1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Prove ||a|-|b|| ≤ |a-b|

  1. Dec 26, 2012 #1
    Hey again! I've got another problem I'd like to check for adequacy. I'm pretty sure I've got all the cases covered, but I want to make sure the work here is satisfactory.

    1. The problem statement, all variables and given/known data

    Prove that for all real numbers [itex]a[/itex] and [itex]b[/itex],
    [itex]||a| - |b|| \leq |a - b|[/itex].

    2. Relevant equations

    The book I'm working with uses the following definition for absolute value:

    [itex]|x| = \begin{cases} x, & \mbox{if } x \geq 0 \\ -x, & \mbox{if } x < 0 \end{cases}[/itex]

    And for later use:
    [itex]|a| = \begin{cases} a, & \mbox{if } a \geq 0 \\ -a, & \mbox{if } a < 0 \end{cases}[/itex]
    [itex]|b| = \begin{cases} b, & \mbox{if } b \geq 0 \\ -b, & \mbox{if } b < 0 \end{cases}[/itex]
    [itex]|a-b| = \begin{cases} a-b, & \mbox{if } a-b \geq 0 \\ b-a, & \mbox{if } a-b < 0 \end{cases}[/itex]

    3. The attempt at a solution
    By definition,
    [itex]||a|-|b|| = \begin{cases} |a|-|b|, & \mbox{if } |a|-|b| \geq 0, \mbox{ i.e. } |a|\geq|b| \\ -(|a|-|b|) = |b|-|a|, & \mbox{if } |a|-|b|< 0, \mbox{ i.e. } |a|<|b| \end{cases}[/itex]

    Suppose [itex]|a| = |b|[/itex]. (I initially let a = b, but using |a|=|b| is more fitting, right?)
    Then, [itex]||a|-|b|| = ||a|-|a|| = |0| = 0[/itex]. So, [itex]0 \leq |a-b|[/itex].
    If [itex]a=b[/itex], then [itex]a-b=0[/itex] and we have [itex]0 \leq 0[/itex], which is true. If [itex]a \not= b[/itex], then [itex]|a-b|>0[/itex] and we have [itex]0 \leq |a-b|[/itex], which is true.

    For this reason, I modify the definition of [itex]||a|-|b||[/itex] to be
    [itex]||a|-|b|| = \begin{cases} |a|-|b|, & \mbox{if } |a|>|b| \\ |b|-|a|, & \mbox{if } |a|<|b| \end{cases}[/itex]
    (I hope there aren't any problems with this step. I thought it would simplify the later sub-cases.)

    Case 1: Suppose [itex]|a|>|b|[/itex].

    Case 1(a): Suppose [itex]a,b \geq 0[/itex]. Then [itex]|a|=a[/itex] and [itex]|b|=b[/itex].
    We thus have [itex]||a|-|b|| = |a-b| \leq |a-b|[/itex], which is true.

    Case 1(b): Suppose [itex]a,b<0[/itex]. Then [itex]|a|=-a[/itex] and [itex]|b|=-b[/itex].
    [itex](|a|>|b|) \; \Rightarrow \; (-a>-b) \; \Rightarrow \; (a-b<0) \; \Rightarrow \; (|a-b|=b-a)[/itex].
    We have [itex]||a|-|b|| = |a|-|b| = -a-(-b) = b-a[/itex], and so we have [itex]b-a \leq |a-b| = b-a[/itex], which is true.

    Case 1(c): Suppose [itex]a\geq 0, b<0[/itex]. Then [itex]|a|=a[/itex] and [itex]|b|=-b[/itex].
    [itex](a\geq 0, b<0) \; \Rightarrow \; (a>b) \; \Rightarrow \; (a-b>0) \; \Rightarrow \; (|a-b|= a-b = a+(-b) = |a|+|b|)[/itex].
    We have [itex]||a|-|b|| = |a-(-b)| = |a+b|[/itex], which means we have
    [itex]|a+b| \leq |a|+|b|[/itex], which is true (by the Triangle Inequality Theorem).

    Case 1(d): Suppose [itex]a<0, b\geq 0[/itex]. Then [itex]|a|=-a[/itex] and [itex]|b|=b[/itex].
    [itex](a<0, b\geq 0) \; \Rightarrow \; (b>a) \; \Rightarrow \; (a-b<0) \; \Rightarrow \; (|a-b|= b-a = b+(-a) = |a|+|b|)[/itex].
    We have [itex]||a|-|b|| = |-a-b| = |-(a+b)| = |a+b|[/itex], and so we have
    [itex]|a+b| \leq |a|+|b|[/itex], as in Case 1(c).

    Case 2: Suppose [itex]|a|<|b|[/itex].

    Case 2(a): Suppose [itex]a,b \geq 0[/itex]. This case is identical to Case 1(a).

    Case 2(b): Suppose [itex]a,b<0[/itex]. Then [itex]|a|=-a[/itex] and [itex]|b|=-b[/itex].
    [itex](|a|<|b|) \; \Rightarrow \; (-a<-b) \; \Rightarrow \; (a>b) \; \Rightarrow \; (a-b>0) \; \Rightarrow \; (|a-b|=a-b)[/itex].
    We also have [itex]||a|-|b|| = |b|-|a| = -b-(-a) = a-b[/itex], and so we have
    [itex]a-b \leq |a-b| = a-b[/itex], which is true.

    Case 2(c): Suppose [itex]a\geq 0, b<0[/itex]. This case is identical to Case 1(c).

    Case 2(d): Suppose [itex]a<0, b \geq 0[/itex]. This case is identical to Case 1(d).​

    Thus, for all real numbers [itex]a[/itex] and [itex]b[/itex],
    [itex]||a| - |b|| \leq |a - b|[/itex].

    (I say the sub-cases under case 2 are identical to those in case 1 since they don't utilize the modified definition.)

    If I'm missing anything important, please do tell! Thanks
  2. jcsd
  3. Dec 26, 2012 #2


    User Avatar
    Education Advisor

    Just a quick read, it looks right, but tedious. There is a much much much easier way to do this. If you're interested, just tell me and I'll help you out.
  4. Dec 26, 2012 #3
    I had a feeling there was, but I wanted to be thorough. What can I omit? I would think some of the algebra isn't all that necessary.
  5. Dec 26, 2012 #4


    User Avatar
    Education Advisor

    If you can prove lx|-|y| less than or equal to |x - y| then it's pretty simple if you think about it the right way.
  6. Dec 26, 2012 #5
    It looks like I partially proved that in cases 1(b) and 2(b).
    Would it have anything to do with the fact that they're the only cases that actually consider the assumptions made in "Case 1:..." and "Case 2:..."?
  7. Dec 26, 2012 #6


    User Avatar
    Education Advisor

    More along the line that if you make a 'clever' change in variables, then the equality just appears. (By the way, if you're really not interested in learning an alternative proof, this convo isn't important. I just like short simple proofs. I reread your proof and I haven't found a mistake, besides some minor math grammar things, but other than that it looks good. It's the way I would've done it when I was learning how to write proofs.)
  8. Dec 26, 2012 #7
    Ah, well, at least I'm on the right track. Thanks for the input, though, much obliged!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook