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Induction Proof

  1. Oct 4, 2008 #1
    What is wrong with this inductive proof that all protrons have the same mass?
    1. if you have one protron then it obviously has mass A
    2. Assume that it is true for a set of n protrons, i.e. that any set of n protrons has the mass n*A.
    3. Now look at a set of n+1 protrons, remove one then it has mass n*A by our hypothesis. This is true any of the protrons are removed, thus all protrons in the set must have the same mass A. Since there are n+1 protrons in the set the mass must then be (N+1)*A.

    We have completed all three steps required for an induction proof so now we know that all protrons must have the same mass no matter what speed they are moving at.
  2. jcsd
  3. Oct 5, 2008 #2
    When you add the (n+1)th proton, you are assuming it has mass A.
    Since this is not a special proton, you are assuming any proton has mass A.
    In other words, the step 3 is true only if any proton has mass A.
    And then, you've finished a proof that
    "Every group of N protons, each proton with mass A, has a total mass N*A"

    Unfortunately, you didn't prove that "all protons have the same mass"...:-)
    Last edited: Oct 5, 2008
  4. Oct 5, 2008 #3

    In short: Step 3 works when going from 2 protons to 3, from 3 to 4, ... but not from 1 to 2.
  5. Oct 5, 2008 #4
    We assumed that all protrons from any set have the same mass. However, you can't prove that by looking at a single member of a set. So no proof of the therom was made for n = 1 and step one has yet to be done. You simply cant talk of doing step three when step one has not been done. Your saying that step three works for going from sets of 2 to 3 protrons, from 3 to 4, ... is like saying that if there are 2 protrons of equal mass then any set of two, three, four ... protrons all have the same mass which again is the same as saying that all protrons have the same mass (which as you admit has not been proven).
    Last edited: Oct 5, 2008
  6. Oct 6, 2008 #5


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    Your base case and inductive step don't match. Your base case is "this particular proton has mass A" and your inductive assumption is "a group of any n protons has mass nA". To use that inductive assumption you'd need the stronger base case "any group of protons of size 1 has mass A", that is, what you're trying to prove!
  7. Oct 6, 2008 #6
    Let me try to be clearer. I will divide your Step 3 in smaller steps:

    3.1. Let S be a set of n+1 protons.
    3.2. Label any of the protons in S as a.
    3.3. Label any (other than a) of the protons in S as b.
    3.4. Let R ("the rest") be the set of all other protons in S other than a or b, that is, R = S - {a,b}.
    3.5. Consider now the set S - {a} = {b} U R; it has n protons, and by the inductive hypothesis all have the same mass.
    3.6. Now consider the set S - {b} = {a} U R; it also has n protons, and by the inductive hypothesis all have the same mass.
    3.7. Both a and b were proven (in 3.5 and 3.6) to have the same mass as any other proton in R; being equal to a third proton, they must be equal to each other. Thus a has the same mass as b.
    3.8. Since a,b were any arbitrary pair of protons in S, and they are equal in mass, we conclude that the set S is made of protons of the same mass.

    Now, the problem is:
    when you go from 1 proton to 2, the set R is empty, and 3.7 fails (there is no such thing as "a third proton", thus a and b are not forced to be equal).
    Last edited: Oct 6, 2008
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