## Die Suche ergab 82 Treffer

13.11.2017 09:48
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

So we have g>1, and we haven't sqr(2)/2, sqr(2)/3, thus we can't say that we have integer value in denominator of fraction. And all the squares of all integer values are integer, but we have an irrational sqr(2), why? Who told you that d,e,f, as face diagonals in situ, form a cuboid? If you only ta...
11.11.2017 11:46
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

Of cause AD/OH' is an integer (2) and DC'/OH is an integer (2). So OH we would carry out to the middle of DC' and OH' we would carry out to the middle of distance, wich parallel and equal to edges AB=AB'. So we analogically can to carry out any perpendicular to the middle of any edge or to the midl...
11.11.2017 10:03
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

You don't see that DC'/OH - integer, so AD'/OH' must be integer too? Your implication only shows that AD/OH' is an integer (2, to be precise), but nothing on AD'/OH'. This is why I said that you have no proof, only wishful thinking, wishfully disregarding an apostrophe. And I am not making a gramma...
11.11.2017 07:25
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

Carefully look at the picture. Good luck! You think that is a proof? You didn't even explain why your first statement is correct. Yes, both face-diagonals are integers, but that does not mean their ratio is also an integer. No matter how carefully I look at the picture, I only see ideas from someon...
10.11.2017 22:46
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

the numbers m and n are different numbers, since c and b are the legs, if the legs are equal, then the face diagonals will be irrational. Do not forget that m and n are different numbers, and, b, c are the legs in the Pythagorean triples. It is necessary to prove. If all variants for the existing c...
10.11.2017 20:25
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

@Zak Are you a troll? No one with some training in math and in his/her right mind will say anything like that. First, it is possible that g=a*sqrt(1+m+n) is an integer for a integer and m,n non-integer. Take a=1, m=9/4, n=7/4. So the last step of your proof is invalid. Since you claim it on general ...
10.11.2017 07:57
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

And may I add that, if you imagine that you can settle this problem, which has resisted the attack of some of the best mathematicians in the history, with a few lines of middle school math, then you are virtually insulting the effort of many hard-working people, and to a certain extent, the intellig...
10.11.2017 07:52
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

@Zak On Equation (5) and (6), your argument that n and m are integer is totally unsound. An integer multiplied by a non-integer can still be an integer. Take a=3, b=2, n=4/9 for instance. Please, please learn more math. Fractions are taught in primary schools, and various closure conditions on integ...
05.11.2017 23:21
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

I came to this conclusion: g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2 if we have a sum of three even values equal to square then sqrt of sum of 3 even values will give an irrational solution thus d ^ 2/2, e ^ 2/2, f ^ 2/2 not may to accept event value each of them, it's clear. g ^ 2, e ^ 2, f ^ 2 - integg...
05.11.2017 21:15
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

1. because d^2 / 2 never will odd, thus d^2 / 2 will be even and integger, it is followed from d - is integger value (face diagonal) It does not follow. d^2 / 2 can be non-integer, which is neither odd nor even. 2. because g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2 here, if we have a non-integger summ, we...
05.11.2017 20:55
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

O.K. So. We have an equation: g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2 we may see that every summand in amount of right side must give an integger solution by dividing on 2, if we want to get integger g^2. Thus we have a sum of three even values equal to square. So sqrt of sum of 3 even values will give...
05.11.2017 20:34
Forum: Fehler, Wünsche / Bugs, Wishes
Thema: [sub-project] Perfect cuboid
Antworten: 256
Zugriffe: 145303

### Re: [sub-project] Perfect cuboid

I have, d=e=1, f=4. Let's insert yours values into expressions of this equation: g ^ 2 = d ^ 2/2 + e ^ 2/2 + f ^ 2/2 So. d ^ 2/2=1^2/2=0.5 - is not odd, cause not corresponds to 2k+1, were k belongs Z https://en.wikipedia.org/wiki/Parity_(mathematics) It is followed that your statement "Not really....