KJW

I prefer to think of the critical point as where the liquid and gas become indistinguishable (the meniscus between them disappears). Interestingly, critical points can only occur between phases that have the same symmetry (e.g. liquid and gas).

Yes, that's the point I'm making. But the vapour pressure of a liquid at a given temperature does not depend on the local atmospheric pressure.

I should add that it is the higher kinetic energy water molecules in the liquid phase that escape to the gas phase. But once the higher kinetic energy water molecules in the liquid phase have overcome the potential energy barrier to escape the surface of the liquid, their kinetic energy is reduced to that of the average in the liquid. However, it is not clear to me that the kinetic energy of the water molecules in the liquid phase that escape to the gas phase corresponds to 100°C. 100°C is the boiling point of water at normal atmospheric pressure, whereas the phenomenon described above is independent of the atmospheric pressure.

No, the gas is still at 20°C. For example, if one placed a flask containing water at 20°C under a high vacuum, sealing the flask after some of the liquid water had boiled away and all of the air had been removed, then after the temperature of the contents of the flask has been allowed to return to 20°C, both the liquid water in the flask, and the gaseous water above it will be at 20°C. The gaseous water will exert a pressure of 2.3388 kPa (0.0231 atm)¹. And since all the air in the flask has been removed, the entire contents of the flask is water (liquid and gas), and the total pressure inside the flask will be 2.3388 kPa (0.0231 atm). Furthermore, if the flask had not been evacuated, then the partial pressure of the gaseous water molecules (the pressure the water molecules themselves exert within the water-air mixture) would also be 2.3388 kPa (0.0231 atm). That is, the presence of air in the flask does not affect the pressure of the gaseous water molecules in equilibrium with the liquid water at a given temperature. ¹ Source: https://en.wikipedia.org/wiki/Vapour_pressure_of_water

Perhaps I misinterpreted your original question concerning the field (0, 1). Anyway, I was focusing on the notion of Bring radicals because I thought they were interesting in terms of solving general quintic (and perhaps higher) equations.

[math]x^5 + x + 1[/math] In general, the root of [math]x^5 + x + a[/math] is called a Bring radical of a, denoted BR(a). Bring radicals extend the notion of radicals with regards to the Abel-Ruffini theorem that states that there is no solution in radicals to general polynomial equations of degree five or higher with arbitrary coefficients. By extending the notion of radicals, quintic equations that have no solution in ordinary radicals have solutions in Bring radicals. Note that asymptotically, [math]\text{BR}(a) \sim -a^{1/5}[/math] for large a.

[math]\big(\sqrt{7} + \sqrt{5}\big)\big(\sqrt{7} - \sqrt{5}\big) = 2 \\ \big(\sqrt{7} - \sqrt{5}\big) = \dfrac{2}{\big(\sqrt{7} + \sqrt{5}\big)} \\ \sqrt{5} = \dfrac{1}{2}\Big(\big(\sqrt{7} + \sqrt{5}\big) - \big(\sqrt{7} - \sqrt{5}\big)\Big) \\ = \dfrac{1}{2}\Big(\big(\sqrt{7} + \sqrt{5}\big) - \dfrac{2}{\big(\sqrt{7} + \sqrt{5}\big)}\Big) \\ = \dfrac{\big(\sqrt{7} + \sqrt{5}\big)}{2} - \dfrac{1}{\big(\sqrt{7} + \sqrt{5}\big)}[/math]

I thought the section https://en.wikipedia.org/wiki/Resource_curse#Examples_in_biology_and_ecology to be interesting. Whereas one might think that the resource curse is an absurdity of human nature (or of the capitalist system), the appearance of something similar in natural systems seems to indicate that the cause is more fundamental.

I don't think so. By the time Trump and his ship of fools are done with America, the current strong growing economy will be a distant memory.

At the time of my initial reply, I had not proved that the map from G to {1, –1} is a homomorphism. And instead of focusing on that proof, I digressed to a discussion about the alternating group. It was later that I devised a proof of the homomorphism based on Cayley tables. However, I subsequently discovered a direct proof that any subgroup of order half that of the group is normal in that group. The proof is that the left and right cosets of the subgroup must be equal, a defining property of normal subgroups.

This is drawing a longbow. Choosing examples where cognition is important to survival isn't sufficient to show that cognitive abilities are essential in natural selection. What about examples where cognition plays no role in survival? For example, what cognitive abilities do trees have? And yet trees are also evolutionarily successful. The various ways that organisms can become evolutionarily successful is unlimited. For example, animals such as cattle, sheep, pigs, and chicken have become evolutionarily very successful simply by being good food for humans. I doubt that is a choice those animals would have made cognitively.

Hmmm. I did not realise THAT was what I had proven. From what I said above, proving that a map from G to {1, –1} is a homomorphism completes the proof that a simple group cannot have subgroups of order half that of the group.

This actually seems like a bad idea to me.

I can't speak about all groups G, but if there is a homomorphism that maps G to {1, –1}, then the kernel of the homomorphism (the subgroup of G that maps to 1) is a normal subgroup of G that contains exactly half the elements of G. In particular, the alternating group An of all even permutations of n objects, is a normal subgroup of the symmetric group Sn of all permutations of n objects. It should be noted that Cayley's theorem states that every group is isomorphic to a subgroup of a symmetric group. It occurred to me that the alternating group An where n is greater than or equal to 5 is a simple group of even order. That is, it has no normal subgroups other than the trivial group and the group itself. Thus, if the alternating group has any subgroups of order half that of the group, then the answer to your question is no. But there might not be any subgroups of order half that of the group.

And now that I'm an adult, I wonder why couples in TV shows always cover themselves when getting out of bed after having sex.

I chose not to look at the linked video, but I was curious, nevertheless.

What is the purpose of this thread? When I saw first saw it, I thought it might have been about censorship, or perhaps the notion of privileged knowledge. But then I thought it might be about preserving treasures. For example, an art museum is not going to loan out the Mona Lisa to the general public.

Then k and n are relatively prime.

No, the group generated by another element is a subgroup of G, not necessarily G itself. k is a factor of n (or perhaps a multiple of a factor of n).

The importance of normal subgroups is the following (which I won't explain at this time): Let G be a group and N be a normal subgroup of G. Then there exists a quotient group or factor group G/N and a homomorphism that maps G to G/N, and under this homomorphism, N maps to the identity of G/N. Conversely, for every homomorphism of group G, there exists a normal subgroup N such that N maps to the identity of the image of the homomorphism, and a quotient group or factor group G/N that is isomorphic to the image of the homomorphism. g1 = h1 g h1–1 g2 = h2 g h2–1 h2–1 g2 h2 = g g1 = h1 g h1–1 = h1 h2–1 g2 h2 h1–1 = (h1 h2–1) g2 (h1 h2–1)–1 Note that in general, (ab)–1 = b–1a–1

Yes. And taken over all elements h of the group, the resulting set is an equivalence class called a conjugacy class. For a given h, the mapping g –> h g h–1 is called an inner automorphism. Subgroups of the group that are invariant to all inner automorphisms of the group are called normal subgroups of the group. Normal subgroups are very important in group theory. Note that all subgroups are normal in abelian groups.

In group theory, the elements h g h–1 for each and every element h of the group form an equivalence class for element g of the group. If g belongs to the centre of the group, then its equivalence class is g alone. Otherwise, the equivalence class contains multiple elements. Thus, although the equivalence classes of a group partitions the group, the equivalence classes are not necessarily equal in size.

Can you provide a proof? I can't at this time provide a counterexample. EDIT: I think you may be correct. If one chooses a single non-identity element as a generator, then if it is a finite generator, the subset generated must contain the identity element and all the inverses of the elements of the generated subset, and hence be a group. But if the generator is not finite, then the subset generated is not necessarily a group.

No. The subset also must contain the identity as well as the inverses for all elements of the subset. The associative property is however inherited from the group G.

Humans use technology to modify the genetics of various organisms. Is this part of evolution or separate from it?