Talk:High School Mathematics Extensions/Mathematical Proofs

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

Contents 1 comment 1 2 Hum?? 3 M5? 4 Incorrect Proof

[edit] 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 1² + 2² + ... + n² multiplies by 4 and adds 1 when n is increased by one. No idea? Think about how many squares there are on a 16×16 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 16×16 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:

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)

[edit] 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

has an explicit formula in terms of n for all j < k (**), we aim to prove that

also has an explicit formula.

Indeed, consider

I think you may have consider ^k instead of ^(k+1). Also, looking at it from another way

as you have mentioned the i^k+1 cancels, leaving smaller powers, rearraging to get

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... (Shouldn't there be some binomial coefficient when expanding out?)

I thought the principle would be like the below:

(Huh, I think I'll have to write another sub-chapter on the nested/2-dimensional sum notation operation...)

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)

[edit] 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)

[edit] 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)