert x - (-y) \rvert \leq \lvert x \rvert + \lvert y \rvert$$.

Solution 2:

If $$x, y \geq 0$$, then $$\lvert x+y \rvert = x+y, \lvert x \rvert + \lvert y \rvert = x + y$$, so $$\lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert$$.

If $$x, y \leq 0$$, then $$\lvert x+y \rvert = -x -y, \lvert x \rvert = -x, \lvert y \rvert = - y$$, so $$\lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert$$.

If $$x \geq -y \geq 0$$, then $$\lvert x+y \rvert = x+y, \lvert x \rvert = x, \lvert y \rvert = -y$$ so $$\lvert x+y \rvert = x+y \leq x -y = \lvert x \rvert + \lvert y \rvert$$. The same argument applies for $$y \geq -x \geq 0$$.

If $$x \leq -y \leq 0$$, then $$\lvert x+y \rvert = -x - y , \lvert x \rvert = -x, \lvert y \rvert = y$$, so $$\lvert x+y \rvert = -x - y \leq -x + y = \lvert x \rvert + \lvert y \rvert$$. The same argument applies for $$y \leq x \leq 0$$.

Note by Calvin Lin
3 years, 8 months ago

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