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Homework Help: Question about proof

  1. Aug 14, 2013 #1
    The explanation given in the paint document that I copied from a book does not seem complete.

    They are saying that the sum of two numbers is least when the two numbers are equal.

    Here is my explanation for why this is not always true.

    Let a and b be two positive numbers.

    I will denote the sum S1 as 2a and the sum S2 as a + b

    Then I can say that S1>S2 if a>b and therefore S1 = S2 for b = a which is greater than S2 for a>b.

    Second explanation: Let a and b be a positive number and I will define the sum S1 to be the sum of a finite number of n terms therefore
    S1 = na = M = M + (b-b) = M + 0
    If I allow M = na to equal the second term in S1 then I have na = 0 or a = 0, however I defined n to be positive and therefore a must equal itself. and therefore I can conclude that
    na>0 or S1> n*0 = 0 where RHS is equal to the sum of n zeros.

    Am I misinterpreting what they are saying in the paint document?

    Attached Files:

    • Sum.jpg
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    Last edited: Aug 14, 2013
  2. jcsd
  3. Aug 14, 2013 #2

    Ray Vickson

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    It never said anything like what you claim. It says that if the *sum is given*, the *product* is maximized when the two numbers are equal, and if the *product is given* the *sum* is minimized when the two numbers are equal.
  4. Aug 14, 2013 #3
    Give me an example
  5. Aug 14, 2013 #4
    Wouldn't it be easier to explain with an example. You could say the sum is equal to 6
    S = 6 = 1+5 = 2+4 = 3+3
    1*5 = 5
    2*4 = 8
    3*3 = 9
  6. Aug 14, 2013 #5
    Let's look at a concrete example.

    Suppose a + b = 6. Here I will only consider a few possibilities. We could have (a, b) = (1, 5) or (2, 4) or (3, 3) or (3, 3), or (4, 2) or (5, 1). Then we could write a = 6 - b. The product of these two numbers can be written P = ab = (6-b)b = 6b - b^2 or 0 = b^2 - 6b + P (a quadratic in b). To show that there is a max (or min) when b = a is easy if we can differentiate with respect to b and set the derivative to 0 and solve for b (we should find b = 3 for this case). Indeed, then P = 9 which is, indeed the max (you could check this for a few values for yourself in order to "convince" yourself of its validity without rigour. That being said, you should try to prove it with more rigour than I've provided in this one example of its validity.

    A similar but "reverse" idea can be applied to show that if P is given S is least when a = b.

    EDIT: As you said, it's easier to 'explain' by using an example. However, the example has only shown to be true for that specific value (ie. a+b = 6). We would prefer to show it is true in general for any positive a and b. Another thing is to note that you only considered whole numbers - what about every other combination of numbers - 1.1 and 4.9, does it hold true for them? (Yes.)

    But what if I asked you to show it for every combination of positive numbers that sums to 6? You would have to explicitly show me an infinite number of calculations before I could truly be 100% convinced of its validity.
  7. Aug 14, 2013 #6
    Last edited: Aug 14, 2013
  8. Aug 15, 2013 #7

    Ray Vickson

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    No need; the paint document he attached proved it in general.
  9. Aug 15, 2013 #8
    Oops, I guess I completely misread that. However, if you were to try to prove this yourself it would not suffice to provide one example in which it is true.
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