neat piece of 'trick' maths

[dave]

i saw something similar to the regular argument used to prove 1=2 the other day, and i thought i'd share it in here since these forums seem a little dead on the maths side sometimes :) it goes like this:

(cos(x))^2 = 1 - (sin(x))^2
then, 1 + cos(x) = 1 + sqrt(1 - (sin(x))^2)
squaring, (1+cos(x))^2 = (1+sqrt(1 - (sin(x))^2))^2

now when x = 2*pi/3:

(1-1/2)^2 = (1 + sqrt(1-3/4))^2
1/4 = (1+sqrt(1/4))^2

therefore 1=9

i haven't looked at it much, as mainly i find these things tedious, but i thought it was quite neat so here you are. enjoy :)

[JaKiri]

It'll be an incorrect sign on a sqrt.

It almost always is.

[zakfab]

Maths 'tricks' are stupid, they only work on people who don't understand maths.

(cos(x))^2 = 1 - (sin(x))^2

This is correct.
Now you have square rooted both sides, and added one.

then, 1 + cos(x) = 1 + sqrt(1 - (sin(x))^2)

Nope, the lhs is 1 + sqrt(cos(x)^2)

The 1 is added to make the lhs positive, as you can't sqrt a negative number (unless you want to go into imaginary numbers).

So when the -0.5 is squared it becomes +0.25. Then is square rooted to be +0.5. The one is added and it all works out.

2.25 = 2.25

[dave]

Originally posted by zakfab
Maths 'tricks' are stupid, they only work on people who don't understand maths.

No they aren't. Sure, some might be quite simple, but it encourages people to actually think about why things are wrong. When I was 16, my maths teacher showed me a bit of 'trick' maths that was interesting because you can't get it to work without explaining why the equation x^3 = 1 has 3 different answers, and this encourages people to continue with mathematical studies - and hence help them understand the mathematics.

[NavajoEverclear]

1 does not = 2 or 9. one is one. I have one head, only one, not two, and certainly not 9. If the equation says that (even if i could understand it i dont care to think about it, its summer) then it is a flaw in the mathmatical system, because 1 is one, this is physically backed everywhere. Look at your computer monitor as you read this. You have one computer monitor, go buy 8 more and tell me this is not different. How does 1=9? Tricks are all nice and good, and it maybe funny to show that our system is flawed, but it is that, proof of flaw, not proof of nonsense. Of coarse if you know me i think proof is impossible to proove anyway, but i think most people would agree with my faith that one is one, not 2 or 9. Unless you can explain how (i am open to listening), i'm sticking to that belief.

[NavajoEverclear]

excuse for that somewhat offended/offending reply, you probably meant it just as a humorous way showing our flawed system. I shouldn't have been dumb enough to believe you believed it yourself. Unless you can explain it logically (im not really in the mood to read mathmatics equations, but i guess i will try if neccesary) as some concept.

[zakfab]

NavajoEverclear: I already showed where the 1 = 9 trick goes wrong.

[NavajoEverclear]

so that means there is not a flaw in the system? but you dont even have to figure it out to know that the only two possibilities are error in use of the system or flaw of the system itself. I know one is not the same thing as 9 unless someone show me some complex concept that prooves it so.

[JaKiri]

There isn't. It's a flaw in the use of the system.

[alt_f13]

I once went through two pages of algebra to find out V = V


It was a life changing experiance.

[NSX]

Originally posted by dave


When I was 16, my maths teacher showed me a bit of 'trick' maths that was interesting because you can't get it to work without explaining why the equation x^3 = 1 has 3 different answers,

What are the 3 answers?
1 is an answer...
What are the other 2?

Originally posted by alt_f13
I once went through two pages of algebra to find out V = V


It was a life changing experiance.


lolz
I know what you mean...

I thought those were bad, then I learned about proofs
:embarass:

[dave]

z^3 = 1
z^3 - 1 = 0
(z - 1) is a factor (z = 1 is a root)

(z - 1)(z^2 + Az +1) = 0
equating co-efficients in z^2 gives A = 1

therefore the other roots (by solving the other equation) are (-1 +- sqrt(3)*i) / 2

[NSX]

dave said in post #12 (http://www.scienceforums.net/forums/showthread.php?s=&postid=19463#post19463):
z^3 = 1
z^3 - 1 = 0
(z - 1) is a factor (z = 1 is a root)

(z - 1)(z^2 + Az +1) = 0
equating co-efficients in z^2 gives A = 1

therefore the other roots (by solving the other equation) are (-1 +- sqrt(3)*i) / 2

ah...very clever...cools

Thanks dave

:)

[Peter Dunn]

I'm going for a sqrt