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difficult problem


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Posted 02 October 2001 - 05:58 PM

prove for a,b,c>0 that: a^2/((a^2+b^2)^0.5) +b^2/((b^2+c^2)^0.5) +c^2/((c^2+a^2)^0.5)>= (a+b+c)/(2^0.5) send answers to gershon_kats@hotmail.com

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Posted 02 October 2001 - 06:04 PM

THAT''S SO EASY....For one thing you just combine the 3 fractions and make them have common denomitors and then you factor out the some stuff and it simplifies to the answer.....

Do your own homework from now on...especially easy crap

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Posted 03 October 2001 - 09:17 AM

it is not that simple as you think. iam trying to solve it for month.who solved it-write a full answer.

gershon kats

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Posted 03 October 2001 - 09:51 AM

I was able to solve this in about five minutes. Prove to me that this is not a homework assignment and I will post my solution.

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Posted 03 October 2001 - 10:43 AM

Hi,

Could anyone please help me with this homework problem:
prove that there exists no solutions for:

X^n + Y^n = Z^n where n > 2

soon please coz I need to hand this in next week.




/Mankind gave birth to God.

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Posted 03 October 2001 - 11:08 AM

quote:
Original post by silvren
Hi,

Could anyone please help me with this homework problem:
prove that there exists no solutions for:

X^n + Y^n = Z^n where n > 2

soon please coz I need to hand this in next week.




/Mankind gave birth to God.


You silly person you. I don''t have 300 years to try and prove that.

---- --- -- -
Blue programmer needs food badly. Blue programmer is about to die!


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Posted 03 October 2001 - 12:28 PM

Ahh, I remember seeing a show about that very problem once^^
I think some guy did end up proving it, but they never told exactly what the solution was.



-Deku-chan

DK Art (my site, which has little programming-related stuff on it, but you should go anyway^_^)

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Posted 03 October 2001 - 06:33 PM

X^n + Y^n = Z^n where n > 2 and X, Y, Z are all > 0.

This is Fermat''s Last Theorem. The proof was a tad long for a homework question...

#9 Anonymous Poster_Anonymous Poster_*   Guests   

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Posted 03 October 2001 - 07:15 PM

about 150 pages long

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Posted 03 October 2001 - 11:15 PM

Never mind.
I came up with a proof.

/Mankind gave birth to God.

Edited by - silvren on October 4, 2001 6:16:07 AM

#11 Anonymous Poster_Anonymous Poster_*   Guests   

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Posted 04 October 2001 - 04:00 AM

I have some ideas that involve elliptical equations and modular forms, but I''m not sure it''ll work. I''ll get back to you in eight years or so.

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Posted 04 October 2001 - 04:23 AM

quote:
Original post by Beer Hunter
X^n + Y^n = Z^n where n > 2 and X, Y, Z are all > 0.

This is Fermat''s Last Theorem. The proof was a tad long for a homework question...


Didn''t he prove that x3+y3 does NOT = z3



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Posted 04 October 2001 - 04:48 AM

This question is similar to your other thread "two problems". How is this related to game development?

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

Edited by - grhodes_at_work on October 4, 2001 11:49:27 AM

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Posted 04 October 2001 - 05:12 AM

quote:
Didn't he prove that x3+y3 does NOT = z3


Euler is credited with proving X^n + Y^n != Z^n where n > 2
for n=3 and subsequent cases for specific values of n have been proven by other mathematicians throughout history.

However, it was proven for all cases of n>2 by Andrew Wiles (in 1994 I think, could be wrong as this is from memory) by proving the the Shimura-Taniyama-Weil conjecture.

Just a irrelevent (but highly interesting) piece of information for any budding mathematicians...

[Edited: Sorry Dobbs. My mistake. You are completely correct.]
-
Supercytro



Edited by - supercytro on October 5, 2001 4:56:48 AM

Edited by - supercytro on October 5, 2001 5:15:25 AM

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Posted 04 October 2001 - 06:31 AM

Supercytro, you are just plain wrong. Fermat''s theorem was there are NO non-zero integer solutions to x^n + y^n = z^n where n>2, so python_regious was right, x^n + y^n != z^n for n>2.

There are however solutions to x^n + y^n + z^n = c^n for n=3 and n=4. Maybe this is what you meant?

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Posted 04 October 2001 - 02:27 PM

Hope you have at least nine years to prove this!

Andrew Wiles, a Cambridge mmathematician, showed that x^n + y^n != z^n for all n > 2.

Elliptical equations and modular forms are understood by few mathematicians anyway. Good luck!

Regards,
Mathematix.

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Posted 04 October 2001 - 10:01 PM

Bad bad mistake by me earlier. Edited my post. Thanks Dobbs for catching that. I guess my test for whether anyone was paying attention paid off ;-)

Next time I'll actually read what I write before posting.
Whilst programming, I've found "!=" and "==" are a common cause of quite a few logical errors.

Apologies to python_regious for misreading your post as well.

quote:
x^n + y^n + z^n = c^n for n=3 and n=4

I knew this but unfortunately it wasn't what I meant. It was just a stupid mistake unfortunately, and not a misunderstood post by me.

Edited by - supercytro on October 5, 2001 5:07:43 AM

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Posted 09 October 2001 - 02:53 AM

I''m closing this thread and several others because the question appears to be a school homework assignment from a math class. The purpose of homework is to teach students to build their comprehension of a subject and their problem-solving skills, possibly with the assistance of other students in the same class or teachers of the class. Especially for math problems such as the one posed here, it is absolutely NOT appropriate to seek the answers from folks outside one''s class or school.

These forums are to be used for assistance in game development activities only.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.




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