Genady

alive and conscious.

No, it is not.

Problem A. Draw two arbitrary points on a sheet of paper. Call them point A and point B. Find the location of point C such that AB = BC. Use only compass. Here is the drawing:

No. I have explained what I wanted in the post which is two posts above this one. Here is what I have explained there:

?

I don't know how you got the problem you describe. I've never mentioned anything like that. I have mentioned two other problems earlier in the thread. Problem A: Problem B: Emphasis is on "compass only". No straightedge, no straight lines, nil, nada, niet! The question in the thread was NOT how to solve these problems. I've solved both of them. (Without drawing any straight lines.) My initial solution used a point where two circles constructed on an intermediate step, touch each other. I was not sure, if it is legitimate to rely on a point that was constructed this way. However, I've found a way how to construct such a point by constructing another circle which crosses the two tangent circles at the point where they touch each other. This shows that using a point where two circles touch is a legitimate shortcut of this a bit longer but definitely legitimate construction. So, the question in the title of the thread is answered. If you want to play with them, try the Problems A and B yourself. But remember, no straight lines are allowed.

1. I don't know which problem you are solving. 2. It should be easy to describe a construction in words if the circles and points are named. Let's try. a. Let's call the given circle, C1. b. You say to "Use an arbitrary radius that is slightly bigger than the radius of the given circle." I set the compass to some radius like that. c. "make 4 arcs with the compass from each quadrant." With the radius set in (b), I make circles C2, C3, C4, and C5 with the centers in A2, A3, A4, and A5, respectively. They intersect in several points with C1 and among themselves. d. "This will give you the center." Where is the center? The center of what?

It'd be more meaningful to discuss something with my dinner table...

The problem does not allow to use straightedge to draw a straight line either.

There are physical laws describing how matter moves and interacts. If life interacting with matter causes the matter to move and to interact differently than required by these physical laws, then life causes violation of the physical laws. Do you mean that life violates laws of physics? If not, then life does not have any effect on matter, IOW, life does not interact with matter.

No. I am doing constructions using only compass. The problem specifically states NOT to use straightedge.

Thank you for the clarification. No, I don't recall

Which object do you have in mind?

Yes, in my mind the difference depends on the area of interest: massive vs. massless, relativistic vs. non-relativistic, fermions vs. bosons, ... All this I think is irrelevant to the question of abiogenesis.

Why do you ask? You know very well that both of them involve matter - particles, waves, systems with Hamiltonians.

When I studied ecology (as a course in biology program) many years ago, several models were presented for a variety of such events in different circumstances. All of them were built upon known chemical and physical principles, which are properties of matter.

Are there currently examples of laws or principles of nature like this, i.e., laws that do not involve matter? Aren't such principles rather mathematical than natural?

For example ...?

... "latkes."

Prove that groups of even order contain at least one element (which is not the identity) that squares to the identity. In case of a cyclic group, it is easy. Such group consists of {e, g, g2, g3, ..., gn-1} and contains element h = gn/2. This element, h2 = (gn/2)2 = gn = e. But how to prove it when the group is not cyclic? P.S. Oh, got it. Just count the pairs, element and its inverse.

I.e., show that A4 has a normal subgroup. I did it by brutal force and would like to know if there is a more elegant way. My solution: A4 consists of 4!/2=12 permutations: e = identity, 8 permutations of the kind (1 2 3)=(1 2)(2 3), (1 3 2)=(1 3)(3 2), etc., and 3 permutations with separated cycles: a = (1 2)(3 4) b = (1 3)(2 4) c = (1 4)(2 3) Because of the separation, the cycles in a, b, and c commute, and thus a2 = b2 = c2 = e. I've checked manually that ab = ba = c, ac = ca = b, and bc = cb = a. So, {e, a, b, c} is an abelian group. An abelian subgroup is normal. Thus, A4 is not simple. Let me know if any of the above need elaboration.

Show that Sn is isomorphic to a subgroup of An+2. I will demonstrate my idea for n=3. S3 has 3!=6 permutations, 3 odd: (1 2), (1 3), (2 3); and 3 even: identity, (1 2 3)=(1 2)(2 3), (1 3 2)=(1 3)(3 2). Let's consider them separately. Put elements 4 and 5 in A5 aside. Identify even permutations in S3 with permutations in A5 with the same cycles as in S3 while the elements 4 and 5 are fixed, e.g., (1 2)(2 3) in S3 ↔ (1 2)(2 3) in A5. Identify odd permutations in S3 with permutations in A5 with the same cycles as in S3 plus the cycle (4 5), e.g., (1 2) in S3 ↔ (1 2)(4 5) in A5. I think, it is obvious how to generalize it for any n, right?

Show that An for n≥3 is generated by 3-cycles, i.e., any element can be written as a product of 3-cycles. My solution is quite simple, but I wonder if I've missed something: Any element of an alternating group can be written as a product of even number of 2-cycles. Let's consider pairs of 2-cycles in such expression. They can be of two forms: (a,b)(b,c) and (a,b)(c,d). The first is immediately a 3-cycle: (a,b)(b,c)=(a,b,c). The second can be made into 3-cycles like this: (a,b)(c,d)=(a,b)(b,c)(b,c)(c,d)=(a,b,c)(b,c,d). So, each pair of 2-cycles can be converted to one to two 3-cycles.

The word "love" appeared there in a translation and by necessity an interpretation of something written in a dead language.

My point is that you pick an interpretation as you like. They are only words. Humans make a meaning.