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Proof on Gauß

  1. Mar 17, 2013 #1
    Hello math&physic fans,

    I might found a proof on Gauß formula (n²+n/2)=1+2+...n). Have anoyone seen it before?
    I´m sorry for my bad english and the bad handwriting.
    (I wrote n#, just as an abbreviation, instead of 1+2+3...+n)
    Last edited: Mar 17, 2013
  2. jcsd
  3. Mar 17, 2013 #2
    Hi, Lls27,
    your attempt at a proof is a bit messy (not only the handwriting, but with that many corrections I think some signs are not properly carried through) - for the next time, do take the time to make a clean copy. "The devil is in the details"!

    Ignoring these issues, there are two things I'd like to comment. First is that you did a clever reorganization in the second part of your "proof" (this was good - I will elaborate on this later). And second, unfortunately, you began from an observation in a sketch, from the two cases[tex]\begin{align*}
    (3 + 4) - (1 + 2) &= 2^2 \\
    (4 + 5 + 6) - (1 + 2 + 3) &= 3^2
    \end{align*}[/tex]but these are only two cases; this does not constitute a proof that the general rule[tex]\big( (n+1) + (n+2) + ... + 2n \big) - \big(1 + 2 + ... + n) = n^2[/tex]is true for any [itex]n[/itex], not just for 2 or 3. It is an astute observation, but not conclusive; and, since the rest of the proof is based on this unproven statement, the proof is not valid. (Actually, the easiest way I can think of proving this initial statement is by using Gauss' formula as the starting point!)

    Now, I'd like you to notice that the reorganization you did afterwards (or, part of it) could have been used to prove Gauss' formula, without using the initial sketch. Let me clear it up for you.

    For the purpose of this explanation I'll follow your (unorthodox) notation,[tex]n\# = 1 + 2 + 3 + ... + n[/tex](but please notice that the # symbol is already used in number theory for something else - google for "primorial").

    At some moment, you did something like[tex]\begin{align*}
    n\# &= 0 + 1 + 2 + ... + n \\
    &= (n-n) + ... + (n-2) + (n-1) + n \\
    &= \underbrace{(n+n+n+...+n)}_{n \scriptsize\mbox{ times}} + (-n - ... -2 -1) + n \\
    &= n^2 - n\# + n
    \end{align*}[/tex]So you have arrived at [itex]n\# = n^2 - n\# + n[/itex]. From here it's just a small step to Gauss' formula.
    Last edited: Mar 17, 2013
  4. Mar 17, 2013 #3
    Hello dodo,

    first of all I´m very thankful for your long and helpful comment.
    The first two calculations were just suome examples. The hole paper was intended for a friend. So I wasn´t very eager on a nice paper (sorry for that) and that´s why I used the n# abbrevation.
    In my "proof" I already assumed the equaltion ((n+1)+(n+2)+...+2n)−(1+2+...+n)=n^2 is true.
    I just had this idea (randomly) and that was my result. I just wanted to share it and see whether it´s right and if somebody noticed this before. That´s one of my first attempts and my basic knowledge is unfortunately (with 16 years) not very advanced.
    But am I right if I say that I can prove the formula ((n+1)+(n+2)+...+2n)−(1+2+...+n)=n^2 with the sum formula?!

    Thanks again for the responding :)
  5. Mar 17, 2013 #4
    Yes. Let me introduce some 'proper' notation: define the function[tex]f(n) = 1 + 2 + \ldots + n[/tex], and Gauss formula gives you[tex]f(n) = \frac {n(n+1)} 2[/tex]The formula you want to prove has a subtraction on the left-hand side of the equation; the two terms subtracted are "the sum of consecutive integers from (n+1) to 2n", that is, [itex]f(2n)-f(n)[/itex], minus "the sum of integers from 1 to n", which we called [itex]f(n)[/itex]; so, in this notation, you want to prove[tex]\big( f(2n)-f(n) \big) - f(n) = n^2[/tex]
    Just apply Gauss' formula and expand the parentheses properly, and you'll have your proof.
  6. Mar 17, 2013 #5

    thanks very much :D
  7. Mar 17, 2013 #6
    one more question
    it could be also a proof for Gauß, by using the form of Reductio ad absurdum?
  8. Mar 17, 2013 #7
    Using a contradiction to prove a statement is a valid method, but I don't understand how do you want to apply it in this case. You'll have to describe what you want to do a little more.

    P.S.: I suppose you could assume that 1 + 2 + 3 + ... + n is not equal to n(n+1)/2, and then work on that until you find a contradiction. This is possible, in principle, but it can get a bit long; normally you'd use "reductio ad absurdum" when (a) you already have an idea of the contradiction you plan to arrive to, or (b) when your initial statement looks like a "negative" statement, maybe of the form "prove that no integer exists such that ...". But even this is not really 100% true. Proof by contradiction is often a "last resort", when there is nothing else to try. The moral is, try something else first.

    A typical method to prove Gauss' formula is by "mathematical induction"; if you're interested, you may want to see this video,
    or this post

    . . .

    Just to make previous things clear (though I believe you already understood this), post#1-2 and post#4 are two separate proofs. One proof is using ((n+1)+(n+2)+...+2n) - (1+2+...+n) = n^2 to prove Gauss' formula, and the other is using Gauss' formula to prove that ((n+1)+(n+2)+...+2n) - (1+2+...+n) = n^2. Obviously, these two arguments cannot be part of the same proof (or you'd have a circular argument: using A to prove B, then using B to prove A, which would prove nothing). Just to be clear.

    (P.S.: Of course, you could use the method at the end of post#2 to prove Gauss' formula on its own, then proceed from here to prove the larger formula.) (Sorry for the extra mess.)
    Last edited: Mar 17, 2013
  9. Mar 18, 2013 #8
    Alright thanks again
    Very helpful !
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