1 + 1 = ?
nah probally finding all the numbers for pi...
2007-11-05 15:08:57
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answer #1
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answered by $bAyArEaBaLLeR$ 3
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The ancient Greeks began stewing over the following problem: Using only a compass and unmarked straightedge, divide a 60û angle into three equal parts. In other words, construct a 20û angle. No protractors allowed.
It is a deceivingly simple problem, says Scheinerman. But it was not until 1837, after generations of mathematicians had attempted in vain to solve it, that a French bridge and highway engineer named Pierre Louis Wantzel finally cracked the angle trisection problem. The answer: It cannot be done.
"It is impossible," declares Scheinerman. "But the hard part is understanding why you cannot do it."
Would he demonstrate why?
"No," responds Scheinerman. "It would take a semester."
But the friendly professor agrees to attempt a general explanation. "It boils down to taking a cube root," he says. A compass and straightedge can be used to add, subtract, multiply, divide, and calculate a square root of a given length. But angle dissection ultimately requires taking a cube root. "And that cannot be done with these tools. It is no more possible than adding together two even numbers to get an odd number."
Wantzel's proof involves showing that the algebraic powers of a compass and straightedge--adding, subtracting, multiplying, etc.- -cannot be used to produce cube roots. (There are many hard math problems, notes Scheinerman. Angle trisection may not be the most difficult, but it certainly took the longest.)
"In some sense, the Greeks never had a chance," says Scheinerman. The abstract algebra required to prove that the angle trisection problem cannot be solved was not developed until Wantzel's time. "It took a lot of understanding of the interplay between numbers and geometry."
Nevertheless, some stubborn souls continue to attempt a solution. Every year several amateur mathematicians submit a new "solution." Mathematician Underwood Dudley has compiled a collection of these "proofs," called A Budget of Trisections (Springer-Verlag, 1987). The only problem, says Scheinerman, is that they are all erroneous.
--Melissa Hendricks
2007-11-05 23:10:10
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answer #2
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answered by Clayton J 2
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the hardest math problems are those that have not been solved, yet.
Look at these (Millennium Prize Problems) problems at this website:
www.claymath.org/millennium/
It is very hard to say which is the hardest since they involve different areas of mathematics.
2007-11-09 13:04:54
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answer #3
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answered by mr green 4
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Some of the hardest math in the world deals with set theory and logic. There are people that make up problems that they spend their entire lives trying to figure out. Look up set theory and logic sometime, it'll blow your mind.
2007-11-05 23:11:02
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answer #4
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answered by cthulau 2
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there is no hardest math problem in the world
2007-11-05 23:09:10
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answer #5
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answered by valleygirl 2
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The Riemann hypothesis
2007-11-05 23:09:23
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answer #6
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answered by Mr. Smith 5
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To me, 9 x 8 = I had a teacher that embarassed me so badly, I still can't do any math. I just have a huge mental block. I could have had a much better life if I wasn't ignorant in this area.
2007-11-05 23:12:24
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answer #7
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answered by Anonymous
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those that connect every thing, i saw a movie on this, the toughest problem and only two people in the world have solved it, centuries ago and two years ago. I think its relativity or something, its like the answer to almost everything and is chalkboards long
2007-11-05 23:10:59
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answer #8
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answered by Jr. 3
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Prove that 0 divided by 0 is 1
2007-11-05 23:11:35
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answer #9
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answered by bigbaddom1989 3
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how to prove that 1 + 1 = 3
2007-11-05 23:09:45
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answer #10
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answered by Nick name 2
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