Wednesday, August 17, 2011

Google's Search Engine for the Day

Leave it google to not include an important part of Fermat's Last Theorem:



It appears they left off the part where X,Y,Z are non zero positive integers.

5 comments:

James R said...

It seems they got the non zero positive part in, but you seem to be right about the integers. I wonder what the by Fermat's original 'doodle' looked like?

Big Dave said...

They said that n>2, where n is the exponent. That's not the same as saying x,y, and z are positive non zero integers. N has to be greater than 2 because with 2 there are certain pythagorean triples that satisfy the equation. For example: 3,4,and 5.

Peter H of Lebo said...

Actually, Google was right. Google was celebrating Fermat's Birthday and Google was pretty historically accurate in the search box, since Fermat was super lazy and never finished the theorem nor published it...the doodle was in the margins of a ancient Greek math textbook, the doodle was in latin stating, "It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain." No integers nor non zero positives. It wasn't until 1994 that is was solved by Andrew Miles with the help of 3 centuries of math.

http://www.csmonitor.com/Science/2011/0817/Why-Pierre-de-Fermat-is-the-patron-saint-of-unfinished-business

Big Dave said...

That's true Pete. So maybe some mathematician proved a theorem slightly different from what he is actually thinking and the original theorem was incorrect, but people have always translated his theorem to mean:

In other words, a^n + b^n can never equal c^n , as long as a, b, and c are positive integers and as long as n is greater than two.

That is not what google has on its doodle.

Big Dave said...

If you want a combination that disproves Google's picture:

Take X=0, y=-3, z=3 and n=4. That is a combination that satisfies:

X^n + Y^n = Z^n for when n>2.