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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

    MarneMath

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    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

    MarneMath

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    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

    MarneMath

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    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!
     
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