Ever wondered why ?
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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
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In all my years of math I was never shown this. Awesome.
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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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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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That should be obvious from the moment you learn how to multiply a 2 digit number by itself.
Did you ever see history portrayed as an old man with a wise brow and pulseless heart, waging all things in the balance of reason? Is not rather the genius of history like an eternal, imploring maiden, full of fire, with a burning heart and flaming soul, humanly warm and humanly beautiful? --Zachris Topelius Training a telescope on one’s own belly button will only reveal lint. You like that? You go right on staring at it. I prefer looking at galaxies. -- Sarah Hoyt
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I wonder if there's a simple visual demonstration of why (for right triangles): a2 + b2 = c2
There are many visual proofs to Pythagoras' theorem. For instance: clicky for vimeo video[^] Mind you, simply having the last frame of the video is probably enough for the visual proof. Also, there is a whole wikipedia article on the subject of the Pythagoras' Theorem, and it includes some visual proofs, though I personally find the one in the video to be the most elegant.
Φευ! Εδόμεθα υπό ρηννοσχήμων λύκων! (Alas! We're devoured by lamb-guised wolves!)
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There are many visual proofs to Pythagoras' theorem. For instance: clicky for vimeo video[^] Mind you, simply having the last frame of the video is probably enough for the visual proof. Also, there is a whole wikipedia article on the subject of the Pythagoras' Theorem, and it includes some visual proofs, though I personally find the one in the video to be the most elegant.
Φευ! Εδόμεθα υπό ρηννοσχήμων λύκων! (Alas! We're devoured by lamb-guised wolves!)
I like it. :thumbsup: