Advanced Absolute Value

This is a continuation of Absolute Value.

Properties of the absolute value:

  1. Non-negative: x0 \vert x \rvert \geq 0 for all real values.

  2. x=x2 \lvert x \rvert = \sqrt{ x^2 } and x2=x2\lvert x \rvert^2 = x^2 for all real values.

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

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

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

  6. Distance to origin: x \lvert x \rvert measures the geometric distance of the real number xx to the origin 00. 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 z = a + bi . This distance from z z to the origin is given by the distance formula: distance(a,b)(0,0)=(a0)2+(b0)2 \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 z = a+bi :

z=a2+b2 \left| z \right| = \sqrt{a^2 + b^2}

Worked Examples

1. Find all real values xx satisfying x3=5 \lvert x - 3 \rvert = 5 .

Solution 1: If x30 x-3 \geq 0 , then we need to solve x3=5 x-3 = 5 , which gives x=8 x = 8 . This satisfies the original condition of x30 x - 3 \geq 0 , hence is a valid solution. If x3<0 x-3 < 0 , then we need to solve (x3)=5 -(x-3) = 5 , which gives x+3=5 -x+3 = 5 or x=2 x = -2 . This satisfies the original condition of x3<0 x-3 < 0 , hence is a valid solution.

Solution 2: Using property 2, we obtain (x3)2=x32=52 (x-3)^2 = \lvert x-3\rvert ^2 = 5^2, or x26x+9=25 x^2 - 6x + 9 = 25 , which reduces to 0=x26x16=(x8)(x+2) 0 = x^2 -6x - 16 = (x-8)(x+2) . This has solutions x=8,2x=8, -2 . We can verify that both of these are solutions.

 

2. Find all real values of xx satisfying x21=x2 \lvert x^2 -1 \rvert = \lvert x-2 \rvert ?

Solution: Let's approach this problem by considering the different regions. We have x210x1,x1 x^2 -1 \geq 0 \Leftrightarrow x\leq -1, x \geq 1 . Also, x20x2 x -2 \geq 0 \Leftrightarrow x \geq 2 .

If x1x \leq -1 then we have x21=2x x^2 -1 = 2 - x , or x2+x3=0 x^2 + x - 3 = 0 . This has roots 1±124×1×(3)2 \frac{-1 \pm \sqrt{ 1^2 - 4 \times 1 \times (-3) }} {2} , of which only the choice of - satisfies x<1 x < -1 . There is one solution in this case.

If 1x1 -1 \leq x \leq 1 , then we have 1x2=2x 1 - x^2 = 2 - x , or x2x+1=0 x^2 -x + 1 = 0 . This has roots 1±124×1×12 \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 1x2 1 \leq x \leq 2 , then we have x21=2x x^2 -1 = 2 - x , or x2+x3=0 x^2 + x - 3 = 0. We check that the root 1+132 \frac{ -1 + \sqrt{13} } { 2} is within this domain. Hence, there is one solution in this case.

If 2x 2 \leq x , then we have x21=x2 x^2-1 = x - 2 , or x2x+1=0 x^2 - x + 1 = 0 . This has no real root.

 

3. Prove the sub-additive property:

x+yx+y. \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 O = 0 , A=x A = x , and B=y B = -y on the real number line. Then ABOA+OB AB \leq OA + OB , implying x(y)x+y \lvert x - (-y) \rvert \leq \lvert x \rvert + \lvert y \rvert .

Solution 2:

If x,y0 x, y \geq 0 , then x+y=x+y,x+y=x+y \lvert x+y \rvert = x+y, \lvert x \rvert + \lvert y \rvert = x + y , so x+yx+y \lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert .

If x,y0 x, y \leq 0 , then x+y=xy,x=x,y=y \lvert x+y \rvert = -x -y, \lvert x \rvert = -x, \lvert y \rvert = - y , so x+yx+y \lvert x+y \rvert \leq \lvert x \rvert + \lvert y \rvert .

If xy0 x \geq -y \geq 0 , then x+y=x+y,x=x,y=y \lvert x+y \rvert = x+y, \lvert x \rvert = x, \lvert y \rvert = -y so x+y=x+yxy=x+y \lvert x+y \rvert = x+y \leq x -y = \lvert x \rvert + \lvert y \rvert. The same argument applies for yx0 y \geq -x \geq 0 .

If xy0 x \leq -y \leq 0 , then x+y=xy,x=x,y=y \lvert x+y \rvert = -x - y , \lvert x \rvert = -x, \lvert y \rvert = y, so x+y=xyx+y=x+y \lvert x+y \rvert = -x - y \leq -x + y = \lvert x \rvert + \lvert y \rvert . The same argument applies for yx0 y \leq x \leq 0 .

Note by Calvin Lin
5 years, 6 months ago

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