when to use 3.16.. as pi and 3.14.. Golden Ratio ; Specialness of 10

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Date: Wed, 25 Dec 2013 11:32:23 -0800 (PST)
From: Archimedes Plutonium
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Subject: when to use 3.16.. as pi and 3.14.. Golden Ratio ; Specialness of
10 and 10, 100, 10^3,10^4 . . 8th ed.: TRUE CALCULUS #106

when to use 3.16.. as pi and 3.14.. Golden Ratio ; Specialness of 10 and 10, 100, 10^3,10^4 . . 8th ed.: TRUE CALCULUS #106
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me (Archimedes Plutonium change)



1:32 PM (less than a minute ago)




when to use 3.16.. as pi and 3.14.. Golden Ratio ; Specialness of 10 and 10, 100, 10^3,10^4 . . 8th ed.: TRUE CALCULUS #106

Now, what I am doing is showing how sqrt10 serves as the alternative pi, and what I mean by that is that in True Math, there are no curves but only straight line segments compiled together so that we imagine a curve but that we just cannot see how tiny they are to make us realize they are just straightlines joined together.

Now I am not saying pi = 3.14.. is replaced by 3.16..
the sqrt10, but saying that pi has an upper limit. And pi is bounded below by a polygon and bounded above by a polygon, both polygons governed by sqrt10, the smaller polygon inside the 3.14.. and the outer polygon are made up of 3.16.. rotation.

Let me show the reader what I mean with this illustration provided by this excellent website:

http://www.cut-the-knot.org/do_you_k...denRatio.shtml

--- quoting from ---

http://www.cut-the-knot.org/do_you_k...denRatio.shtml


Also, in a 2011 article, Jo Niemeyer, offered a beautiful way of constructing the Golden Ratio with three equal segments, their midpoints and a pair of perpendicular lines.
--- end quote ---

Now, what we do is continue that construction by adding four more equal segments and so we form, roughly a flattened semi-pentagon.

We place one of those segments at the midpoint of B3A3 and laying parallel to the x-axis, and with the other three segments we do a mirror image in reverse.

So that if we had a scissors and trimmed off the B2A1 and the entire x-axis and B3A2 and all other
such segments dangling off of the squashed pentagon shaped figure, what we end up with is a semi-pentagon squashed. And a Regular-semi-pentagon squashed.

So, what I am doing and saying, is that in True Math, no curves existed at all. But all figures were compilations of straightline segments. They all have vertices but we just cannot see them for they are so tiny.

Now that means that pi as 3.14... could never really deliver to us the Golden Ratio number of 1.61... with its sqrt5 involved. Pi as 3.14... is not evenly divisible by sqrt5 at infinity. At infinity, pi is evenly divisible by 120 as pi is 3.14159..32000.

So, what that means is every time we use circles and pi to find the Golden Ratio, what we are really doing is finding the enclosed polygon and the larger enclosing polygon (both based on 3.16...) while the pi is 3.14...

Another example from that same website where he shows a semicircle and three colored smaller circles of green, red, yellow and where he says "the ratio of the radius of the latter to the diameter of the small circles is phi."

In reality, what we have there are three small polygons inside a larger semi-polygon. These polygons are governed by 3.16... and not 3.14...
Their points of tangency maybe either a side of the polygon or a vertex of the polygon.

The point is that in Old Math, using 3.14.. as pi with the make-believe of having curves when no curves exist, is unable to ever deliver a number such as (1+sqrt5)/2 when 3.14... does not have that number, but sqrt10 = 3.16.... does have that number.

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Archimedes Plutonium