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Matrix
vbVeryBeginner wrote: a^2 + b^2 = c^2 Pithagorasz already did http://wwwgap.dcs.stand.ac.uk/~history/Mathematicians/Pythagoras.html Last edited by Matrix on 16 Oct 2004, 08:16; edited 1 time in total 

16 Oct 2004, 07:10 

vbVeryBeginner
either i prove it, or i disprove it
( a^n + b^n + 2(ab) ) c^(n2) = c^n Code: i try n = 3 a^3 + b^3 = c^3 assume a = 88, b = 99, c == 88 + 99 = (187) final value: a^3 = 681472 b^3 = 970299 c^3 = 6539203 the formular from a^2 + b^2 = c^2 show that a^2 + b^2 + 2(ab) = c^2 {a^2 + b^2 + 2(ab)} (a+b or c) = c^3 (7744 + 9801 + 17424) * 187 = c^3 this solve the n=3 for n = 4 {a^2 + b^2 + 2(ab)} c = c^3 {a^2 + b^2 + 2(ab)} c^2 = c^4 1222830961 so ( a^n + b^n + 2(ab) ) c^(n2) = c^n this would solve n = any integer > 2 sincerely, sulaiman chang 

16 Oct 2004, 14:47 

Jaques
urs looks too complex for a kid like me


16 Oct 2004, 15:21 

vbVeryBeginner
Jaques wrote:
are you really sure it is complex? i thought it would be the easiest proof compare with other proofs for FLT! 

16 Oct 2004, 17:48 

Eoin
vbVeryBeginner, in your proof for a^2 + b^2 = c^2 you are actually showing that (a + b)^2 = a^2 + 2ab + b^2. And technically what you've done isn't a proof. Let me try to explain.
If you want to prove that there exists three numbers a, b & c for which a^2 + b^2 = c^2 hold then one way is to just find them. For example a=3, b=4 & c=5 works, so does a=5, b=12 and c=13. But thats not what Pythagoras was trying to do. He was refeering to right angled triangles and the fact that the sum of the squares of two sides (b & b) equals the square of the hypothenuse (side a), a robbed image should explain nicely; Now the people of the day knew this and whenever they measured sides on rightangled triangles it always worked out and they quite happily "assumed" it would be true for every rightangled triangle. But Pythagoras wasn't happy assuming it would be true, he wanted to know it would always be true for every rightangled triangle. So he proved it, for examples of numerous proofs see here. With his proof we now know that we will never find a rightangled trianlge which doesn't obey the rule. A proof is an abstract notions. You start at axioms, a set of ground rules which are so obvious that they don't need to be proved, then working in logical steps you develop proofs. For example there is a proof for 0 != 1, and another proof for 1 > 0. More complex proofs are built up on existing ones and this process just continues on and on. Jaques, I understand your worry of being robbed but I would be very interested in reading your proof if you trust me enough . 

16 Oct 2004, 20:26 

Jaques
well i may post in a while but ill tell you this
in the proof of the second u have to divid by 1 dim in the proof of the third u have to divid by 1/2 dim in the proof of the third u have to divid by 1/3 dim and 1/2 in the proof of the third u have to divid by 1/4 dim and 1/3 and 1/2 the way to get the plynomial into a formula in them of using a and k equals b is recursice and from there its all but getting down to the first irrational step 

16 Oct 2004, 20:55 

vbVeryBeginner
thank you for your explaination eion, i am still bit blur, but i would love to understand it this thread just make me feel a little bit interested with numbers
so, what should i find in order to prove it? explain to me in simple words plez i really wish to find it em, does this mean i got to find a formular to get a, b & c which equal to a^2 + b^2 = c^2? or , we need to find a rightangled trianlge which doesn't obey the rule. a^2 + b^2 = c^2 

16 Oct 2004, 23:27 

Jaques
I mean that im using all integers sorry if i cant be specific... my diction is not very good.


17 Oct 2004, 01:59 

Eoin
Jacques, honestly I'm confused, but very interested. I can't seem to grasp your ideas so the bits you mention simply throw me. I look forward to a fuller explanation, I'd like to think I'm reasonably good at math so I hope once I can see where you're coming from I'll understand. I hope .
vbVeryBeginner, thats really cool that this thread has you interested in numbers (as a side issue I'm trying to set up a forum for discussing AI and math and I wonder if people would be interested in it? If there is an interest then let me know, the infrastructre is already in place.). If you really want to understand then let's forget for the moment the a^2 + b^2 = c^2, lets go back to (a + b)^2 = a^2 + 2ab + b^2. What I mean in English is can you prove for me that for any two numbers a & b the the square of thier sum is equal to the sum of their square plus twice their product. As an example given a=2 and b=3 then (a + b) = 5, so (a + b)^2 = 25. Also a^2 = 4, b^2 = 9 and 2ab = 12, so 4+9+12 = 25. So that proves it for 2 & 3, Now we could pick any two numbers and they would probably work, but try and see if you can prove they will always work. That means don't work with numbers, but instead stick with the letters a&b and prove it using them. 

17 Oct 2004, 02:13 

Jaques
well the proof itself is insane but lets try to explian
first set some basic facts if a funtion uses basic arithmetic and all variables / constants are rational then return value will be rational, and if a rational number is dived by irattional resutl is irational cosider k to be the difference between b and c c^n = (b+k)^n so a^n = (b+k)^n  b^n for the first power it is easy to realize k = a for second you use only the fist power (a^2k^2)/2k = b for third you divide by the sqrt of 3 and in 4th u need a cubic root and you find the recursion at a point thus reaching the answer; (b+k)^n  b^n b*(b+k)^(n1)+k*(b+k)^(n1) b^2*(b+k)^(n2)+2*b*k*(b+k)^(n2)+k^2*(b+k)^(n2) this shows basic recursion of the proof im sorry if im not too elouent _________________  My ignorance is now your problem. 

17 Oct 2004, 04:08 

Matrix
yeah,
and here comes trigonomertic: Code: sin alfa = a/c cos alfa = b/c tg alfa = a/b ctg alfa = b/a cos^2(alfa) + sin^2(alfa) = 1 (xu)^2+(yv)^2=r^2 ; you know this, the circle where x,y = centre the next important thing is the MATRIX 

17 Oct 2004, 04:53 

vbVeryBeginner
Eion wrote:
i am interested Eion wrote:
thanks, i am getting more confused : ? a^2 + b^2 + 2ab = (a+b)^2 i don't understand what do you mean by stick with the letter? eg. Code: 70^2 = 50^2 + 20^2 + 2(50*20) = 4900 (70^2) 70^2 = 49^2 + 21^2 + 2(49*21) = 4900 (70^2) and so on.... with whatever number on "a" and "b" so 48^2 + 22^2 = 4900 47&2 + 23^2 = 4900 so if u give me a value let say 8888888888888^2 so, it would be equal to 4444444444444^2 + 4444444444444^2 + 2(4444444444444 * 4444444444444) = 8888888888888^2 Quote:
i don't see how they would fail to work? 

17 Oct 2004, 09:01 

vbVeryBeginner
how about this?
sincerely, sulaiman chang 

17 Oct 2004, 10:25 

Jaques
vbVeryBeginner i do not understand what you are saying.... what does this have to do with the proof???


17 Oct 2004, 16:25 

vbVeryBeginner
sorry, sorry jaques
i think i will get insane soon if i go deep with maths :p 

17 Oct 2004, 17:44 

roticv
Eoin,
How's life? Can you give me the url of your maths and AI forum? It sounds interesting. Hehe soon I will be visiting countless forums *smiles* vbVeryBeginner, You are not proving anything. 

17 Oct 2004, 18:21 

Eoin
Well guys, as I said the infrastructre is in place, but noone visits it yet I'll post the URL tomorrow after I've had a chance to clean it up. But amn't really expecting too many frequneters as those particular topics don't get dicsussed much.
vbVeryBeginner I'll give the proof to show what I mean by using the letters Code: Prove (a + b)^2 = a^2 + 2ab + b^2 Starting at left and show its equal to the right. We know a^2 = a*a Call this (1) So (a + b)^2 = (a + b)*(a + b) Using (1) = a*a + a*b + b*a + b*b = a^2 + a*b + b*a + b^2 Using (1) going backwards So we're almost there, but we need to remember one of the properties of multiplication, its commutative. This means a*b = b*a Call this (2) So back to (a + b)^2 = a^2 + a*b + b*a + b^2 and using (2) = a^2 + a*b + a*b + b^2 = a^2 + 2a*b + b^2 = out righthand side And thats it, the proof shows that the left and right hand side are the same thing. And since it show'd it without involving specific numbers it means it will hold true for all numbers, not just the ones you can check (even if you can check millions or billions you can't check them all ). 

17 Oct 2004, 18:58 

Jaques
Mathmatics is all simple mathmatics folded onto itself...
What is the relation between Primes, Diprimes, Triprimes, ect 

17 Oct 2004, 22:22 

vbVeryBeginner
roticv wrote:
yup, sorry for not proving anything, don't angry lah :p to: eion thank you for spending your time explaining to me, i sincerely appreciate it Code: A . \  \  \  \  \ +\ C B em, could i know, is there any other method to obtain the length of AB? without using the theorem pithogoras? AC^2 + BC^2 = AB^2 

17 Oct 2004, 22:39 

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