Challenge Math Challenge - July 2019

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I understand my mistake: I missed the [itex]\alpha \in K[/itex] requirement. We don't have [itex]\alpha\in\mathbb Q(\beta)[/itex], hence argument that followed is gibberish.
Put [itex]\alpha := \sqrt{1-\sqrt{3}}[/itex] and [itex]\beta := \sqrt{1+\sqrt{3}}[/itex]. The polynomial [itex]x^4-2x^2 -2[/itex] is irreducible by Eisenstein, hence minimal polynomial of [itex]\beta[/itex], which yields [itex][\mathbb Q(\beta),\mathbb Q] = 4[/itex] with a basis of [itex]\{1,\beta,\beta ^2,\beta ^3\}[/itex]. Observe that [itex]\alpha^2 = 2-\beta ^2\in\mathbb Q(\beta)[/itex] so it suffices to have [itex]\{1,\alpha\}[/itex] to generate [itex]\mathbb Q(\alpha,\beta) / \mathbb Q(\beta)[/itex] and they are linearly independent over [itex]\mathbb Q(\beta)[/itex]. Thus
[tex]
[\mathbb Q(\alpha,\beta),\mathbb Q] = 8.
[/tex]
Per definition, the Galois group is the group of automorphisms on [itex]\mathbb Q(\alpha,\beta)[/itex] that pointwise fix [itex]\mathbb Q[/itex]. If [itex]\tau[/itex] is such an automorphism and [itex]\alpha[/itex] is a root, then [itex]\tau (\alpha)[/itex] is also a root. An automorphism is therefore determined by how it acts on the roots. This has to be (isomorphic to) a subgroup of [itex]S_4[/itex], one of order [itex]8[/itex] to be exact.

One also notices that the polynomial is even so the roots come in pairs: [itex]\alpha, -\alpha[/itex] and [itex]\beta,-\beta[/itex].
I'm really looking for the exact isomorphism structure of the Galois group. There are only 5 groups of order 8 (up to isomorphism). Which one is it?
 
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I'm really looking for the exact isomorphism structure of the Galois group. There are only 5 groups of order 8 (up to isomorphism). Which one is it?
The resolvent cubic is reducible and there are two complex roots, so it must be the dihedral group.
 
For #14,

Consider the limiting cases. With P and Q 0 away from the vertices, we have 50% red and 50% blue. As they move away from the vertices, the yellow area increases, so that when P and Q are both 1 away (the side of the rectangle), the diagram is all yellow. Yellow takes away from both red and blue, and the diagram is all straight lines, so, the rate at which yellow replaces blue and at which yellow replaces red are both consistent and constant. Therefore, it does not matter how far down the sides the points P and Q are. In every case, the ratio between red and blue is 1:1.
 

fresh_42

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The answer is correct, but I do not follow you on your central statement.
Yellow takes away from both red and blue, and the diagram is all straight lines, so, the rate at which yellow replaces blue and at which yellow replaces red are both consistent and constant.
Why couldn't we have e.g. a parabola, with equal values at the extremes, but a maximum in the middle? Straight lines alone isn't sufficient. Of course it is in the end, but the constant rates have to be shown, either by analytic methods (easier) or by geometric theorems.
 
I began in my mind by considering the triangles formed by P and Q and the opposite sides of each, and I noted that no matter the placement of P or Q, the area of the triangle remains constant. Now I warn that I am a mere beginner in approaching Calculus, but, I think I grasp a little of it anyway, intuitively. The formalization of what I think I see needs improvement.

Anyway, this -- watching the triangles -- was the most obvious example of a parameter that does not change at all, in other words whose graph has slope zero. Here is the intuitive leap. I reasoned as follows:

Calculus is the mathematics of change. And it does not matter whether the change is in time or in space -- in this case it is change along the spatial dimension of the line segment. The rate of change is the derivative of the change itself. So I reasoned that, since everything we are looking at is straight lines, including the vector describing the path of P or Q as it transitions along the segment, their derivatives cannot but be horizontal lines, i. e. a constant rate of change. Two quantities simultaneously moving from 0.5 to 0 at constant rates of change must necessarily remain equal at every point.

If you can flesh that out with actual formalism, go right ahead. At the present juncture I don't know how to do that.
 

fresh_42

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I began in my mind by considering the triangles formed by P and Q and the opposite sides of each, and I noted that no matter the placement of P or Q, the area of the triangle remains constant. Now I warn that I am a mere beginner in approaching Calculus, but, I think I grasp a little of it anyway, intuitively. The formalization of what I think I see needs improvement.

Anyway, this -- watching the triangles -- was the most obvious example of a parameter that does not change at all, in other words whose graph has slope zero. Here is the intuitive leap. I reasoned as follows:

Calculus is the mathematics of change. And it does not matter whether the change is in time or in space -- in this case it is change along the spatial dimension of the line segment. The rate of change is the derivative of the change itself. So I reasoned that, since everything we are looking at is straight lines, including the vector describing the path of P or Q as it transitions along the segment, their derivatives cannot but be horizontal lines, i. e. a constant rate of change. Two quantities simultaneously moving from 0.5 to 0 at constant rates of change must necessarily remain equal at every point.

If you can flesh that out with actual formalism, go right ahead. At the present juncture I don't know how to do that.
Well, as already mentioned, your idea is correct. Here we simply can calculate the areas and do not need to consider the rate of change:

246673

Let's first label the areas and call the area of the square ##X##.

Since height and baseline of both triangles equal the side length of the square, their area is half of ##X##:
$$
A+E+H = \frac{1}{2}X = A+F+G = B+C+D+F+G = B+C+D+H+E
$$
This means ##\dfrac{A}{B+C+D}= \dfrac{A}{\frac{1}{2}X -F-G}=\dfrac{A}{A}=1##
 
Actually, your last bit seems unnecessary. If A + F + G = B + C + D + F + G then we can subtract (F + G) from both sides and get A = B + C + D, or red = blue. Thank you for your help!
 
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