Ever wondered why ?
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I doubt there is. If it was something that was easily recognizable, then it could be turned into an algorithm and there aren't any of those. The only things that I could think of would require an infinite dimensional drawing, so not very useful. I haven't read either of those books, but they are now on my list.
Curvature of the Mind now with 3D
Here are a few articles on point about autistic savants and prime numbers. http://www.integra.pt/textos/autism.pdf[^] http://goertzel.org/dynapsyc/yamaguchi.htm[^] http://www.scientiareview.org/pdfs/122.pdf[^] All are fascinating.
Tom Clement Serena Software, Inc. www.serena.com articles[^]
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There is a problem with that though. This proof is based on geometry. And geometry, as other branches of maths, are built upon basic postulates. Postulates are just assumptions, that could go wrong. Did you know that there is a strange kind of geometry, in which some of the basic postulates of common (Euclidean) geometry is left out. see this.[^] Actually, I don't know if you can prove this thing in those other geometry or not. :)
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Very nicely explained. I've never seen that before. Give that guy a medal! His accent makes it all the more entertaining.
If your actions inspire others to dream more, learn more, do more and become more, you are a leader." - John Quincy Adams
You must accept one of two basic premises: Either we are alone in the universe, or we are not alone in the universe. And either way, the implications are staggering” - Wernher von BraunIn South Africa we would say give that man a Bells(as in the whiskey) :) But yeah that's very cool
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lewax00 wrote:
(a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2
You don’t need this part, it’s more clean without it. Anyway have a five.
There is only one Vera Farmiga and Salma Hayek is her prophet! Advertise here – minimum three posts per day are guaranteed.
Logically it seems cleaner to me without that as well. My first reaction is a*a, not a*(a+b).
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I wonder if there's a simple visual demonstration of why every even integer greater than 2 can be expressed as the sum of two primes.
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much easier to disprove smallest prime 2 so smallest sum of 2 primes = 2+2=4 3 is an integer > 2 3<4 by hypothesis 3 cannot be an integer > 2 contradiction therefore hypothesis is false
"even integer", not just "integer".
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"even integer", not just "integer".
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Pictures of math don't help me understand it better, I'm better at reasoning through it: (a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2 So, (a+b)^2 = a^2 + 2ab + b^2 But I guess that's just how I learn. I'm generally better at objective subjects (math, physics, etc.) than subjective subjects (English, history, etc.) as a result.
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There is a problem with that though. This proof is based on geometry. And geometry, as other branches of maths, are built upon basic postulates. Postulates are just assumptions, that could go wrong. Did you know that there is a strange kind of geometry, in which some of the basic postulates of common (Euclidean) geometry is left out. see this.[^] Actually, I don't know if you can prove this thing in those other geometry or not. :)
With geometry, you can easily prove that Sqrt(2) = 2.
| (assume these lines are 1 unit high and 1 unit along) | |____________|
The distance between the open ends is sqrt(2) [From Pythagoras's theorem: sqrt(1^2 + 1^2)] 1st approximation of the diagonal:
\_\_\_\_ | | Length of diagonal = verticals (1/2 + 1/2) + horizontals (1/2 + 1/2)______| | = 2
| |
|___________|2nd approximation of the diagonal:
\_\_ \_\_| | Length of diagonal = verticals (4 \* 1/4) + horizontals (4 \* 1/4) \_\_| | = 2__| |
|___________|3rd approximation of the diagonal:
\_,-| Length of diagonal = verticals (8 \* 1/8) + horizontals (8 \* 1/8) \_,-' | = 2_,-' | (Assume ' represents a small vertical line)
'___________|(At this stage, I have reached beyond the capability of ASCII art) No matter how many times you do the better approximations of the diagonal, even until the verticals and horizontals are smaller than an atom, the total horizontal distance = 1 and the total vertical distance is also 1, so the diagonal is 2. Therefore, using geometry we have proved that (using Pythagoras's theorem) Sqrt(2) = 2.
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georani wrote:
First you should explain why: (a+b)(a+b) = a(a+b) + b(a+b)
Exactly. I learned it without this step. They called it the 'FOIL' method. First Outer Inner Last. (a+b)(a+b) First: a*a Outer: a*b Inner: b*a Last: b*b a^2 + 2ab + b^2
The world is going to laugh at you anyway, might as well crack the 1st joke! My code has no bugs, it runs exactly as it was written.
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georani wrote:
First you should explain why: (a+b)(a+b) = a(a+b) + b(a+b)
Exactly. I learned it without this step. They called it the 'FOIL' method. First Outer Inner Last. (a+b)(a+b) First: a*a Outer: a*b Inner: b*a Last: b*b a^2 + 2ab + b^2
The world is going to laugh at you anyway, might as well crack the 1st joke! My code has no bugs, it runs exactly as it was written.
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With geometry, you can easily prove that Sqrt(2) = 2.
| (assume these lines are 1 unit high and 1 unit along) | |____________|
The distance between the open ends is sqrt(2) [From Pythagoras's theorem: sqrt(1^2 + 1^2)] 1st approximation of the diagonal:
\_\_\_\_ | | Length of diagonal = verticals (1/2 + 1/2) + horizontals (1/2 + 1/2)______| | = 2
| |
|___________|2nd approximation of the diagonal:
\_\_ \_\_| | Length of diagonal = verticals (4 \* 1/4) + horizontals (4 \* 1/4) \_\_| | = 2__| |
|___________|3rd approximation of the diagonal:
\_,-| Length of diagonal = verticals (8 \* 1/8) + horizontals (8 \* 1/8) \_,-' | = 2_,-' | (Assume ' represents a small vertical line)
'___________|(At this stage, I have reached beyond the capability of ASCII art) No matter how many times you do the better approximations of the diagonal, even until the verticals and horizontals are smaller than an atom, the total horizontal distance = 1 and the total vertical distance is also 1, so the diagonal is 2. Therefore, using geometry we have proved that (using Pythagoras's theorem) Sqrt(2) = 2.
:doh: What you're saying is true, for EUCLIDEAN GEOMETRY, where you can have the basic assumptions about parallel lines and all that. But what I'm saying is, there are other types of geometry versions other than euclidean geometry (i.e. the normal geometry we know). Would you believe it if I say that the sum of angles inside a triangle is not 2*pi? In euclidean geometry that sum is 2*pi, but that's not always true with other geometries.
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georani wrote:
First you should explain why: (a+b)(a+b) = a(a+b) + b(a+b)
Because it's the law! ...the distributive law. With maybe a little commutative on the side.
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Yes, this is the law. The question is: WHY? Could you show the WHY of this law in a short explanation?
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There is a problem with that though. This proof is based on geometry. And geometry, as other branches of maths, are built upon basic postulates. Postulates are just assumptions, that could go wrong. Did you know that there is a strange kind of geometry, in which some of the basic postulates of common (Euclidean) geometry is left out. see this.[^] Actually, I don't know if you can prove this thing in those other geometry or not. :)
The more detailed representation of that kind of geometric multiplication is the Exterior Algebra[^]. Applying the same concept to curved spaces leads to differential forms[^] and a multidimensional form of the fundamental theorem of calculus[^].
Curvature of the Mind now with 3D
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With geometry, you can easily prove that Sqrt(2) = 2.
| (assume these lines are 1 unit high and 1 unit along) | |____________|
The distance between the open ends is sqrt(2) [From Pythagoras's theorem: sqrt(1^2 + 1^2)] 1st approximation of the diagonal:
\_\_\_\_ | | Length of diagonal = verticals (1/2 + 1/2) + horizontals (1/2 + 1/2)______| | = 2
| |
|___________|2nd approximation of the diagonal:
\_\_ \_\_| | Length of diagonal = verticals (4 \* 1/4) + horizontals (4 \* 1/4) \_\_| | = 2__| |
|___________|3rd approximation of the diagonal:
\_,-| Length of diagonal = verticals (8 \* 1/8) + horizontals (8 \* 1/8) \_,-' | = 2_,-' | (Assume ' represents a small vertical line)
'___________|(At this stage, I have reached beyond the capability of ASCII art) No matter how many times you do the better approximations of the diagonal, even until the verticals and horizontals are smaller than an atom, the total horizontal distance = 1 and the total vertical distance is also 1, so the diagonal is 2. Therefore, using geometry we have proved that (using Pythagoras's theorem) Sqrt(2) = 2.
That is true in the taxi cab metric[^]. I have a little demo page[^] that generates voronoi diagrams in both Euclidean, Taxicab and hyperbolic geometries.
Curvature of the Mind now with 3D
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There is a problem with that though. This proof is based on geometry. And geometry, as other branches of maths, are built upon basic postulates. Postulates are just assumptions, that could go wrong. Did you know that there is a strange kind of geometry, in which some of the basic postulates of common (Euclidean) geometry is left out. see this.[^] Actually, I don't know if you can prove this thing in those other geometry or not. :)
Also if you are interested in some models of hyperbolic geometry complex dynamics [^] is a great site for that.
Curvature of the Mind now with 3D
