For those budding mathmeticians

M

Munkey-

Guest
stuck as fuck on these. Pythagoras theorum = A(squared)+B(squared)=C(squared)

so given these numbers as lenght sides

A B C

3 4 5
5 12 13
7 24 25
9 40 41

blah blah blah. already worked out that A is +2 every time. B is x4+4 thingy everytime. and c is b+1.

now how the hell do i prove this???

*kills maths*
 
M

Munkey-

Guest
well it goes.

B

123(4)56789 10 11 (12) 13 14 15 16 17 18 19 20 21 22 23 (24)


4-8-12-16 etc
 
D

Damini

Guest
Munskster, I don't get it. Maybe write out the actual question? A string of numbers is no good :( Well, not to me.

If I could understand what the hell you are after then I will help :)
 
M

Munkey-

Guest
something like "pythagorean triplets" or something
 
X

xane

Guest
Oh I get you

3 4 5 = 3 x1 + 1 = 4
5 12 13 = 5 x2 + 2 = 12
7 24 25 = 7 x3 + 3 = 24
9 40 41 = 9 x4 + 4 = 40
11 60 61 = 11 x5 + 5 = 60

etc

right

a b c are original sides
A B C are new sides

a^2 + b^2 = c^2

A^2 + B^2 = C^2

A = a + 2
B = A * ( ( A - 1 ) / 2 ) + ( ( A - 1 ) / 2 )
C = B + 1

I don't like the look of B, someone else take over ...
 
X

xenon2000

Guest
Originally posted by xenon2000
This is GCSE coursework innit? I remember doing it; got info from here:
http://www.friesian.com/pythag.htm
Hope it helps :)
Click to expand...
To cut to the chase;
The fact that the difference between successive squares is successive odd numbers suggests comparison to another Pythagorean discovery, that every square is the sum of two successive triangular numbers, where the triangular numbers are the successive sums of all integers: 0 + 1 = 1, 0 + 1 + 2 = 3, 0 + 1 + 2 + 3 = 6, etc
Click to expand...
and...
-Every even number is the a side of a Pythagorean triplet.
-The b side of such a Pythagorean triplet is simply (a/2)2 - 1.
-The c side is b + 2
Click to expand...
I *think* that was what I used... it was a while ago mind.
 
M

Munkey-

Guest
thanks xenon2000. staring at that site for past 1/2 hour
 
M

Munkey-

Guest
damini. trying to PROVE it. sorry if it was confusing but had to explain it somehow as forgotten what the dakmn things were called

edit: xenon. could you explain this bit pweez

The fact that the difference between successive squares is successive odd numbers suggests comparison to another Pythagorean discovery, that every square is the sum of two successive triangular numbers, where the triangular numbers are the successive sums of all integers: 0 + 1 = 1, 0 + 1 + 2 = 3, 0 + 1 + 2 + 3 = 6, etc
 
X

xenon2000

Guest
Damn, if some maths genius comes in I'm f0rked... I know it involves *something* to do with the triangular number sequence (factorials - 3!=6 5!=15 etc). I have a pen and paper infront of me right now... bear with me while I try and work it out :)

Basically, you just need to know a formula for a, b and c in terms of n, yeah?
 
M

Munkey-

Guest
i cant remember :( he confiscated the sheets. but i go with what you say xenon
 
X

xane

Guest
Summary then, as the link gave a nice breakdown for B

a^2 + b^2 = c^2
A^2 + B^2 = C^2

A = a + 2
B = ( A^2 - 1 ) / 2
C = B + 1

and

C = ( A^2 + 1 ) / 2

so

A = ( a^2 + 2 )
B = ( a^2 + 3 ) / 2
C = ( a^2 + 5 ) / 2

go from there ...
 
X

xenon2000

Guest
hang on... just noticed you've gotten to the formula part already, nevermind - just the proof job needed :/

edit - camazotz got it there; I would explain what I've just worked out but the phone rang and I've forgotten it all now :(
 
O

old.Rostam

Guest
While at it could u also prove fermat's theorum. I don't understand the equations.
 
O

old.Rostam

Guest
Sorry wij, should have clarified it before.

The one before his last one.
 
X

xenon2000

Guest
I think I've cracked it; here goes...

OK, just to give some background, triangular numbers are numbers of a certain value. They equal the total of that value plus all numbers below it down to one. They are also called 'factorials' of a certain value, and are written in the form "n!", where "n" is the number... EG:

1! = 1
2! = 3 (2+1)
3! = 6 (3+2+1)
4! = 10 (4+3+2+1)
5! = 15 (5+4+3+2+1)

They can be quickly obtained by this formula:

n! = [n(n-1)/2] +n

In the case of the Pythagorean Triplets, it can be seen that "B" is always 4 times the factorial of the triplet value (n). For instance, the "B" value for the 2nd triplet can be found by multiplying 2! (which equals three) by four, to get 12. Do this for all the triplets investigated just to proove that it is true. It can be summed up as so:

B = 4(n!).

Substituting "n!" with the formula above, it can be compared to that of "B" in terms of n...

B = 4([n(n-1)/2]+n)

Further subsituting of "B" with its formula gives:

[(2n+1)^2 -1]/2 = 4([n(n-1)/2]+n)

This equation can then be simplified down as proof.
-------------------------

Exactly why B=4(n!) I'm unsure about. It's probably worth looking into it as further work if your after an A* (level 8). This diagram could be of use:

pythag-3.gif


...it proves that n! + (n+1)! always equals a square number. Something could probably be drawn from that as to the reason "B" can calculated by multiplying the factorial by four. Maybe...

Hope you have a long time till hand-in date, that alone took me ages to figure out :)
 
O

old.mrmen

Guest
I don't wanna burst your bubble Xenon but I am afraid factorials don't work like that. A factorial number is generated in this method.

Lets take the number 4 for example.

4! = 4 x 3 x 2 x 1 = 24

Try it on your calculater and you will find that works (or maybe its just mine :) ).

So if we take a indefinate number like Z then its factorial can be calculated by:

Z! = Z x (Z-1) x (Z-2) ... until (Z - n) = 1


On a more related topic, I don't see how these pythagorean triplets are related to factorial numbers. I'll try and work it out later but I've got my own A Level Maths work to do for tomorrow. :p

Edit : I got A* at GCSE level and didn't do anything like this. Just ask your teacher to help (or in other words tell) you how to do it.
 
X

xenon2000

Guest
fuck :/

OK, replace every occurence of the word "factorial" with some other word that means the same thing as "triangle number of"... I just used the wrong term...
 
M

Munkey-

Guest
what about finding the n 'th term?


for B and C. already figured out A

n=[n-1]*2+3
 
M

Munkey-

Guest
spent about 3/4 hour trying to figure these out but no luck :(
 
X

xenon2000

Guest
I refuse to attempt to help anymore with the mrmen on the loose :(
 
O

old.mrmen

Guest
Originally posted by xenon2000
I refuse to attempt to help anymore with the mrmen on the loose :(
Click to expand...

Sorry :rolleyes:

I'll work it out for you at earliest friday afternoon, munkey. Got loads of IT to hand in on friday and then I will be free.... well until the next deadline.
 
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