1. Not finding help here? Sign up for a free 30min 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!

Prove the formula for the maximum of two numbers

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    The maximum of two numbers x and y is denoted by max(x,y). Thus max(-1,3) = max (3,3) = 3. Prove that:

    [tex]
    \mathrm{max}(x,y) = \frac{x+y+|y-x|}{2}[/tex]

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I have no idea where to begin. I've thought about it for a long time, I swear! Any hints on how to get started?
     
  2. jcsd
  3. Sep 16, 2009 #2
    You'll need to consider two cases: x>y and y>x. Then use that with the definition of absolute value. That should show the right results.
     
  4. Sep 16, 2009 #3
    Here's an attempt:

    Assume x > y .

    [tex]
    \begin{align*}
    \mathrm{max}(x,y) &= \frac{x+y+x-y}{2}\\
    \mathrm{max}(x,y) &= \frac{x+y+|x-y|}{2}\\
    \mathrm{max}(x,y) &= \frac{x+y+|y-x|}{2}
    \end{align*}
    [/tex]

    But I haven't proved it yet...
     
  5. Sep 16, 2009 #4
    Okay, so I'm on the right track with that last post.

    Assume y > x .
    [tex]
    \begin{align*}
    \mathrm{max}(x,y) &= \frac{x+y-x+y}{2}\\
    &= \frac{x+y+|y-x|}{2}
    \end{align*}
    [/tex]
     
  6. Sep 16, 2009 #5
    So is that a sufficient proof? Because it could just as well be

    [tex]
    \mathrm{max}(x,y) = \frac{x+y+|x-y|}{2}
    [/tex]

    couldn't it?
     
  7. Sep 16, 2009 #6
    With each case, you can show that the formula leaves you with one value x, or y, that should be enough for the proof.
     
  8. Sep 16, 2009 #7
    Ah, thank you very much.
     
  9. Sep 16, 2009 #8
    (1) Assume x > y .

    [tex]
    \begin{align*}
    \mathrm{max}(x,y) &= \frac{x+y+|y-x|}{2}\\
    &= \frac{x+y+(x-y)}{2}\\
    &= \frac{2x}{2}\\
    &= x
    \end{align*}
    [/tex]


    (2) Now assume y > x .

    [tex]
    \begin{align*}
    \mathrm{max}(x,y) &= \frac{x+y+|y-x|}{2}\\
    &= \frac{x+y+(y-x)}{2}\\
    &= \frac{2y}{2}\\
    &= y
    \end{align*}
    [/tex]


    Therefore:
    [tex]
    \mathrm{max}(x,y) &= \frac{x+y+|y-x|}{2}
    [/tex]

    is true for all values of x,y provided x,y are real numbers.

    (Although I'm still not thoroughly convinced that it's sufficient...)
     
  10. Sep 16, 2009 #9
    Since x and y are arbitrary real numbers, I don't see why it wouldn't be considered a strong enough proof. You can make an argument about how you come to that particular formula, but I don't think it's necessary.
     
  11. Sep 16, 2009 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    There is the third possibility that y=x. Include and your proof is complete.

    ehild
     
  12. Sep 17, 2009 #11
    Thanks very much.
     
  13. Sep 18, 2009 #12
    you're reading from the spivak textbook are you? your post on x^2 + xy + y^2 > 0 was also an exercise from the textbook
     
  14. Sep 18, 2009 #13
    yes i am. it's such an awful textbook, it gives practically no examples.

    or maybe i should say that it's a textbook that require a bit more thinking...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook