Math, anyone?

Well Jrod here is one to put on your next homework hand in:

For A^n + B^n = C^n there are no integer solutions for n >= 3
I have discovered the most elegant proof of this but the margin of this note book is too small for it.

Too bad i am no longer bothered with math homework. Maybe putting a whole bunch of these on one page and distrubuting them to the math department would be something to do if I wasnt so lazy.

Well Jrod, I really wasn't serious about doing what I recommended it my last post - it might get you into a little trouble.

Originally posted by jagdflieger
Train 1 speed = 256 mph
Train 2 speed = 145 mph
Relative closure = 401 mph
Initial distance between trains = 4*401 = 1604 miles
Bee at mid point = 802 miles from start
Travel distance for Train 1 untill collision = 1024 miles
Travel distance for Train 2 untill collision = 580 miles
Assuming bee does not evade trains and flies towards the slower train (since the slower train will never reach his current location by the crash time):
The relative speed of Train 1 and the bee is 255 mph (256 - 1)
By using the relative speed we can simplify and perform the computation using the bee's current location as if the bee were motionless:
So the time for Train 1 to reach the current location of the bee (using the relative speed) would be
802/255 = 3.14509... hours
The bee could fly 3.14509... miles before Train 1 overtakes it and squishes it.

Very nice try, but I'm afraid your wrong.

The answer is a integer btw.

Originally posted by Saucerat

Originally posted by jagdflieger
Train 1 speed = 256 mph
Train 2 speed = 145 mph
Relative closure = 401 mph
Initial distance between trains = 4*401 = 1604 miles
Bee at mid point = 802 miles from start
Travel distance for Train 1 untill collision = 1024 miles
Travel distance for Train 2 untill collision = 580 miles
Assuming bee does not evade trains and flies towards the slower train (since the slower train will never reach his current location by the crash time):
The relative speed of Train 1 and the bee is 255 mph (256 - 1)
By using the relative speed we can simplify and perform the computation using the bee's current location as if the bee were motionless:
So the time for Train 1 to reach the current location of the bee (using the relative speed) would be
802/255 = 3.14509... hours
The bee could fly 3.14509... miles before Train 1 overtakes it and squishes it.

Very nice try, but I'm afraid your wrong.

The answer is a integer btw.

Ok, we both did it slightly different ways. You did it more logically while i did it more algebraically, yet he says we are both wrong.

If it is an integer value then it must be a trick.

Well greenkoolaid since we are all online this very instant perhaps saucerat will enlighten us with the answer.

just had my math test on f(x) Y^2=x y=1/x and inverse functions pretty easy stuff, just repititive

Alright kids, the answer to the problem.

Are you ready?

The answer is 4 miles.

Now, your probably thinking, WTF?? Okay, look at it htis way, since it takes four hours for the trains to collide, the bee is freely to fly around for four hours before he is killed. So:

Distance = Rate * Time
D = 1mph * 4 hr
4 = 1mph * 4 hr

Okay, sorry, trick question.

Saucerat, you remind me of some of the engineering management creeps I used to work for. Give bad information and then wonder why the engineering team does not come up with the correct results. Since the bee is at the mid-point (mile 802), train 1 will reach that position in 802/256 hours or 3.132825 hours. Your information states that the bee is squashed by a train. No way that the bee can be squashed by the collision since the collision will happen some 222 miles from his initial location. Since you stated that the bee was squashed by a train, then the bee must be overtaken by one of the trains while in linear flight. (The assumption is that the bee makes no attempt to evade a train, or it would not be squished.) Hence the two answers. You have a bright carrer adhead of you in engineering management.

Well, you are right. I just noticed that right after I posted the answer. I posted the problem from memory, so I apologize for not getting it right. However, your answer is also correct assuming it gets killed before the collision in midflight. I don't have much interest going into the engineering field, but you are absolutely right, some people just give you the wrong info. I'm not sure what I'll pursue in the future actually. I used to want to be in the airforce, but my eyesight sucks (not anymore though, I have contacts) so I don't think I can do that. Before that, I wanted to be an astronant when I was really young, but then I saw Apollo 13, so I changed my mind.

Well, now that's done with, does anyone else have any intriguing problems?

Shoot, I haven't visited this thread in a couple of days and missed all the fun!

We need a new problem, huh?

Originally posted by Gazrok
Yep, as 0 has the unique property of...

0-0=0
0+0=0
0x0=0
0/0=0

0/0=Impossible !

But I have a strange theory (for the day I will ask or my nobel prize) :

1/0.0001=1000
1/0.00001=10000
1/0.000001=100000 ...etc

So

1/0 = infinite

----------------

a+b = b (No ! a+b = 2a = 2b !)


So :

a+b = b
b+b = b
2a=2b=2
a = b=1

It's with the boolean operator and binaries maths that you can do

1+1=1 or Yes + Yes = Yes

a = b
a^2 = ab WHY ?! It's in the problem ?
a^2 - b^2 = ab - b^2 = 0
(a+b)(a-b) = b(a-b) NO !! (a+b)(a-b) = a�- b� (Pythagore). and if a=b then (a+b)(a-b)=(2a)(0)=0
a+b = b WHY ?! if a=b then a+b =2a or 2b
b+b = b (The same than above)
2=1 (I'm sorry, but I know that maths are mind streching, but I'm not sure that in maths you can demonstrate than 2=1, try in biology or in human sciences...)



[Edited on 22-10-2003 by Nans DESMICHELS]

Originally posted by Valhall
Shoot, I haven't visited this thread in a couple of days and missed all the fun!

We need a new problem, huh?

Vahall, would you like to work with me ? We can get the nobel prize of maths if we demonstrate that :

---------------------------------------------------------

For

1/x=y

When x tend to zero
Then y tend to infinite

And so that :

1/0 result is not impossible but infinite.


Ok for the deal ?

Bah, you people are STILL working on zero?

Well, anyways, here's a quickie problem:

Find the number that best completes teh sequence:

2, 3, 4, 6, 16, 12 256, ?

I havent really examined any of the wacky proofs on here so I dont know if any are wrong.

2,3,4,6,16,12, 256,24.....

Originally posted by jagdflieger
Train 1 speed = 256 mph
Train 2 speed = 145 mph
Relative closure = 401 mph
Initial distance between trains = 4*401 = 1604 miles
Bee at mid point = 802 miles from start
Travel distance for Train 1 untill collision = 1024 miles
Travel distance for Train 2 untill collision = 580 miles
Assuming bee does not evade trains and flies towards the slower train (since the slower train will never reach his current location by the crash time):
The relative speed of Train 1 and the bee is 255 mph (256 - 1)
By using the relative speed we can simplify and perform the computation using the bee's current location as if the bee were motionless:
So the time for Train 1 to reach the current location of the bee (using the relative speed) would be
802/255 = 3.14509... hours
The bee could fly 3.14509... miles before Train 1 overtakes it and squishes it.

No offense, but your point is?

ok wtf?!?!?! you have lost me and i've been lost since clickin on this thread

Originally posted by jrod
I havent really examined any of the wacky proofs on here so I dont know if any are wrong.

2,3,4,6,16,12, 256,24.....

Yep, you got it.