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Average Proof

  1. Jan 15, 2014 #1


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    1. The problem statement, all variables and given/known data
    Obviously this is not a homework question,but I put this here as this looks like one.
    Lets see.
    The average of something is [itex]\frac{Sum of all the numbers}{Amount of the numbers.}[/itex]
    For example,1,2,3,4,5,6
    The average is 3.5 which is also the middle of the sequence.
    If the numbers were 1,4,8,9,11
    Average is 6.6
    Which is not the middle of course.
    Anyway,Why does the average give the middle number?

    3. The attempt at a solution
    Let x and y be integers.
    [itex]\frac{x+y}{2}[/itex] =Average
    But I don't see the logic behind this.

    Also let x and y be integers.and y is the largest number
    First y-x to find the range.
    So [itex]\frac{y-x}{2}[/itex] gives the middle number between y and x range
    So we add the middle number to the x to get the middle number between y and x.
    Rearranging,this is same as the average formula.
    So is there another way of logicing(1) this?
    P.S(1) Couln't think of another word
  2. jcsd
  3. Jan 15, 2014 #2


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    Staff: Mentor

    No. This gives you the distance between x and the middle number. To get the middle number, you need to add to the distance the starting point. Hence
    \frac{y-x}{2} + x = \frac{x+y}{2}
    which is indeed the average.
  4. Jan 15, 2014 #3


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    Read a little further and you'll see that's what he meant.

    A physical interpretation I like is that the average will give you the center of mass. Take a plank of wood and distribute n equal weights amongst it. If you treat the plank of wood as a ruler (distance values from 0-100 say) and then the average of all the values that the weights are located at will be the position where the plank would balance if held up by a pivot.
  5. Jan 15, 2014 #4


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    As Dr. Claude said, this does not give the middle number.

    This makes no sense. If you already have the middle number to add to the x, adding x will give you something else, not the middle number again.

    I think what you are trying to say is that y- x is the distance between x and y so that [itex]\frac{x- y}{2}[/itex] is half the distance from y to x, not "the middle number". Suppose x and y were the distances from your present location to two cities, X and Y, along a straight road, X being closer than Y (x being less than y). Then [itex]\frac{y- x}{2}[/itex] would be half the distance between them.

    To go from your present location to the point halfway between them you must go to X, then half way between X and Y. That is you must go distance x first, then distance (y- x)/2. That gives, as you say,
    [tex]x+ \frac{y- x}{2}= \frac{2x}{2}+ \frac{y- x}{2}= \frac{2x+ y- x}{2}= \frac{x+ y}{2}[/tex]
  6. Jan 15, 2014 #5


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    ok ok..It's not the middle number,the distance from x to middle number.Sorry.
    From the posts of all,I did not see any other way of proving that (x+y)/2 gives the middle number(or median) between them.
    How was this formula derived?
  7. Jan 15, 2014 #6


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    How about

    y - (y-x)/2

  8. Jan 15, 2014 #7


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    That is not a proof.
    That is equal to saying the proof of: X*Y gives XY: is YX
  9. Jan 15, 2014 #8


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    I disagree!

    And the only "proof" given to you so far was that it is equivalent to x+(y-x)/2, hence you're satisfied with the proof that X*Y = XY. Now I'm showing you a whole other world with X*Y = YX :biggrin:

    Another idea I have is essentially given in the formula of the mean, but it might lead to some insight that you could have missed.

    If we have two numbers x and y, then we're looking for a number z such that x+y = z+z.


    x+y = z+z
    x+y = 2z
    (x+y)/2 = z

    EDIT: I realize this didn't answer your question. So how can we show that z is in the middle of x and y? Well, all I can think of is to say that y-z = z-x must be satisfied. Given what we found z to be earlier, this equality does indeed hold, hence it's the middle number.
    Last edited: Jan 15, 2014
  10. Jan 15, 2014 #9

    Ray Vickson

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    If ##x < y## (integers or not--it doesn't matter), any number ##z## between ##x## and ##y## can be written in the form
    [tex] z = (1-t) x + ty, \;\; 0 \leq t \leq 1[/tex]
    When ##t=0## we have ##z=x## and when ##t=1## we have ##z=y##. We can also write this as
    [tex] z = x + t(y-x) [/tex]
    which means that ##t## is the "fraction of the way from ##x## to ##y##" at the point ##z##. When we put ##t=1/2## we will be at the mid-point of the segment between ##x## and ##y##. Of course, we can re-write the corresponding ##z## as ##z = x + (1/2)(y-x) = (x+y)/2##.
  11. Jan 15, 2014 #10


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    I see.Thank you all
    I will think about this further tomorrow. :)))
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