Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Properties of the Absolute Value

  1. Jul 25, 2008 #1
    Just wanted to say hi before I start my post! :smile:

    As you may know there is a property of the absolute value that states; for [tex]a, b \in R[/tex];

    [tex]|ab| = |a||b|[/tex]

    Well, my friend asked me if I knew a proof for this... but I don't know...
    How can we prove this statement/property? I know there is a proof for the triangle inequality but for this I really don't know but I'm curious.

    I'd appreciate it if anyone could show me any kind of proof or send me some links etc. Thanks!
  2. jcsd
  3. Jul 25, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    You prove that |ab|= |a||b| the same way you prove any such elementary statement: use the definitions.

    The simplest definition of |x| (there are several equivalent definitions) is that |x|= x if x is positive or 0, -x if x is negative.

    Now break it into "cases":

    case 1: x and y are both positive: |x|= x and |y|= y. xy is also positive so |xy|= xy= |x||y|.

    case 2: x is positive while y is negative: |x|= x and |y|= -y. xy is negative so |xy|= -xy= x(-y)= |x||y|.

    case 3: x is negative while y is positive: |x|= -x and |y|= y. xy is negative so |xy|= -xy= (-x)y= |x||y|.

    case 4: x and y are both negative: |x|= -x and |y|= -y. xy is positive so |xy|= xy= (-x)(-y)= |x||y|.

    case 5: x= 0 and y is positive: |x|= 0 and |y|= y. xy= 0 so |xy|= 0= 0(y)= |x||y|.

    case 6: x= 0 and y is negative: |x|= 0 and |y|= -y. xy= 0 so |xy|= 0= (0)(-y)= |x||y|.

    case 7: x is positive and y is 0: |x|= x and |y|= 0. xy= 0 so |xy|= 0= x(0)= |x||y|.

    case 8: x is negative and y is 0: |x|= -x and |y|= 0. xy= 0 so |xy|= 0 = (-x)(0)= |x||y|.

    case 9: both x and y are 0: |x|= 0 and |y|= 0. xy= 0 so |xy|= 0= (0)(0)= |x||y|.

    There are simpler ways to prove that but I thought this would be conceptually clearest.
  4. Jul 25, 2008 #3
    First, we have to understand that the absolute value is a function defined by:

    [tex]|x| = \begin{cases}
    x & \text{if } x\geq 0 \\
    -x & \text{if } x<0


    [tex]|ab| = \begin{cases}
    ab & \text{if } ab\geq 0 \\
    -ab & \text{if } ab<0

    Now, let's see what |a||b| is:

    [tex]|a||b| = \begin{cases}
    ab & \text{if } a\geq 0 \wedge b\geq0 \\
    (-a)(-b) & \text{if } a\leq 0 \wedge b\leq0 \\
    (-a)b & \text{if } a> 0 \wedge b<0 \\
    a(-b) & \text{if } a<0 \wedge b<0
    \end{cases} \Leftrightarrow

    [tex]|a||b| = \begin{cases}
    ab & \text{if } a\geq 0 \wedge b\geq0 \\
    ab & \text{if } a\leq 0 \wedge b\leq0 \\
    -ab & \text{if } a> 0 \wedge b<0 \\
    -ab & \text{if } a<0 \wedge b<0
    \end{cases} \Leftrightarrow

    Notice that you have ab if a and b have the same sign and that you use -ab otherwise.

    Now, if a and b have the same sign, [tex]ab\geq0[/tex]. If they have opposite signs (and are different than zero), [tex]ab<0[/tex].

    Using this,

    [tex]|a||b| = \begin{cases}
    ab & \text{if } ab \geq 0 \\
    -ab & \text{if } ab<0
    \end{cases} = |ab|

    Quod erat demonstrandum :tongue2:
    Last edited: Jul 25, 2008
  5. Jul 28, 2008 #4
    let a = mcis(kpi) where m => 0, k is integer
    and b = ncis(hpi) where n => 0, h integer
    (clearly a,b are real)

    |a||b|= mn
    |ab|=|mcis(kpi)*ncis(hpi)| = |mncis[pi(k+h)]| = mn

    as required
  6. Jul 28, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    This would make more sense if you had said that "mcis(hpi)" is
    [tex]m (cos(h\pi)+ i sin(h\pi))[/tex]
    That is much more an "engineering notation" than mathematics.

    If you really want to go to complex numbers, why not
    [tex]x= r_xe^{i\theta_x}[/tex]
    [tex]y= r_ye^{i\theta_y}[/tex], then
    [tex]|xy|= |r_xe^{i\theta_x}r_ye^{i\theta_y}|= |(r_xr_y)e^{i(\theta_x+\theta_y)}|[/tex]

    But for any [itex]z= re^{i\theta}[/itex], |z|= r, so
    [tex]|xy|= r_x r_y= |x||y|[/tex]
  7. Jan 8, 2011 #6
    [tex]\Huge |ab|=\sqrt{(ab)^2}=\sqrt{a^2b^2}=\sqrt{a^2}\sqrt{b^2}=|a||b|[/tex]


    [tex]\large |a+b|=\sqrt{(a+b)^2}=\sqrt{a^2+2ab+b^2}\leq \sqrt{a^2+|2ab|+b^2}=\sqrt{(|a|+|b|)^2}=||a|+|b||[/tex]

    where I utilize the following fact:


    À bientôt
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?