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Since bisecting an angle boils down to bisecting a line segment, can't you trisect an angle by following the same process as to bisect it, but then trisecting the segment (which is easy) rather than bisecting it? --75.41.187.26 (talk) 21:14, 28 November 2009 (UTC)[reply]
I am not sure this formula is relevant - the trisecting subsegments are not chords. It shouldn't be difficult to come up with the appropriate formula, and it will show that this approach only works for bisection. -- Meni Rosenfeld (talk) 06:15, 29 November 2009 (UTC)[reply]
You are given two (half-)lines that intersect with a given angle. You set your pair of compasses to an arbitrary distance, put the centre at the intersections of the lines and draw a circle. You draw a line segment between the points where the lines intersect the circle. That line segment is a chord and it is that line segment that the OP wanted to trisect. However, trisecting that line wouldn't help because the length of a chord is given by the formula I quoted, rather than being proportional to the angle (as the length of an arc is). It doesn't matter what the formula is, just that the length isn't proportional to the angle. --Tango (talk) 15:43, 29 November 2009 (UTC)[reply]
The line segment that the OP wants to trisect is indeed a chord. The 3 congruent segments that trisect it are not chords. The OP asserts that the angle on these segments is one third the original angle; the formula doesn't in any way refute this assertion. Indeed, the assertion works in the bisection case, but by your argument it wouldn't either. -- Meni Rosenfeld (talk) 19:39, 29 November 2009 (UTC)[reply]
It's useful to consider the example of trying to trisect an angle that is close to 180 degrees. Let AB and BC meet at B at an angle close to (but less than) 180 degrees. Suppose that AB and BC are both of length 1. So the length of AC, the line segment the OP proposes trisecting, has length close to 2. Since B is very close to AC, if you draw in the actual angle trisectors BD and BE, you will see that they are very short. BDE is an equilateral triangle, and DE is also very short. In particular, it should be clear that the length of DE is much shorter than one third of AC. So trisecting the line segment and trisecting the angle work out quite differently. Ctourneur (talk) 03:37, 1 December 2009 (UTC)[reply]
You don't. You define it to be true. 2 is defined as the successor of 1 and addition is defined by the fact that adding one takes you to the successor of the number you started with. (See Peano axioms.) --Tango (talk) 22:10, 28 November 2009 (UTC)[reply]
Also see number system. You can choose different representations of numbers, but unless otherwise specified, it is usually safe to assume you are talking about the standard decimal-based natural number system, or the real number system, where as Tango has specified, the definitions are valid. You can learn about more complicated ways to define sets of numbers, such as the construction of the real spectrum, in various articles. Our number theory article may be a good place to start. Note that some of the math articles are deceptively advanced and may require a lot of heavy math-theory background. Nimur (talk) 22:15, 28 November 2009 (UTC)[reply]
It depends on the way you defined the natural numbers. One simple way is to base the natural numbers on the classes of sets having bijections between them (sadly, this is not detailed in Natural_number). In this case, proving that 1+1=2 consists in proving that there is a bijection between the union of two single-element sets and a single two-element set. Jerome.Abela (talk) 16:41, 30 November 2009 (UTC)[reply]
Natural number contains the sentence "Arguably the oldest set-theoretic definition of the natural numbers is the definition commonly ascribed to Frege and Russell under which each concrete natural number n is defined as the set of all sets with n elements." There's not much detail, though. Algebraist18:47, 30 November 2009 (UTC)[reply]
Isn't it redundant because it follows directly from the law of cosines, since the law of cosines gives the unique possible value for the third side of an angle given two sides and the angle between them? --75.41.187.26 (talk) 22:18, 28 November 2009 (UTC)[reply]
Can you prove the Pythagorean theorem without using that axiom? For an axiom to be redundant you need to be able to prove it using just the other axioms in the system. --Tango (talk) 23:01, 28 November 2009 (UTC)[reply]
That's a picture. It isn't a sequence of logical statements each implied by the last and starting with the other axioms. It's my favourite proof of Pythag, but it is a proof in a specific model of those axioms, not one based on the axioms themselves. --Tango (talk) 23:56, 28 November 2009 (UTC)[reply]
Here's a way to make the graphical argument rigorous:
Given a right triangle ABC, rotate it 180 degrees to produce rectangle ACBC'. Then, reflect the rectangle across a line at 45 degrees to AC' and BC' to create another rectangle A''C''B''C'. Extend BC and B''C'' to intersect at D, and extend AC and A''C'' to intersect at E. The area of CDC''E can be written as (a+b)2 since it is a square with side length a+b; it can also be written as a2+b2+4[ABC] by considering the areas of each section separately.
Then, construct BB'C' such that C, B, and B' are collinear and BCB' is a right angle; repeat this construction two more times to produce a square with side length a+b and therefore area (a+b)2. However, this area can also be written as c2+4[ABC] by considering this area separately.
There's so much just assumed there. That theorem comes at the end of a large amount of other work. Just taking the very first statement for instance; why would two similar right angle triangles form a rectangle? That's saying that the angles of a triangle add up to two right angles. Try it with a right angle triangle on a sphere and you find they don't. You need to build up carefully from the axioms. Dmcq (talk) 08:28, 29 November 2009 (UTC)[reply]
Hilbert's axioms specifically describe Euclidean geometry, so the cases of spherical and hyperbolic geometry can be disregarded. ABC and BAC' form a rectangle because all four angles are right angles. ACB is a right angle by definition; AC'B is a right angle because triangle BAC' was formed by a rotation of triangle ABC, and rotations produce congruent figures; CAC' is a right angle because ABC+BAC=90 and ABC=BAC'; and CBC' is a right angle because ABC+BAC=90 and BAC=ABC'. --70.129.187.89 (talk) 14:24, 29 November 2009 (UTC)[reply]
How do you prove that rotations produce congruent figures? How do we know that ABC+BAC=90? You are using your personal knowledge of geometry, not the axioms. --Tango (talk) 14:38, 29 November 2009 (UTC)[reply]
We can disregard something because it doesn't happen in our intended model? That's not axiomatics. That's not even proof. That's just plain assuming the answer. Algebraist14:43, 29 November 2009 (UTC)[reply]
Hilbert's axioms deal strictly with Euclidean geometry. Therefore, it's irrelevant that triangles may have different total angle measures in spherical or hyperbolic geometry, since Hilbert's axioms were never made to describe those geometries, and in fact they don't. Rotated figures are congruent because a rotation can be described as two successive reflections across non-parallel lines, and the angle of rotation is defined as the angle between these two lines. ABC+BAC=90 because ABC+BAC+ACB=180. Even if this doesn't follow directly from the other 19 axioms, the statement that the sum of the angles of a triangle is 180 is a much weaker, and thus more preferable, statement than SAS congruence. In fact, making "The sum of the angles of a polygon is 180° if and only if it is a triangle." an axiom also eliminates the need for the Parallel Postulate. --70.129.187.89 (talk) 16:10, 29 November 2009 (UTC)[reply]
Do you have an idea what a proof is, or how Hilbert-style formal systems work? You don't exclude possibilities by asserting that they aren't what you meant, you do it by proving from your axioms that those possibilities cannot occur. From Hilbert's axioms in toto it can be proved that no weird non-Euclidean possibility occurs; from his axioms without SAS it cannot. When building a subject axiomatically, you can't assume anything that is not an axiom. For example, just to blithely refer to a right angle as 90, as you do, requires the theorem that all right angles are equal. Hilbert proves this using his SAS axiom; how do you intend to prove it? Algebraist16:27, 29 November 2009 (UTC)[reply]
Hilbert's axioms describe Euclidean geometry. Therefore, non-Euclidean geometry is irrelevant. All right angles are equal because a right angle is defined as an angle with a measure of 90 degrees. --76.211.91.170 (talk) 17:22, 29 November 2009 (UTC)[reply]
So you don't have any idea how formal systems work, and you haven't read the text of Hilbert you constantly refer to either (it defines right angles very differently, and does not use angle measures at all). Solving the second of these problems may help with the first. Algebraist17:27, 29 November 2009 (UTC)[reply]
"Hilbert's axioms are a set of 20 assumptions (originally 21) proposed by David Hilbert in 1899 as the foundation for a modern treatment of Euclidean geometry." Therefore non-Euclidean geometry is irrelevant. --76.211.91.170 (talk) 19:24, 29 November 2009 (UTC)[reply]
Just because Hilbert wanted the axioms to be a foundation for Euclidean geometry doesn't mean that they are (and it doesn't help if he closes his eyes and wishes really hard). You need to actually prove that statements like "the sum of angles in a triangle is 180°" follow from the axioms.
Of course, proof of this particular statement is only possible once you've defined what an angle measure is, and by Algebraist's account (I haven't read the text myself), Hilbert did no such thing. Just because you defined "right angle is 90 degrees" in high school geometry doesn't mean it's the definition used by Hilbert, and just because you've learnt all sorts of statements, or they seem intuitively obvious in your everyday geometry, doesn't mean they follow from the axioms.
The non-Euclidean geometry issue was only meant as an example to demonstrate your confusion about the concept of rigorous proof. You would do well to take the example to heart rather than insisting on how irrelevant it is. -- Meni Rosenfeld (talk) 19:49, 29 November 2009 (UTC)[reply]
(outdent) Let's not pile on quite so hard, but explain the missing point. To IP 76.etc: Your source says that Hilbert proposed the axioms as a foundation for Euclidean geometry. But he could do this only because they are sufficient to prove all of Euclid's theorems, and Hilbert had to show this to show that his proposal was reasonable. If you take away any one of the 20, the remaining collection is no longer strong enough to prove all of Euclid's theorems, so you don't know a priori which of the facts we all know from Euclid still follow from the reduced set of axioms. In particular, if you take away SAS, you can't prove just from the others that the angles of a triangle sum up to two right angles, so you can't use that familiar fact from Euclidean geometry to prove SAS from the reduced set of axioms. (I hope that's any clearer -- sorry if not.) Elphion (talk) 04:17, 2 December 2009 (UTC)[reply]
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