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>> No.6516419 [View]
File: 90 KB, 1920x1158, constructing sqrt2 3.png [View same] [iqdb] [saucenao] [google]
6516419

>>6516369

So far you'll note that we have a Field of rational numbers, closed under addition and multiplication.

The next technique is kind of a special case. You construct a straight line. Mark the unit length of 1 using your unit circle. Then proceed to mark some more unit lengths on your line. In this picture we've drawn 1 and then 4 more (the total length of our line is 5). We then find the center point of our line (use two large circles of the same radius at either edge, draw a line connecting their two points of intersection). At that center point we draw a large circle so that our line cuts it in half. Now we draw a perpendicular line (construction is the same as finding the midpoint) at the point 1. Circumscribe a triangle in our half-circle using the intersection of our new vertical line and the half-circle. I know this is confusing so just look at the picture as it will make things clearer.

Thanks to Thales Theorem we know that our triangle circumscribed in the half-circle is a right triangle. Furthermore we know that the two smaller triangles that make up the larger one are also right triangles. It's then somewhat easy to show that all of the triangles are similar by Angle-Angle-Angle (hint: each of the smaller triangles shares two angles with the larger one, but because the angles in a triangle must sum to 180 degrees then the third angle in each is determined). Now because they are similar their sides must be in ratio and we can determine the height of the vertical line. It will always be the square root of however many units we went to the right of the initial 1. In this picture it will be square root of 4. A theorem for the construction of square root of n (where n is constructable) easily follows.

There are a handful of other ways to construct square roots but this is the best, imo.

(cont.)

>> No.5640223 [View]
File: 90 KB, 1920x1158, constructing sqrt2 3.png [View same] [iqdb] [saucenao] [google]
5640223

>>5639572

You're looking at it from a more limited greek perspective. It's true that we're talking about ratios, but by talking about those ratios as numbers then we can work out what things can and can't be constructed. You can't ever expect to work out what polygons are constructable and which aren't by talking about areas.

Here is how you construct the square root of any number. It's a superior method than the right triangle of sides 1 and 1. This is sqrt(4). You define a line segment to be length one, then you repeat said line segment with the compass in order to create 4 equal line segments to the right of it all on the same line. Now you should have a line segment 5 units long. You bisect it and create a circle that uses it as the diameter (the bisection point as the center). Now you create a line orthogonal to your first and second segment and intersecting between them by similarly bisecting them. Extend this vertical line until it reaches the semicircle. Now connect the new intersection with the other two intersections between the line segments and semicircle. This will produce a right triangle by thales theorem and the vertical line will cut this triangle into two other right triangles both similar to the large one and therefore similar to each other. By using some geometry it's easy to prove that this vertical line will have a ratio of sqrt(4) to your original line segment.


You can combine methods as well in order to create field extensions and shit.

>> No.5029679 [View]
File: 90 KB, 1920x1158, constructing sqrt2 3.png [View same] [iqdb] [saucenao] [google]
5029679

>>5029677
forgot my picture

>> No.4948619 [View]
File: 90 KB, 1920x1158, constructing sqrt2 3.png [View same] [iqdb] [saucenao] [google]
4948619

>>4948563
Yeap, also if you continue the spiral in >>4948572 for one more step then you'll see that the next hypotenuse is also sqrt(5).

Here is a more detailed diagram showing the sqrt(5) using the other method I posted. Note how counting is done with the compass and how forming the perpendicular line is accomplished by the intersection of two circles. It's a shame that you actually see lot more of this type of stuff in autocad courses than you do in geometry courses.

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