Talk:High School Mathematics Extensions/Mathematical Proofs

Could anybody prove that $$n-m = -1*(m-n)$$? r3m0t 14:44, 13 Dec 2003 (UTC)

comment 1
Uhhh... Take the right hand side and expand
 * -m - (-1)n

which is equivalent to
 * -m+(-1)(-1)n

which is
 * -m+1n
 * n-m

which is the desired result.

Dysprosia 00:03, 14 Dec 2003 (UTC)


 * *grin* Cool. Now I really know it's true! ;) r3m0t 00:32, 14 Dec 2003 (UTC)

I found a good example for the visual proofs section - prove that the series $$1^2+2^2+...+n^2$$ multiplies by 4 and adds 1 when n is increased by one. No idea? Think about how many squares there are on a 16&times;16 grid. 212.135.1.53 08:29, 5 Jan 2004 (UTC) (r3m0t)


 * Okay, good work. Let me think why it is so... Xiaodai 09:49, 5 Jan 2004 (UTC)
 * Whoops. Never even tried it out on paper, I was so convinced. No, that isn't even true *grins*. Ah well, in which case, a good recursion exercise is finding out how many squares (of any size) there are on a 16&times;16 grid. The answer happens to be 1,496 (or aleph-one, depending on how you interpret the question...) I will however try and investigate how to fix that times 4 and plus 1 rule, but chances are the outcome will be pretty complicated. r3m0t 19:11, 5 Jan 2004 (UTC)

Is the following line correct, it would seem more logical to prove v=w instead of v = u: 2. Prove using only the axioms if u + v = u + w then v = u (subtracting u from both sides is not accepted as a solution) I can't find the solution to this problem, so it would be great if you could give me a hint Problemset nr 1 seems to be wrong as well, the greater than sign should be turned around, otherwise simply trying n=2 will show you that it's not trueMmartin 19 july 2004
 * Thanks I corrected the error. As an hint to this type of problems, we always use 0 e.g. a = a + 0. If you are still scratching your head, the solutions is as follows:

\begin{matrix} v &=& v + 0\\ &=& v + (u - u)\\ &=& (v + u) - u\\ &=& (w + u) - u\\ &=& w + (u - u)\\ &=& w + 0 \\ &=& w \end{matrix} $$
 * as required. Thank you very much for you question. I love getting questions, it assures me that somebody out there is reading my work. Xiaodai 08:42, 20 Jul 2004 (UTC)

Hum??
I've trouble on doing question 3 of the first exercise. Can you give me some hints? (Also it seems the question itself has some minor typing mistakes) -lemontea


 * Thanks for pointing it out lemontea, fixed. Xiaodai 04:11, 18 Dec 2004 (UTC)


 * Now I'm completely stucked. I tried doing binomial expansion of that (i-1)^k and cancel out the highest power i^k, leaving only smaller power behind so I can employ induction, but this method failed. How did you do it? (Though I have an idea on making use of the generating function, but them this can expand to a project which the size is comparable to the project in the first chapter) Lemontea 18:11, 26 Dec 2004 (GMT+8)


 * I hope you dont mind, i will write out the complete solution here. You may have been confused by the questions. Anyway..


 * It's clear that 11 + 21 + ... = (n+1)n/2. So we can start the induction from there.
 * Suppose that
 * $$\sum_{i=1}^ni^j$$
 * has an explicit formula in terms of n for all j < k (**), we aim to prove that
 * $$\sum_{i=1}^ni^k$$
 * also has an explicit formula.


 * Indeed, consider
 * $$\sum_{i=1}^n[i^{k+1} - (i-1)^{k+1}] = n^{k+1}$$
 * I think you may have consider ^k instead of ^(k+1). Also, looking at it from another way
 * $$\sum_{i=1}^n[i^{k+1} - (i-1)^k] = \sum_{i=1}^n[i^{k+1} - i^{k+1} + i^k - i^{k-1} + ... \ 1]$$
 * as you have mentioned the ik+1 cancels, leaving smaller powers, rearraging to get
 * $$\sum_{i=1}^ni^k$$
 * on the left hand side and using the induction hypothesis (**) then we are done. Hope this has been helpful. Xiaodai 00:13, 28 Dec 2004 (UTC)

Thanks! That helped. (Oh no, I found I was just a few steps away from the end of the solution, a ghost must have covered my eyes, ahhhhhhhhh...) However, it seems the ending part could be written more clearly, and the above did contain some minor error... $$\sum_{i=1}^n[i^{k+1} - (i-1)^{k(??)}]$$ $$ = \sum_{i=1}^n[i^{k+1} - i^{k+1} + i^k - i^{k-1} + ... \ 1]$$(Shouldn't there be some binomial coefficient when expanding out?)

I thought the principle would be like the below: $$\sum_{i=1}^n[i^{k+1} - (i-1)^{k+1}] = n^{k+1}$$ $$\sum_{i=1}^n[i^{k+1} - \sum_{j=0}^{k+1} {k+1 \choose j} i^j] = n^{k+1}$$ $$\sum_{i=1}^n[i^{k+1} - {k+1 \choose k+1} i^{k+1} - \sum_{j=0}^k {k+1 \choose j} i^j] = n^{k+1}$$ $$\sum_{i=1}^n[\sum_{j=0}^k {k+1 \choose j} i^j] = n^{k+1}$$ $$\sum_{j=0}^k[\sum_{i=1}^n {k+1 \choose j} i^j] = n^{k+1}$$(Huh, I think I'll have to write another sub-chapter on the nested/2-dimensional sum notation operation...) $$\sum_{j=0}^k[{k+1 \choose j} \sum_{i=1}^n i^j] = n^{k+1}$$ I think things should be clear now :)

However, after doing this question I feel the urge to feedback my strong opinion on this solution. The exsistence of a formula is usually to help people do calculation faster and easier, but like above, the result is that the formula relies on combining all of the pervious formula in that series, although the formula is still a finite calculation, but when k increases to somewhere like 10, the so called "formula" will become so complex that it lost its elegence, and that it doesn't serve the function of reliefing difficulties in calculations anymore. Lemontea 10:37, 28 Dec 2004 (GMT+8)
 * You are right about the coefficient, i just forgot. Good wook. You comment about the solution is correct, may i remind you that the point of the question is to test induction skills, although you are absolutely right about the deficiency of the question. Thanks. Xiaodai 03:01, 28 Dec 2004 (UTC)

I know that for mathematicians, it is considered an art to say just enough....to find a balance between not giving every single step, but yet making the point clear to a fellow mathematician. However, non-mathematicians don't necessarily think this way. I know we want the reader to think and ask, but we don't also want them to think they're stupid and give up. It seems that one of the beauties of this sort of media is that perhaps we can provide the extra explanation, yet bury it with hyperlinks, etc. The above discussion is one such example.Lizmaemom (discuss • contribs) 23:56, 3 July 2011 (UTC)

M5?
M5:for every a there exist b such that ab=1

What about 0? Does a inverse of 0 exist? --Lemontea 13:42, 1 Mar 2005 (UTC)


 * See, that's what I mean about messy (I don't mean the omission, that's an easy mistake to make, but the way the thing was structured). There's a better way of saying this (which I'll just say now, so you'll see what I mean). Dysprosia 22:25, 1 Mar 2005 (UTC)

Incorrect Proof
A lot can be learned by being asked to find the mistaken logic in an incorrect proof, of which many can be found online. One is as follows for Goldbach's conjecture:
 * step 1: Any even number >=6 can be described by 6Q +b, b={-2,0,+2}

(thus any even number is either 2 behind, 2 ahead, or coincident with, a multiple of 6) Thus, we can write E = 6Q +b
 * step 2: a recognized required (but not sufficient) condition for any prime>=5 is that they are of the form 6N +/1 (it is 1 ahead or behind a multiple of 6)


 * step 3: rewrite step 1 as  (any) even = 6L+/-1 + 6M +/-1

We can substitute step 2 into step 3 and get:
 * step 4: (any >3) even  = prime1 + prime2

the above essentially "proves" GC - where is the mistake in the proof?--Billymac00 (talk) 02:41, 10 March 2008 (UTC)

Feedback
This section is fascinating. I hope you can improve it. Some examples are tough to understand and I hope you can explain them better. --200.120.252.46 (talk) 01:08, 5 May 2010 (UTC)

Shouldn't math mode be used for writing mathematical statements? --ranade

Additive Inverse ?
Here is axiom three in the Axioms and inference section : A3: For every a, there exist b (written -a), such that a + b = 0 (existence of an additive inverse) Not that I am mathematician or anything but I think inverse is not the right word, -a is the opposit of a, right?

Would you have a place for this approach? A scientific approach to mathematics - Observation of patterns then proving them

Mathematics is a hobby of mine and on more than one occasion, I've reinvented the wheel. This paper is one such example. Therefore, even though it is original work for me, Bernoulli was the first to use this approach. At least that is what a Washington University professor told me. This also means that it can be published on Wiki without violating the policy against original work. It also means that I am free to offer it as public domain. I have solved problems that only the greats in mathematics are given credit, but I lack a formal education in mathematics, so I would never be able to convince any journal to publish my work. Yet, I've been able to impress others with my knowledge and ability to invent new approaches to solve problems. Being in a position without the credentials, I always provide irrefutable evidence that my approaches are correct. Most Ph.D professors can't say the same. I don't need credit for any of my work in math nor do I need any money. I'm making a comfortable living using my mind in the sciences. But, I would be greatly rewarded if I could share some of my understanding with young people. I'd like to get other people to think outside the box. The traditional lessons are important, but I would also like to see young people learn the value of creativity. Mathematics is also tainted with an image of logic and theory instead of as a true science. The paper that I'm offering here is a good example of how a problem can be solved by starting with observation.