A question for everybody except everybody {For Todd's Eyes only}
- Thread starter Todd
- Start date
You stupid Creationist Creation Cristian Todd.
If you gave up beleiveing in dinasuars and man walking together and that stupid fundaMENTAList literal interpretation of that stupid Bible the answer would be so clear to you.
Oh and I am not gonnat ell you the answer either, so there Todd.
If you gave up beleiveing in dinasuars and man walking together and that stupid fundaMENTAList literal interpretation of that stupid Bible the answer would be so clear to you.
Oh and I am not gonnat ell you the answer either, so there Todd.
KillerMuffin
Seraphically Disinclined
- Joined
- Jul 29, 2000
- Posts
- 25,603
It's the pythagoran theorem if n is two. Lookee here buddy boy. Christians are going to be persecuted. Accept it, deal with it, live with it and thank your everlasting toe cheese that the whole feeding Christians to the lions thing went out with the Romans. You are on a porn board. Okay? This is a hotbed of sinners, most of whom are NOT going to be Christian. In this part of the world Christianity has been literally forced on us like it or not. Right, wrong, or indifferent, a lot of people are pissed off at the general hypocrasy that the average Christian indulges themselves in.
Get over it.
Get over it.
Dea Artia said:Todd, Sweety. Great formula, but you need to know at least one other number, either x, y or z. preperably z.. then maybe i can solve it!
not really
x{to the 2} + y{to the 2} = z{to the 2}
You can ask, what are the whole number solutions to this equation, and you can see that:
3{to the 2} + 4{to the 2} = 5{to the 2}
and
5{to the 2} + 12{to the 2} = 13{to the 2}
Pyper
Lurking
- Joined
- Sep 19, 2000
- Posts
- 12,211
That is incorrect, Todd. I may be a writing major, but I aced the AP Calculus and AP Physics tests. As previously stated, x squared + y squared= z squared is the Pythagoran Theorem. It cannot be solved without at least one number. If you make n bigger, you cannot solve it without at least two numbers. As far as I can see, you did not solve the equation at all, you simply gave the solutions.
Dea Artia said:OIC! Neat. Hmm.. if i think of one i'll post it for you Todd. I'm not much into math, but i like playing games with my mind, you know..
have fun with it Dea Artia its a bit of a trick question.
Its actually Fermat's Last Theorem creater around 1630 no answer has yet been found
Pyper said:
solve \Solve\, n. A solution; an explanation. [Obs.] --Shak.
Source: Webster's Revised Unabridged Dictionary, © 1996, 1998 MICRA, Inc.
~~~~~
solve v 1: find the solution to (a problem or question):"did you solve the problem?"; understand the meaning of; "did you get it?"; "Did you get my meaning?" [syn: work, figure out, puzzle out, lick] 2: find the solution to; "solve an equation" [syn: resolve]
Source: WordNet ® 1.6, © 1997 Princeton University
~~~~~
A solution came by solving? So solving the theorem came the solution. Thats what I am asking with solve:
x{to the n} + y{to the n} = z{to the n} where n is bigger than 2
Pyper
Lurking
- Joined
- Sep 19, 2000
- Posts
- 12,211
Todd. Listen to me buddy. You did not solve the equation simply by telling us the answers. The equation is unsolvable as it is.
Solving an equation looks like this:
x(squared)+y(squared)= z(squared) What is y where x=3 and z=5?
3(squared)+y(squared)= 5(squared)
9+ y(squared)= 25
y(squared)= 25-9
y(squared)= 16
y= (squared root of 16)
y= 4
That is called solving an equation. Now try to do that with no numbers. See? You can't.
Solving an equation looks like this:
x(squared)+y(squared)= z(squared) What is y where x=3 and z=5?
3(squared)+y(squared)= 5(squared)
9+ y(squared)= 25
y(squared)= 25-9
y(squared)= 16
y= (squared root of 16)
y= 4
That is called solving an equation. Now try to do that with no numbers. See? You can't.
DarlingBri
Brit Wit
- Joined
- Mar 7, 2001
- Posts
- 1,262
Fermat's last theorum...
is that there are no positive integers such that x^n + y^n = z^n when n>2.
An equation like this CAN be solved with no variables defined other than n>2, as Todd demonstrated with his Pythagoran examples. It just can't be solved algebraically.
However, there's no solution to THIS equation, which merely proves Fermat's last theorum.
is that there are no positive integers such that x^n + y^n = z^n when n>2.
An equation like this CAN be solved with no variables defined other than n>2, as Todd demonstrated with his Pythagoran examples. It just can't be solved algebraically.
However, there's no solution to THIS equation, which merely proves Fermat's last theorum.
Pythagoran Theorem is simply:
a{squared} + b{squared}= c{squared}
no numbers are neeed to come up with a solution
you can inject whatever numbers you like into it
do the same with
a{to the ngreater than 2} + b {to the ngreater than 2} = c{to the ngreater than 2}
Use what ever numbers you want to my dear, I am not restricting your choice of numbers the only thing that is asked is n be greater than 2, you can cube, quad, quint, sext, octop what ever you like.
You can show your work if you like or use a calculator just come up with a number for a, b, c, and n>2 any numbers you want
a{squared} + b{squared}= c{squared}
no numbers are neeed to come up with a solution
you can inject whatever numbers you like into it
do the same with
a{to the ngreater than 2} + b {to the ngreater than 2} = c{to the ngreater than 2}
Use what ever numbers you want to my dear, I am not restricting your choice of numbers the only thing that is asked is n be greater than 2, you can cube, quad, quint, sext, octop what ever you like.
You can show your work if you like or use a calculator just come up with a number for a, b, c, and n>2 any numbers you want
Pyper
Lurking
- Joined
- Sep 19, 2000
- Posts
- 12,211
Still confused. The Pythagoran Theorem only works with certain sets of numbers. For instance, while a=3, b=4, c=5 does work, a=3, b=4, c=7 does not work. Therefore, to solve for all the unknowns, one would need some numbers.
And as far as I can tell, if I do the same thing, except just cube the buggers, it still does not work except for certain sets of numbers.
And as far as I can tell, if I do the same thing, except just cube the buggers, it still does not work except for certain sets of numbers.
DarlingBri
Brit Wit
- Joined
- Mar 7, 2001
- Posts
- 1,262
OK
I'll try
In theory, it can be solved, but not using algebra.More or less, you just plug in numbers and try them. Say I wanted N to be 3, OK? So I cube an infinite number of integers:
1^3 = 1
2^3 = 8
3^3 = 27
4^3 = 64
... and on and on. You then simply see if *any* two smaller numbers add up to a larger one: does 1 + 8 = 9? No. Does 8 + 27 = 64? No. Etc, etc, etc. There are computers whose full time job it is to try numerical combinations like that, first with n = 3, then with n = 4, etc.
In theory, you can solve any valid algebraic equation this way; we cannot actually use algebra to do this, as there is no way to solve for any single variable and then in turn use that number to solve for other variables, but we can just "try" random numbers.
However, since the actual theory of FLT is that the equation is NOT solveable - there are NO numbers that will fit - you cannot solve Todd's equation at all.
[Edited by DarlingBri on 04-14-2001 at 02:16 PM]
I'll try
In theory, it can be solved, but not using algebra.More or less, you just plug in numbers and try them. Say I wanted N to be 3, OK? So I cube an infinite number of integers:
1^3 = 1
2^3 = 8
3^3 = 27
4^3 = 64
... and on and on. You then simply see if *any* two smaller numbers add up to a larger one: does 1 + 8 = 9? No. Does 8 + 27 = 64? No. Etc, etc, etc. There are computers whose full time job it is to try numerical combinations like that, first with n = 3, then with n = 4, etc.
In theory, you can solve any valid algebraic equation this way; we cannot actually use algebra to do this, as there is no way to solve for any single variable and then in turn use that number to solve for other variables, but we can just "try" random numbers.
However, since the actual theory of FLT is that the equation is NOT solveable - there are NO numbers that will fit - you cannot solve Todd's equation at all.
[Edited by DarlingBri on 04-14-2001 at 02:16 PM]
Pyper said:
Yes but you can randomly guess a and b in the Pyth. Ther. and then see what you need to come up with for c in the Ther.
Go ahead and cube the buggers, come up with a random a and b to establish what c should be.
thats all I was asking myself in the opening post. Surely one such as yourself not deluded with images of a God would have much better capacities to guess a and b to come up with c
