Making use of induction. Suppose a,b,c is a nonzero solution. Due to x^2\equiv 0,1\pmod{3} it must hold that a^2,b^2 are multiples of three. Putting d:= (a,b), if d\mid c holds, then \left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c') would be a solution with (a',b')=1 which is impossible.
Making use of induction. Suppose a,b,c is a nonzero solution. Due to x^2\equiv 0,1\pmod{3} it must hold that a^2,b^2 are multiples of three. Putting d:= (a,b), if d\mid c holds, then \left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c') would be a solution with (a',b')=1 which is impossible.
Could we show d\mid c?
In this method, ##a^2=9x^2, b^2=9y^2## .
This implies ##c^2## is a multiple of 3. Then you can continue with the same method I did. Using contradiction to prove that solutions don't exist.
However, I'm not sure how to prove that d divides c (if I'm not wrong, d is the gcd of a,b).
#53
member 587159
Pi-is-3 said:
Sure. First I'll explain why x is even. Since l is less than equal to m and n, we can divide both the sides by ##2^{2l}## . That in any scenario gives either x is even or x and y are even or all x,y,z are even (when l=m=n) by taking mod 4 both sides. A similar reasoning is followed for z.
As to why it is okay to consider only two cases instead of three (by including the case of m being smaller), considering the case for x is the same as considering the case for y. If it hadn't been the same, I would have to consider 3 cases. I don't know how to explain this one any better. You could think that if (a,b,c) satisfy the equation, then (b,a,c) will also satisfy.
I'm satisfied with your explanations. Well done!
In fact, there is a shorter solution.
You got to ##a^2+b^2 +c^2\equiv 0 \bmod 4##, implying that ##2|a,b,c##. Now, if we take a minimal positive solution ##(a,b,c)## this procedure yields the solution ##(a/2,b/2,c/2)##, contradicting minimality.
I think this is in the lines of what @nuuskur was trying.
I was thinking about it in a "give a counter-example" fashion. The question then is rather easy don't you think? I could put forward any function ##f## where ##f(0) \neq f(1)## for example.
... plus ##f'(x)\neq 0## for all ##x##.
If you have three conditions which are needed to draw a single conclusion, then the violation of any of the conditions together with the opposite of the conclusion is a counterexample. The best solution would have been to list three examples, violating one condition after the other. And a fourth example where the conclusion holds without any of the conditions.
Suppose g\mapsto g^{-1} is an endomorphism and take g,h\in G. Then gh \mapsto (gh)^{-1}. Due to compatibility gh \mapsto g^{-1}h^{-1}, thus
<br />
(gh)^{-1} = g^{-1}h^{-1} = (hg)^{-1} \iff gh = hg.<br />
Suppose G is abelian then for every g,h\in G
<br />
gh \mapsto (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1}.<br />
The map g\mapsto g^{-1}\sigma (g) on G is injective, because if g^{-1}\sigma (g) = h^{-1}\sigma (h), then hg^{-1} = \sigma (hg^{-1}). The only fixpoint is e, thus h=g. The map is also surjective, because G is a finite set.
Now, pick g\in G and write g= h^{-1}\sigma (h), then
<br />
\sigma (g) = \sigma (h^{-1}\sigma (h)) = (\sigma (h))^{-1} (h^{-1})^{-1} = (h^{-1}\sigma (h))^{-1} = g^{-1}.<br />
By 7a G is abelian.
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#57
member 587159
nuuskur said:
Suppose g\mapsto g^{-1} is an endomorphism and take g,h\in G. Then gh \mapsto (gh)^{-1}. Due to compatibility gh \mapsto g^{-1}h^{-1}, thus
<br />
(gh)^{-1} = g^{-1}h^{-1} = (hg)^{-1} \iff gh = hg.<br />
Suppose G is abelian then for every g,h\in G
<br />
gh \mapsto (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1}.<br />
The map g\mapsto g^{-1}\sigma (g) on G is injective, because if g^{-1}\sigma (g) = h^{-1}\sigma (h), then hg^{-1} = \sigma (hg^{-1}). The only fixpoint is e, thus h=g. The map is also surjective, because G is a finite set.
Now, pick g\in G and write g= h^{-1}\sigma (h), then
<br />
\sigma (g) = \sigma (h^{-1}\sigma (h)) = (\sigma (h))^{-1} (h^{-1})^{-1} = (h^{-1}\sigma (h))^{-1} = g^{-1}.<br />
By 7a G is abelian.
Correct! I will give feedback on the measure theory question soon! Bit busy now!
Making use of induction. Suppose a,b,c is a nonzero solution. Due to x^2\equiv 0,1\pmod{3} it must hold that a^2,b^2 are multiples of three. Putting d:= (a,b), if d\mid c holds, then \left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c') would be a solution with (a',b')=1 which is impossible.
Could we show d\mid c?
Write
<br />
a = 3 ^{k_1} \ldots p_n^{k_n} \quad b = 3^{l_1}\ldots p_n^{l_n} \quad c = 3^{r_1} \ldots p_n ^{r_n}<br />
where the powers are non-negative. Then (a,b) = 3 ^{m_1} \ldots p_n ^{m_n}, where m_j = \min \{k_j,l_j\}. The goal is to show m_j\leq r_j. The initial equality can be written as
<br />
3^{2m_1} \ldots p_n^{2m_n} \left ( 3^{2k_1 - 2m_1} \ldots p_n^{2k_n-2m_n} + 3^{2l_1 - 2m_1} \ldots p_n^{2l_n-2m_n}\right ) = 3^{2r_1+1}\ldots p_n^{2r_n}<br />
Suppose, for a contradiction, some m_j>r_j. The case j\neq 1 would run into 2m_j + k = 2r_j, where k\geq 0, which is impossible. So it must be that j=1. But then 2m_1 + k = 2r_1 + 1 for some k\geq 0. By assumption m_1 \geq r_1+1 so
<br />
2m_1 + k = 2r_1 + 1 \Rightarrow 2r_1 + 1 - k \geq 2r_1 + 2 \Rightarrow k\leq -1,<br />
a contradiction. Thus (a,b) \mid c.
I think this is overkill, not sure how to optimise.
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#59
member 587159
nuuskur said:
Let \Sigma be an infinite sigma algebra on a set X, choose \emptyset\subset A_1\subset X and pick for every n\in\mathbb N
<br />
A_{n+1} \in \Sigma \setminus \left ( \left\lbrace\emptyset, X\right\rbrace \cup \left\lbrace\bigcup_{k=1}^mA_k\mid m\leq n\right\rbrace\cup\left\lbrace\bigcap _{k=1}^m A_k^c\mid m\leq n\right\rbrace\right ).<br />
\Sigma is infinite, so we can do this. Put S_n := \bigcup _{k=1}^nA_k\in\Sigma,n\in\mathbb N. Observe that all S_n\subset X.
The sequence S_n^c,n\in\mathbb N, is strictly decreasing. Put
<br />
T_n := S_n^c\setminus S_{n+1}^c\in \Sigma, \quad n\in\mathbb N.<br />
The T_n are pairwise disjoint. The map
<br />
\mathcal P(\mathbb N) \to \Sigma, \quad A\mapsto \begin{cases} \emptyset, &A=\emptyset \\ \bigcup_{x\in A}T_x, &\emptyset\subset A\subset\mathbb N \\ X, &A=\mathbb N \end{cases}<br />
is injective. Note that if A\subset\mathbb N then \bigcup _{x\in A}T_x \subseteq S_1^c \subset X.
What happens when we take ##X =\mathbb{N}, A_1=\{0,1\},A_2=\{0\}##?
Then ##S_1=S_2 = A_1## and ##T_1=\emptyset## so again injectivity is violated.
This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets S_n\in\Sigma, n\in\mathbb N. Then
<br />
\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x<br />
is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct \emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N and put S_n := \bigcup _{k=1}^n A_k\in\Sigma. Then T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N, is a sequence of pairwise disjoint non-empty subsets.
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#61
member 587159
nuuskur said:
This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets S_n\in\Sigma, n\in\mathbb N. Then
<br />
\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x<br />
is injective, because of disjointedness.
Now we are going somewhere. Or course, the question is why such a sequence exists. How can we construct one?
#62
member 587159
nuuskur said:
This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets S_n\in\Sigma, n\in\mathbb N. Then
<br />
\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x<br />
is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct \emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N and put S_n := \bigcup _{k=1}^n A_k\in\Sigma. Then T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N, is a sequence of pairwise disjoint non-empty subsets.
Again, it can happen that ##T_i=\emptyset## for some i. See my last counterexample.
Again, it can happen that ##T_i=\emptyset## for some i. See my last counterexample.
You're right, we need a change of gears ..
Call a subset S\in \Sigma to have property (P) iff the sub sigma algebra \Sigma _S := \{S\cap A \mid A\in\Sigma\} is infinite. Note that
<br />
\Sigma = \sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ).\tag{E}<br />
On the one hand we have by definition
<br />
\Sigma _S \cup \Sigma _{S^c} \subseteq \Sigma \Rightarrow\sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ) \subseteq \sigma (\Sigma) = \Sigma.<br />
Conversely, take A\in \Sigma, then
<br />
A = A\cap (S\cup S^c) = A\cap S \cup A\cap S^c \in \sigma \left ( \Sigma _S \cup \Sigma _{S^c}\right ).<br />
Suffices to show the following: if S has property (P), then there exists a partition S = T\dot{\cup}T' such that T,T'\in\Sigma _S are non-empty and at least one of them has property (P). Since X has property (P) by assumption, we can repeatedly apply this fact and obtain a strictly decreasing sequence of non-empty proper subsets.
Proof of fact. Suppose S\in\Sigma has property (P). Then \Sigma _S is an infinite sub sigma algebra. Pick T \in \Sigma _S \setminus \{\emptyset, S\} Write S= T\cup (S\cap T^c). By (E) at least one of the respective sub sigma algebras must be infinite.
Consider the map
<br />
\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.<br />
the map \varphi (g) is injective, because gx = gy implies x=y and for a fixed h\in G we have \varphi (g) (g^{-1}h) = gg^{-1}h = h. Take g,h\in G, then
<br />
\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)<br />
By Sylow's first theorem there exists a \in G of order two. Note that (g,ga), g\in G, are cycles in the permutation \varphi (a), because
<br />
\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.<br />
Due to injectivity, the cycles are disjoint, thus there are precisely m such cycles, which implies \varphi (a) is an odd permutation. Consider the signum morphism \varepsilon : \mbox{Sym}(G) \to \{-1,1\} where \varepsilon (\sigma ) = 1 iff \sigma is an even permutation. We have \varepsilon \varphi : G\to \{-1,1\} a surjective morphism, because e\mapsto 1 (\varphi (e) has no inversions). By first isomorphism theorem
<br />
G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}<br />
which implies |\mbox{Ker} (\varepsilon\varphi)| = m. Also \mbox{Ker} (\varepsilon\varphi) \triangleleft G.
summer's almost over ..
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#66
member 587159
nuuskur said:
Suppose GG is of order 2m2m, where m>1m>1 is odd.
Consider the map
<br />
\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.<br />
the map φ(g)φ(g) is injective, because gx=gygx=gy implies x=yx=y and for a fixed h∈Gh∈G we have φ(g)(g−1h)=gg−1h=hφ(g)(g−1h)=gg−1h=h. Take g,h∈Gg,h∈G, then
<br />
\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)<br />
By Sylow's first theorem there exists a∈Ga∈G of order two. Note that (g,ga),g∈G,(g,ga),g∈G, are cycles in the permutation φ(a)φ(a), because
<br />
\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.<br />
Due to injectivity, the cycles are disjoint, thus there are precisely mm such cycles, which implies φ(a)φ(a) is an odd permutation. Consider the signum morphism ε:Sym(G)→{−1,1}ε:Sym(G)→{−1,1} where ε(σ)=1ε(σ)=1 iff σσ is an even permutation. We have εφ:G→{−1,1}εφ:G→{−1,1} a surjective morphism, because e↦1e↦1 (φ(e)φ(e) has no inversions). By first isomorphism theorem
<br />
G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}<br />
which implies |Ker(εφ)|=m|Ker(εφ)|=m. Also Ker(εφ)◃GKer(εφ)◃G.
summer's almost over ..
Correct! Well done! Could you maybe give some insight in why you thought about that action?
#67
Not anonymous
143
58
fresh_42 said:
Summary: 1. - 2. posed and moderated by @QuantumQuest
3. - 8. posed and moderated by @Math_QED
9. - 10. posed and moderated by @fresh_42
keywords: calculus, abstract algebra, measure theory, mechanics, dynamical systems
Questions
9. Three identical airplanes start at the same time at the vertices of an equilateral triangle with side length ##L##. Let's say the origin of our coordinate system is the center of the triangle. The planes fly at a constant speed ##v## above ground in the direction of the clockwise next airplane. How long will it take for the planes to reach the same point, and which are the flight paths?
Disclaimer: This spoiler may not be a spoiler at all, as I am no sure if the answer is correct :) Too long since I last used calculus to solve anything other than the most basic and even longer since I last worked on solving a physics problem :)
If the solution is incorrect, please explain which assumption or step is incorrect.
I am unable to draw here the diagram I sketched on paper to derive the solution, but the fundamental observation is that if all the 3 planes fly at the same speed, then at any point in time till they meet, the positions of the 3 planes will continue to form an equilateral triangle and this triangle keeps shrinking in size as time progresses.
Suppose the current distance between 2 planes is ##a## and ##\delta## is the small distance traveled by each of them, along the direction of edges of the current triangle, in a very small instant of time. Let ##b## denote the distance between the 2 planes after this instant. Using trigonometry, I found that the following equation determines the value of ##b##:
##b^2 = a^2 - 3 \delta (a - \delta)##
If ##v## is the constant velocity and ##\Delta t## is the small time instant, then ##\delta = v \Delta t##, yielding ##b^2 = a^2 - 3 v \Delta t (a - v\Delta t)##
If ##D^2## is used as the variable denoting the squared length of side of equilateral triangle, then we see that ##\Delta D^2 = -3 v \Delta t (D - v\Delta t)##.
$$
\lim_{\Delta t \rightarrow 0} {\frac {\Delta D^2} {\Delta t}} = -3 v D \Rightarrow
\frac {dD^2} {dt} = -3 v D \Rightarrow 2 D \frac {dD} {dt} = - 3 v D \Rightarrow \\
\frac {dD} {dt} = - \frac {3} {2} v
$$
So the cumulative change in the value of ##D## over a period of time ##T## would be
$$
\int_0^T - \frac {3} {2} v \, dt = - \frac {3} {2} v T
$$
When the planes meet, the value of D would be 0 whereas its initial value is ##L##, so the cumulative change in ##D## has to be ##-L##. Substituting this in the previous equation gives the value of ##T## to be
I think P6 generalises via induction to the case 2^rm.
Math_QED said:
Correct! Well done! Could you maybe give some insight in why you thought about that action?
I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some \varphi : G\to\mbox{Sym}(G) such that I would get \varepsilon \varphi : G\to \{-1,1\} surjective and apply the first isomorphism theorem. Of course, getting a preimage for 1 is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in \mbox{Sym}(G) which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order 2m gave a clue. A way to get such cycles is to put (g,ga) where a is of order two and at that point the definition of \varphi became self-evident.
#69
member 587159
nuuskur said:
I think P6 generalises via induction to the case 2^rm.
I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some \varphi : G\to\mbox{Sym}(G) such that I would get \varepsilon \varphi : G\to \{-1,1\} surjective and apply the first isomorphism theorem. Of course, getting a preimage for 1 is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in \mbox{Sym}(G) which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order 2m gave a clue. A way to get such cycles is to put (g,ga) where a is of order two and at that point the definition of \varphi became self-evident.
Yes, using actions to find kernels is something that works when all other things seem to fail. You are right btw that this exercise generalises (via induction)
The generalisation is:
If ##G## is a group of order ##2^nm## with ##m## odd and ##G## has a cyclic Sylow 2-subgroup, then ##G## has a normal subgroup of order ##m##.
Disclaimer: This spoiler may not be a spoiler at all, as I am no sure if the answer is correct :) Too long since I last used calculus to solve anything other than the most basic and even longer since I last worked on solving a physics problem :)
If the solution is incorrect, please explain which assumption or step is incorrect.
I am unable to draw here the diagram I sketched on paper to derive the solution, but the fundamental observation is that if all the 3 planes fly at the same speed, then at any point in time till they meet, the positions of the 3 planes will continue to form an equilateral triangle and this triangle keeps shrinking in size as time progresses.
Suppose the current distance between 2 planes is ##a## and ##\delta## is the small distance traveled by each of them, along the direction of edges of the current triangle, in a very small instant of time. Let ##b## denote the distance between the 2 planes after this instant. Using trigonometry, I found that the following equation determines the value of ##b##:
##b^2 = a^2 - 3 \delta (a - \delta)##
If ##v## is the constant velocity and ##\Delta t## is the small time instant, then ##\delta = v \Delta t##, yielding ##b^2 = a^2 - 3 v \Delta t (a - v\Delta t)##
If ##D^2## is used as the variable denoting the squared length of side of equilateral triangle, then we see that ##\Delta D^2 = -3 v \Delta t (D - v\Delta t)##.
$$
\lim_{\Delta t \rightarrow 0} {\frac {\Delta D^2} {\Delta t}} = -3 v D \Rightarrow
\frac {dD^2} {dt} = -3 v D \Rightarrow 2 D \frac {dD} {dt} = - 3 v D \Rightarrow \\
\frac {dD} {dt} = - \frac {3} {2} v
$$
So the cumulative change in the value of ##D## over a period of time ##T## would be
$$
\int_0^T - \frac {3} {2} v \, dt = - \frac {3} {2} v T
$$
When the planes meet, the value of D would be 0 whereas its initial value is ##L##, so the cumulative change in ##D## has to be ##-L##. Substituting this in the previous equation gives the value of ##T## to be
My solution is slightly different, but of course not basically.
The side length of the triangle at ##t=0## is ##L(0)=L.## For the position ##\vec{r}(t)## of the first airplane we have ##|\vec{r}(0)|=r(0)=\dfrac{2}{3}L\cos \dfrac{\pi}{6}=\dfrac{L}{\sqrt{3}}.## The distance between the airplanes are the same at any point in time, because of the symmetry, i.e. the airplanes will always mark the vertices of an equilateral triangle with its center at the origin. Thus the angle between the velocity ##\vec{v}(t)## and the position ##\vec{r}(t)## is always
$$
\sphericalangle (\vec{v}(t),\vec{r}(t))=\psi(t)=\psi(0)=\psi =\pi - \dfrac{\pi}{6}
$$
Thus we have
\begin{align*}
\vec{v}(t)&=\dot{\vec{r}}(t)\\
\vec{r}(t))\vec{v}(t)&=\vec{r}(t)\dot{\vec{r}}(t)\\
r\cdot v \cdot \cos \psi &=\dfrac{1}{2}\dfrac{d}{dt}(\vec{r}(t)\vec{r}(t)\\
r\cdot v \cdot \cos \psi &=\dfrac{1}{2}\dfrac{dr^2}{dt}\\
r\cdot v \cdot \cos \psi &=r\dfrac{dr}{dt}\\
\dfrac{dr}{dt}&= -v \dfrac{\sqrt{3}}{2}\\
r(t)&= \dfrac{L}{\sqrt{3}}-v\dfrac{\sqrt{3}}{2}t
\end{align*}
Hence ##r(t_f)=0 ## implies ##t_f=\dfrac{2L}{3v}.##
Call a subset S\in \Sigma to have property (P) iff the sub sigma algebra \Sigma _S := \{S\cap A \mid A\in\Sigma\} is infinite. Note that
<br />
\Sigma = \sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ).\tag{E}<br />
On the one hand we have by definition
<br />
\Sigma _S \cup \Sigma _{S^c} \subseteq \Sigma \Rightarrow\sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ) \subseteq \sigma (\Sigma) = \Sigma.<br />
Conversely, take A\in \Sigma, then
<br />
A = A\cap (S\cup S^c) = A\cap S \cup A\cap S^c \in \sigma \left ( \Sigma _S \cup \Sigma _{S^c}\right ).<br />
Suffices to show the following: if S has property (P), then there exists a partition S = T\dot{\cup}T' such that T,T'\in\Sigma _S are non-empty and at least one of them has property (P). Since X has property (P) by assumption, we can repeatedly apply this fact and obtain a strictly decreasing sequence of non-empty proper subsets.
Proof of fact. Suppose S\in\Sigma has property (P). Then \Sigma _S is an infinite sub sigma algebra. Pick T \in \Sigma _S \setminus \{\emptyset, S\} Write S= T\cup (S\cap T^c). By (E) at least one of the respective sub sigma algebras must be infinite.
Ok, now that we know..
Pick a strictly decreasing sequence X \supset S_1 \supset S_2 \supset \ldots from the sigma algebra. Then put A_n := S_n\setminus S_{n+1}, n\in\mathbb N, which is a sequence of pairwise disjoint elements and
<br />
\mathcal P(\mathbb N) \setminus \{ \emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} A_x,<br />
is injective, thus an infinite sigma algebra must be at least of the cardinality of the continuum.
#72
member 587159
nuuskur said:
Ok, now that we know..
Pick a strictly decreasing sequence X \supset S_1 \supset S_2 \supset \ldots from the sigma algebra. Then put A_n := S_n\setminus S_{n+1}, n\in\mathbb N, which is a sequence of pairwise disjoint elements and
<br />
\mathcal P(\mathbb N) \setminus \{ \emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} A_x,<br />
is injective, thus an infinite sigma algebra must be at least of the cardinality of the continuum.
Will give feedback tomorrow. Currently busy with abstract algebra and measure theory needs a change of mindset :P.
The fundamental theorem of abelian groups says that for any finite abelian group
<br />
<br />
A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}<br />
Hyperlinking breaks my raw text for some reason. I'll be using Sylow's theorems which I will refer to as theorems 1, 2 and 3 respectively. (https://en.wikipedia.org/wiki/Sylow_theorems)
In the abelian case there are \mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3.
Quick remark. Being a Sylow p-subgroup is invariant with respect to conjugating, thus if H was a unique Sylow p-subgroup, we would have gHg^{-1} = H, g\in G, i.e H would be normal.
Fact. Suppose a non-abelian group G has order pq (primes) with q\equiv 1 \pmod{p}, then there are precisely q Sylow p-subgroups in G.
Proof of fact. Let n_p,n_q be the numbers of Sylow p,q-subgroups respectively. By assumption q>p. By theorem 3 n_q \equiv 1 \pmod{q} and by theorem 1 n_q\mid p, which forces n_q = 1, thus we have a (normal) subgroup of order q, which makes it cyclic, therefore abelian.
For n_p we thus have two choices: n_p = 1,q. If n_p = 1, then G\cong \mathbb Z_p \oplus \mathbb Z_q would be abelian, thus it must be that n_p = q.
Our non-abelian group G therefore contains three Sylow 2-subgroups, call them H_1,H_2,H_3, which by theorem 2 are all conjugate to each other. We have \mbox{Sym}(H_1,H_2,H_3) \cong S_3. We show that G\cong S_3. Define
<br />
\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}<br />
(also called conjugating)
Firstly, if H is a Sylow p-subgroup, then conjugating it gives another Sylow p-subgroup, hence it must hold that gH_ig^{-1} = H_j for some j.
The map \varphi (g) is injective, because if gH_ig^{-1} = gH_jg^{-1}, then
<br />
h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.<br />
The argument is symmetrical, thus H_i=H_j. It is surjective due to finiteness.
Take g,h\in G, then
<br />
\varphi (gh) (H_j) = (gh)H_j(gh)^{-1} = g(hH_jh^{-1})g^{-1} = g (\varphi (h)(H_j)) g^{-1} = \varphi (g) (\varphi (h)(H_j)) = (\varphi (g) \varphi (h))(H_j).<br />
To show it's an isomorphism, it suffices to show it's injective (because |G| = |S_3| = 6). Suppose \varphi (g) = \mbox{id} i.e gH_jg^{-1}=H_j, then by definition g\in N(H_j) the normaliser of H_j. If N(H_j) = G, then H_j would be normal, which would then make it equal to its conjugates, contradicting the fact that there are three Sylow 2-subgroups.
Therefore, all N(H_j) = H_j and g\in H_1\cap H_2\cap H_3 = \{e\} i.e g=e, which makes \varphi an isomorphism.
Would constructing all possible character tables be an acceptable answer? I can get my group theory text and swat this up, but it would take me a couple hours to refresh that stuff since I have not looked at it in some time.
#75
member 587159
DEvens said:
Regarding question 5, a clarification question.
Would constructing all possible character tables be an acceptable answer? I can get my group theory text and swat this up, but it would take me a couple hours to refresh that stuff since I have not looked at it in some time.
A character table does not determine a group uniquely. If I recall correctly, ##D_8## and ##Q_8## are groups with the same character table that are not isomorphic.
The fundamental theorem of abelian groups says that for any finite abelian group
<br />
<br />
A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}<br />
Hyperlinking breaks my raw text for some reason. I'll be using Sylow's theorems which I will refer to as theorems 1, 2 and 3 respectively. (https://en.wikipedia.org/wiki/Sylow_theorems)
In the abelian case there are \mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3.
Quick remark. Being a Sylow p-subgroup is invariant with respect to conjugating, thus if H was a unique Sylow p-subgroup, we would have gHg^{-1} = H, g\in G, i.e H would be normal.
Fact. Suppose a non-abelian group G has order pq (primes) with q\equiv 1 \pmod{p}, then there are precisely q Sylow p-subgroups in G.
Proof of fact. Let n_p,n_q be the numbers of Sylow p,q-subgroups respectively. By assumption q>p. By theorem 3 n_q \equiv 1 \pmod{q} and by theorem 1 n_q\mid p, which forces n_q = 1, thus we have a (normal) subgroup of order q, which makes it cyclic, therefore abelian.
For n_p we thus have two choices: n_p = 1,q. If n_p = 1, then G\cong \mathbb Z_p \oplus \mathbb Z_q would be abelian, thus it must be that n_p = q.
Our non-abelian group G therefore contains three Sylow 2-subgroups, call them H_1,H_2,H_3, which by theorem 2 are all conjugate to each other. We have \mbox{Sym}(H_1,H_2,H_3) \cong S_3. We show that G\cong S_3. Define
<br />
\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}<br />
(also called conjugating)
Firstly, if H is a Sylow p-subgroup, then conjugating it gives another Sylow p-subgroup, hence it must hold that gH_ig^{-1} = H_j for some j.
The map \varphi (g) is injective, because if gH_ig^{-1} = gH_jg^{-1}, then
<br />
h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.<br />
The argument is symmetrical, thus H_i=H_j. It is surjective due to finiteness.
Take g,h\in G, then
<br />
\varphi (gh) (H_j) = (gh)H_j(gh)^{-1} = g(hH_jh^{-1})g^{-1} = g (\varphi (h)(H_j)) g^{-1} = \varphi (g) (\varphi (h)(H_j)) = (\varphi (g) \varphi (h))(H_j).<br />
To show it's an isomorphism, it suffices to show it's injective (because |G| = |S_3| = 6). Suppose \varphi (g) = \mbox{id} i.e gH_jg^{-1}=H_j, then by definition g\in N(H_j) the normaliser of H_j. If N(H_j) = G, then H_j would be normal, which would then make it equal to its conjugates, contradicting the fact that there are three Sylow 2-subgroups.
Therefore, all N(H_j) = H_j and g\in H_1\cap H_2\cap H_3 = \{e\} i.e g=e, which makes \varphi an isomorphism.
Typo
By theorem 3 n_q \equiv 1 \pmod{q} and by theorem 3 n_q\mid p
As a continuation to #73.
In the abelian case there are
<br />
\mathbb Z_8 \qquad \mathbb Z_4 \oplus \mathbb Z _2 \qquad \mathbb Z_2 \oplus\mathbb Z_2\oplus\mathbb Z_2<br />
Suppose G is non-abelian of order 8. If there is an element of order 8, then G is cyclic, thus abelian. Suppose the maximal order is 2 and pick g,h\in G. If gh=e, then they commute. Suppose (gh)^2 = e, then
<br />
gh = geh = g(ghgh)h = (gg)hg(hh) = ehge = hg<br />
and again G would be abelian. Thus we must have an element a of order 4. Its generated subgroup \langle a \rangle =: H is of index 2, therefore normal. Take b\notin H, then we have bab^{-1}\in H due to normality. Notice that
<br />
(bab^{-1})^4 = bab^{-1}bab^{-1}bab^{-1}bab^{-1} = baaaab^{-1} = e.<br />
If also (bab^{-1}) ^2 = e, then baab^{-1} = e would lead to aa=e, contradicting the order of a.
Next, we find the possible values of bab^{-1}.
If bab^{-1} = a^4 = e, then a=e, which is impossible.
If bab^{-1} = a, then ba = ab, but this will make G abelian. Indeed, pick g,h\in G. If they're both in H, then they commute. Suppose g\notin H and write g = ba^m and h = a^n, then
<br />
gh = ba^ma^n = ba^na^m = a^n ba^m = hg.<br />
and if both reside outside H, then writing g = ba^m and h = ba^n would similarly lead to gh=hg. But our group is not abelian.
If bab^{-1} = a^2, then (bab^{-1})^2 = e would contradict the order of bab^{-1}.
Thus, the only possibility is bab^{-1}=a^3 =a^{-1}.
There are two cases to consider.
b is of order 2, then we have G \cong \langle a,b \mid \mbox{ord}(a) = 4, \mbox{ord}(b) = 2, bab^{-1} = a^{-1} \rangle this is the dihedral group D_8.
b is of order 4. We show G is the dicyclic group \mbox{Dic}_2, where
<br />
\mbox{Dic}_2 \cong \langle x,y \mid x^{4} = e, x^2 = y^2, x^{-1} = y^{-1}xy \rangle.<br />
Notice that
<br />
bab^{-1} = a^{-1} \Rightarrow b^{-1}ab = b^{-1}(ba^{-1}b^{-1})b = a^{-1}.<br />
So we have left to show a^2 = b^2. Considering the natural projection G\to G/H\cong\mathbb Z_2, we have b^2 \in H. Consider possible values of b^2.
b^2 = a^4 = e contradicts order of b.
b^2 = a contradicts order of a
b^2 = a^2 is what we want.
b^2 = a^{-1} contradicts order of a^{-1} (hence order of a).
Thus b^2 = a^2 must hold.
The initial average speed before the change of heart is thus $$v_1 = \frac {d+10} {0.25 + \frac{d}{180}},$$
In a similar vein the average speed after the driver decides to lower his speeds is
$$v_2 = \frac {d+10} {0.5 + \frac{d}{160}},$$
We are told that the difference between these two speeds is 40kmh, so we solve the following equation for d
$$\frac {d+10} {0.25 + \frac{d}{180}} = \frac {d+10} {0.5 + \frac{d}{160}} + 40$$
Cleaning up that mess gives a quadratic
$$d^2 - 120d + 3600 = 0$$
which yields a single solution of d=60. Hence the total distance is d+10 = 70.
#80
Not anonymous
143
58
fresh_42 said:
13. David drives to work every working day by car. Outside towns he drives at an average speed of ##180\,\text{km/h}##. On the
##10\,\text{km}## in town, he drives at an average speed of ##40\,\text{km/h}##. As a result, he is often too fast and gets a ticket. Meanwhile he has realized that things can not go on like this and he decides to reduce his average speed by ##20\,\text{km/h}## in town as well as outside. How long is his way to work, if this reduces his average speed by ##40\,\text{km/h}## on total?
The total distance to work is 70 km, including the 10 km inside the town (which is given).
Let ##d## be the distance traveled outside the town when David goes to work. Then his original average speed would be $$s_1 = \frac {(d + 10)} {\frac {d} {180} + \frac {10} {40}}$$. After reducing the speeds inside and outside the towns by ##20 km/h## each, the new average speed becomes $$s_2 = \frac {(d + 10)} {\frac {d} {160} + \frac {10} {20}}$$
Given that the new average speed is lower than the previous average speed by ##40km/h##, we get $$s_1 - s_2 = 40 \Rightarrow \frac {(d + 10)} {\frac {d} {180} + \frac {10} {40}} - \frac {(d + 10)} {\frac {d} {160} + \frac {10} {20}} = 40$$
The above equation reduces to ##(d - 60)^2 = 0##, giving ##d = 60##
P.S.
Thanks for giving 1 simple question that I could solve confidently . But more thanks for the tough questions for which I couldn't not solve - I am keen to learn from the solutions posted by others and from the learning resources.
The total distance to work is 70 km, including the 10 km inside the town (which is given).
Let ##d## be the distance traveled outside the town when David goes to work. Then his original average speed would be $$s_1 = \frac {(d + 10)} {\frac {d} {180} + \frac {10} {40}}$$. After reducing the speeds inside and outside the towns by ##20 km/h## each, the new average speed becomes $$s_2 = \frac {(d + 10)} {\frac {d} {160} + \frac {10} {20}}$$
Given that the new average speed is lower than the previous average speed by ##40km/h##, we get $$s_1 - s_2 = 40 \Rightarrow \frac {(d + 10)} {\frac {d} {180} + \frac {10} {40}} - \frac {(d + 10)} {\frac {d} {160} + \frac {10} {20}} = 40$$
The above equation reduces to ##(d - 60)^2 = 0##, giving ##d = 60##
P.S.
Thanks for giving 1 simple question that I could solve confidently . But more thanks for the tough questions for which I couldn't not solve - I am keen to learn from the solutions posted by others and from the learning resources.
There are only a couple of rules to attack a problem, if there is no direct, computational approach:
simplify it
get rid of what disturbs
find and use symmetries
E.g. problems with rational numbers are also problems with integers, What does it say in the integer case? Integer problems are often solved by looking at the remainders modulo one or more primes: If there is an integer solution, then there is a solution with remainders, too.
##x^2+y^2=z^2 \Longrightarrow x^2+y^2\equiv z^2 \mod 2 \Longrightarrow \{\,x,y,z\,\} \text{ contains at least one even number }## because not all ##x,y,z## can be odd, since this would give ##1+1=1## for the remainders.
If there are disturbing denominators, multiply them off.
If there are symmetries, find other symmetric expressions and investigate whether there are relations.
#82
Not anonymous
143
58
fresh_42 said:
There are only a couple of rules to attack a problem, if there is no direct, computational approach:
simplify it
get rid of what disturbs
find and use symmetries
E.g. problems with rational numbers are also problems with integers, What does it say in the integer case? Integer problems are often solved by looking at the remainders modulo one or more primes: If there is an integer solution, then there is a solution with remainders, too.
##x^2+y^2=z^2 \Longrightarrow x^2+y^2\equiv z^2 \mod 2 \Longrightarrow \{\,x,y,z\,\} \text{ contains at least one even number }## because not all ##x,y,z## can be odd, since this would give ##1+1=1## for the remainders.
If there are disturbing denominators, multiply them off.
If there are symmetries, find other symmetric expressions and investigate whether there are relations.
Hi @fresh_42 , thanks for the tips. In that example, when you say ##x^2 + y^2 \equiv z^2 \mod 2## did you mean to say ##(x^2 + y^2) \mod 2 \equiv z^2 \mod 2##?
#83
Not anonymous
143
58
2. Find the equation of a curve such that ##y''## is always ##2## and the slope of the tangent line is ##10## at the point ##(2,6)##.
Is the answer ##y = x^2 + 6x - 10##?
##y'' = 2 \Rightarrow y' = \int y'' \, dx = \int 2 \,dx = (2x + a)##, where ##a## is some constant
##y = \int (2x + a) \, dx = (x^2 + ax + b)## where ##b## is some constant
Since slope of tangent at ##x = 2## is 10, we get ##y'_{(x=2)} = 2 \Rightarrow 2 \times 2 + a = 10 \Rightarrow a = 6##
Since ##(2, 6)## is a point on the curve, ##y_{(x=2)} = 6 \Rightarrow (x^2 + 6x + b)_{(x=2)} = 6 \Rightarrow 4 + 12 + b = 10 \Rightarrow b = -6##
#84
Not anonymous
143
58
1. The maximum value of ##f## with ##f(x) = x^a e^{2a - x}## is minimal for which values of positive numbers ##a## ?
Is the answer ##a = e^{-2}##?
Here is the calculation I used - ##f'(x) = ax^{a-1} e^{2a - x} - x^a e^{2a - x} = x^{a-1} e^{2a - x} (a - x)##
At the maximal point, ##f'(x) = 0 \Rightarrow x=a## or ##x=0##. I have not worked out a rigorous proof for maximality at ##a##, but it can be seen that since when ##x > a##, ##f'(x) < 0##, ##f(a) > 0## (given ##a > 0##), ##x = a## must be the global maximum.
Therefore, maximum value of ##f## is ##a^a e^a##, which we will now denote by ##g(a)##, a function of variable ##a##. Obviously, for ##a > 0##, this expression will not have a finite maximum, since as ##a \rightarrow \inf##, ##f(a) \rightarrow \inf##. Only a finite minimum would be possible and at the minimal point ##g'(a) = 0##. To simplify differentiation, we may apply logarithm on the expression to find the minimum, which is valid for ##a > 0## since logarithm is a monotonically increasing function. ##h(a) = \ln g(a) = a \ln a + a##. At the minimal point, ##h' = 0 \Rightarrow 2 + \ln a = 0 \Rightarrow a = e^{-2}##
Hi @fresh_42 , thanks for the tips. In that example, when you say ##x^2 + y^2 \equiv z^2 \mod 2## did you mean to say ##(x^2 + y^2) \mod 2 \equiv z^2 \mod 2##?
Yes, except the correct notation is ##a \equiv b \mod 2## or ##a=b \mod 2## The module automatically covers the entire equation. It is basically a function from ##\mathbb{Z}## to ##\mathbb{Z}/\mathbb{2Z}=\mathbb{Z}_2=\{\,0,1\,\}##. So ##\mod n## means to apply this function on both sides, otherwise it wouldn't make sense.
#86
Pi-is-3
49
13
##2x^6+3y^6=z^3##
It suffices to prove that ##2x^3+3y^3=1## has no rational roots (expression obtained by dividing by ##z^3## ).
We see that ##\frac{3y^3}{2} = \frac{1}{2} -x^3##
We see that ##x^3= \frac{1}{2} - t^3## and ##y= \frac{2t^3}{3}##
If t is rational, then y is irrational. If t is irrational, then x is irrational. Hence there are no solutions for x and y in rationals.
It suffices to prove that ##2x^3+3y^3=1## has no rational roots (expression obtained by dividing by ##z^3## ).
Prove it!
#88
Pi-is-3
49
13
Not anonymous said:
1. The maximum value of ##f## with ##f(x) = x^a e^{2a - x}## is minimal for which values of positive numbers ##a## ?
Is the answer ##a = e^{-2}##?
Here is the calculation I used - ##f'(x) = ax^{a-1} e^{2a - x} - x^a e^{2a - x} = x^{a-1} e^{2a - x} (a - x)##
At the maximal point, ##f'(x) = 0 \Rightarrow x=a## or ##x=0##. I have not worked out a rigorous proof for maximality at ##a##, but it can be seen that since when ##x > a##, ##f'(x) < 0##, ##f(a) > 0## (given ##a > 0##), ##x = a## must be the global maximum.
Therefore, maximum value of ##f## is ##a^a e^a##, which we will now denote by ##g(a)##, a function of variable ##a##. Obviously, for ##a > 0##, this expression will not have a finite maximum, since as ##a \rightarrow \inf##, ##f(a) \rightarrow \inf##. Only a finite minimum would be possible and at the minimal point ##g'(a) = 0##. To simplify differentiation, we may apply logarithm on the expression to find the minimum, which is valid for ##a > 0## since logarithm is a monotonically increasing function. ##h(a) = \ln g(a) = a \ln a + a##. At the minimal point, ##h' = 0 \Rightarrow 2 + \ln a = 0 \Rightarrow a = e^{-2}##
One wrong assumption, x=a is not the global maximal for all a. If a is negative or an even natural number, then it is just a critical point. But the answer is correct because it is the global maxima of ##e^{-2}##.
#89
Pi-is-3
49
13
fresh_42 said:
Prove it!
Sorry, I thought it was sufficient to leave it at that.
Let ##x^2=a, y^2=b##.
Then we have ##3a^3+2b^3=z^3##
Let ##\frac{a}{z}=j ## and ## \frac{b}{z}=k##
If x,y,z are rational, then so are j and k.
Finally, if their are no rational solutions of j and k, then their are no rational solutions for x,y,z.
This is why it suffices to prove that ##3j^3+2k^3=1## has no rational solutions.
EDIT: I made one more mistake, I didn't mention that z is not equal to 0. If z=0, then the answers come to be (0,0,0) which is excluded.
Sorry, I thought it was sufficient to leave it at that.
No. It is a bad habit and a common place to hide mistakes.
You have to show that a solution of the general version provides a solution for the restricted one. So let us assume we have a solution ##(r,s,t)## such that ##2r^6 +3s^6=t^3##.
From ##t=0## we obtain the trivial solution only, since the left hand side is nonnegative.
For ##t\neq 0## we get ##2 \left( \dfrac{r}{t} \cdot r \right)^3\cdot 3 \left( \dfrac{s}{t} \cdot s\right)^3=1\,.##
It suffices to prove that ##2x^3+3y^3=1## has no rational roots (expression obtained by dividing by ##z^3## ).
Done in post #89, although the additional variables make it unnecessarily complicated (see post #90).
We see that ##\frac{3y^3}{2} = \frac{1}{2} -x^3##
We see that ##x^3= \frac{1}{2} - t^3## and ##y= \frac{2t^3}{3}##
We do not "see" it. What we see is that you invented out of the blue a certain number ##t##. What you should have done is telling us, that you set ##t:= \sqrt[3]{\frac{3}{2}y} \in \mathbb{R}## such that ##y=\frac{2}{3}t^3##. But now ##x^3\neq \frac{1}{2}-t^3##.
To correct this, we have to set ##t:=\sqrt[3]{\frac{3}{2}}y##. Now ##x^3= \frac{1}{2}-t^3##.
If t is rational, then y is irrational.
How that?
If t is irrational, then x is irrational. Hence there are no solutions for x and y in rationals.
#92
member 587159
nuuskur said:
Proof of fact. Suppose S\in\Sigma has property (P). Then \Sigma _S is an infinite sub sigma algebra. Pick T \in \Sigma _S \setminus \{\emptyset, S\} Write S= T\cup (S\cap T^c). By (E) at least one of the respective sub sigma algebras must be infinite.
First of all, sorry for the late reply. I'm trying to understand this part of your answer. I guess you want to show that either ##\Sigma_T## or ##\Sigma_{S \cap T^c}## has property ##(P)##, right?
Can you please explain me why the following sentence is true?
"By (E) at least one of the respective sub sigma algebras must be infinite."
Can you please explain me why the following sentence is true?
"By (E) at least one of the respective sub sigma algebras must be infinite."
If ##\Sigma _T \cup \Sigma _{S\cap T^c}## was a finite set, then their generated sub sigma algebra, which is ##\Sigma _S##, would also be finite.
I guess you want to show that either ##\Sigma_T## or ##\Sigma_{S \cap T^c}## has property ##(P)##, right?
They can both have that property, too.
Last edited:
#94
Not anonymous
143
58
8. Show that there is no triple ##(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}## such that ##a^2 + b^2 = 3 c^2##.
After spending a few seconds thinking about it, I realized that this problem is much simpler than what I had imagined when I first came across this question in the challenge list. An interesting problem nevertheless for people like me who have never had the opportunity to a course in number theory and the like and want some warm-up questions before moving to tougher problems.
I think there might be a simpler proof than what I came up with.
Since ##a^2 + b^2 = 3 c^2##, it must be the case that ##a^2 + b^2 \equiv 0 \mod 3##. Since ##(x + y) \mod n = x \mod n + y \mod n)##, we consider all possible combinations of ##a \mod 3## and ##b \mod 3## and find their effect on ##a^2 \mod 3 + b^2 \mod 3##. We only need to consider the modulo pairs (0, 0), (0, 1), (0, 2), (1, 1), (1, 2) and (2, 2) for the modulos of a and b, since the remaining cases like (2, 1) are symmetric. And we use the observation that if ##x \equiv k \mod n##, then ##x^2 \equiv k^2 \mod n## (since ##x = pn + k## for some integer ##p## and ##x^2 = (p^2 n^2 + k^2 + 2kpn) \equiv k^2 \mod n##).
So the only case where ##(a^2 + b^2) \mod 3 = 0## is when the LHS modulo pair is (0, 0), i.e. both a and b are multiples of 3, say ##a = 3A, b = 3B## for some non-zero integers ##A, B## (since question states a, b, c are all non-zero). Using this in the original expression, we get ##a^2 + b^2 = 9A^2 + 9B^2 = 3 c^2 \Rightarrow A^2 + B^2 = \frac {c^2} {3}##
Since ##A^2 + B^2## and hence ##\frac {c^2} {3}## is an integer, and ##c## is an integer and 3 is prime, it must be the case that ##c## too is a multiple of 3, say ##c = 3C## for some non-zero integer ##C##. So the above equivalence becomes, ##A^2 + B^2 = \frac {9C^2} {3} = 3C^2##, which is of the same for as the original equation. So as long as the 2 integers on LHS are multiples of 3, we can reduce the expression to another expression of the same form but with the absolute values of the integers becoming smaller. This cannot continue infinitely, implying that we will reach a point where at least one of the 2 integers is no longer a multiple of 3, meaning that the modulo pair is no longer (0, 0). And for any such non-(0, 0) pair, we already showed that the RHS cannot be a multiple of 3, but this contradicts the expression from that required the RHS to be a multiple of 3. Therefore, the only remaining possibility, i.e. of having an LHS module pair of the form (0, 0) that also satisfies the expression form ##a^2 + b^2 = 3 c^2##, too is ruled out. Hence proved.
#95
member 587159
Not anonymous said:
8. Show that there is no triple ##(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}## such that ##a^2 + b^2 = 3 c^2##.
After spending a few seconds thinking about it, I realized that this problem is much simpler than what I had imagined when I first came across this question in the challenge list. An interesting problem nevertheless for people like me who have never had the opportunity to a course in number theory and the like and want some warm-up questions before moving to tougher problems.
I think there might be a simpler proof than what I came up with.
Since ##a^2 + b^2 = 3 c^2##, it must be the case that ##a^2 + b^2 \equiv 0 \mod 3##. Since ##(x + y) \mod n = x \mod n + y \mod n)##, we consider all possible combinations of ##a \mod 3## and ##b \mod 3## and find their effect on ##a^2 \mod 3 + b^2 \mod 3##. We only need to consider the modulo pairs (0, 0), (0, 1), (0, 2), (1, 1), (1, 2) and (2, 2) for the modulos of a and b, since the remaining cases like (2, 1) are symmetric. And we use the observation that if ##x \equiv k \mod n##, then ##x^2 \equiv k^2 \mod n## (since ##x = pn + k## for some integer ##p## and ##x^2 = (p^2 n^2 + k^2 + 2kpn) \equiv k^2 \mod n##).
So the only case where ##(a^2 + b^2) \mod 3 = 0## is when the LHS modulo pair is (0, 0), i.e. both a and b are multiples of 3, say ##a = 3A, b = 3B## for some non-zero integers ##A, B## (since question states a, b, c are all non-zero). Using this in the original expression, we get ##a^2 + b^2 = 9A^2 + 9B^2 = 3 c^2 \Rightarrow A^2 + B^2 = \frac {c^2} {3}##
Since ##A^2 + B^2## and hence ##\frac {c^2} {3}## is an integer, and ##c## is an integer and 3 is prime, it must be the case that ##c## too is a multiple of 3, say ##c = 3C## for some non-zero integer ##C##. So the above equivalence becomes, ##A^2 + B^2 = \frac {9C^2} {3} = 3C^2##, which is of the same for as the original equation. So as long as the 2 integers on LHS are multiples of 3, we can reduce the expression to another expression of the same form but with the absolute values of the integers becoming smaller. This cannot continue infinitely, implying that we will reach a point where at least one of the 2 integers is no longer a multiple of 3, meaning that the modulo pair is no longer (0, 0). And for any such non-(0, 0) pair, we already showed that the RHS cannot be a multiple of 3, but this contradicts the expression from that required the RHS to be a multiple of 3. Therefore, the only remaining possibility, i.e. of having an LHS module pair of the form (0, 0) that also satisfies the expression form ##a^2 + b^2 = 3 c^2##, too is ruled out. Hence proved.
Looks correct. A solution anaologuous to yours was already provided though.
#96
member 587159
nuuskur said:
If ##\Sigma _T \cup \Sigma _{S\cap T^c}## was a finite set, then their generated sub sigma algebra, which is ##\Sigma _S##, would also be finite.
Right, I wrote (E) for S=X and T=S. Perhaps, "similarly to (E)" would have been a better choice of words. The proof is the same up to labelling, regardless. Since \Sigma _T \cup \Sigma _{S\cap T^c} \subseteq \Sigma _S we have
<br />
\sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right ) \subseteq \Sigma _S.<br />
Conversely, for every A\in\Sigma_S, we have
<br />
A = A \cap (T\cup S\cap T^c) = [A\cap T] \cup [A\cap (S\cap T^c)]\in \sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right )<br />
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#98
Not anonymous
143
58
Not anonymous said:
8. Show that there is no triple ##(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}## such that ##a^2 + b^2 = 3 c^2##.
After spending a few seconds thinking about it, I realized that this problem is much simpler than what I had imagined when I first came across this question in the challenge list. An interesting problem nevertheless for people like me who have never had the opportunity to a course in number theory and the like and want some warm-up questions before moving to tougher problems.
I think there might be a simpler proof than what I came up with.
Since ##a^2 + b^2 = 3 c^2##, it must be the case that ##a^2 + b^2 \equiv 0 \mod 3##. Since ##(x + y) \mod n = x \mod n + y \mod n)##, we consider all possible combinations of ##a \mod 3## and ##b \mod 3## and find their effect on ##a^2 \mod 3 + b^2 \mod 3##. We only need to consider the modulo pairs (0, 0), (0, 1), (0, 2), (1, 1), (1, 2) and (2, 2) for the modulos of a and b, since the remaining cases like (2, 1) are symmetric. And we use the observation that if ##x \equiv k \mod n##, then ##x^2 \equiv k^2 \mod n## (since ##x = pn + k## for some integer ##p## and ##x^2 = (p^2 n^2 + k^2 + 2kpn) \equiv k^2 \mod n##).
So the only case where ##(a^2 + b^2) \mod 3 = 0## is when the LHS modulo pair is (0, 0), i.e. both a and b are multiples of 3, say ##a = 3A, b = 3B## for some non-zero integers ##A, B## (since question states a, b, c are all non-zero). Using this in the original expression, we get ##a^2 + b^2 = 9A^2 + 9B^2 = 3 c^2 \Rightarrow A^2 + B^2 = \frac {c^2} {3}##
Since ##A^2 + B^2## and hence ##\frac {c^2} {3}## is an integer, and ##c## is an integer and 3 is prime, it must be the case that ##c## too is a multiple of 3, say ##c = 3C## for some non-zero integer ##C##. So the above equivalence becomes, ##A^2 + B^2 = \frac {9C^2} {3} = 3C^2##, which is of the same for as the original equation. So as long as the 2 integers on LHS are multiples of 3, we can reduce the expression to another expression of the same form but with the absolute values of the integers becoming smaller. This cannot continue infinitely, implying that we will reach a point where at least one of the 2 integers is no longer a multiple of 3, meaning that the modulo pair is no longer (0, 0). And for any such non-(0, 0) pair, we already showed that the RHS cannot be a multiple of 3, but this contradicts the expression from that required the RHS to be a multiple of 3. Therefore, the only remaining possibility, i.e. of having an LHS module pair of the form (0, 0) that also satisfies the expression form ##a^2 + b^2 = 3 c^2##, too is ruled out. Hence proved.
For the last step of my proof, I found a simpler variant, based on same fundamentals however.
We need to prove that ##a^2 + b^2 = 3 c^2## where ##(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}## does not have a solution for the case of ##(a \mod 3, b \mod 3) \equiv (0, 0)## (absence of solution for other modulo combinations was already proven).
Say ##a = 3^{p}z, b = 3^{q}y, c = 3^{r}z##, where ##p, q, r## are the highest powers associated with 3 in the prime factorization of ##a, b, c## respectively. By this definition, ##x, y, z## will all be non-multiples of 3, implying ##x \mod 3 \neq 0, y \mod 3 \neq 0, z \mod 3 \neq 0##. If we assume ##(a \mod 3, b \mod 3) \equiv (0, 0)##, then ##p, q## should be positive integers. Therefore, ##a^2 + b^2 = 3 c^2 \Rightarrow 3^{2p}x^2 + 3^{2q}y^2 = 3 \times 3^{2r}z^2##. If we arbitrarily assume that ##p \ge q##, then LHS becomes ##3^{2p} (x^2 + 3^{2q - 2p}y^2)##. The proof of ##p \ge q## case is symmetric.
Subcase 1: If ##p = q##, then LHS simplifies further to ##3^{2p} (x^2 + y^2)##. Since x, y are non-multiples of 3, ##(x^2 + y^2) \mod 3 \neq 0## as already show earlier, so the exponent of 3 in prime factorization of LHS is ##2p##, an even natural number.
Subcase 2: If ##p \neq q##, then too exponent of 3 in prime factorization of LHS still remains ##2p##, because then in ##(x^2 + 3^{2q - 2p}y^2)##, the 2nd component, ##3^{2q - 2p}y^2## would be a multiple of 3 but ##x^2## would not be (because x is not a multiple of 3 by definition), and hence ##(x^2 + 3^{2q - 2p}y^2) \mod 3 \neq 0##. So here too, the exponent of 3 in prime factorization of LHS is an even natural number.
Whereas, on the RHS we have ##3 \times 3^{2r}z^2 = 3^{2r+1} z^2##, and by definition ##z## is a non-multiple of 3. Hence the exponent of 3 in prime factorization of RHS is ##(2r + 1)##, an odd number.
This means that the exponents of 3 on LHS and RHS cannot match, which contradicts the requirement of LHS = RHS. Hence proved
For the sake of convenience I have put all candidate solutions to this post. If possible, please delete #73 and #76.
The fundamental theorem of abelian groups says that for any finite abelian group
$$
A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}
$$
I'll be using Sylow's theorems which I will refer to as theorems 1, 2 and 3 respectively. (https://en.wikipedia.org/wiki/Sylow_theorems)
Of order ##6## there are
$$
\mathbb Z_6 \qquad S_3.
$$
Of order ##8## there are
$$
\mathbb Z_8 \qquad \mathbb Z_4 \oplus \mathbb Z _2 \qquad \mathbb Z_2 \oplus\mathbb Z_2\oplus\mathbb Z_2\qquad D_4 \qquad \mbox{Dic}_2.
$$
Of order ##12## there are
$$
\mathbb Z_{12}\qquad \mathbb Z_6 \oplus \mathbb Z_2 \qquad A_4 \qquad \mathbb Z_3 \rtimes_\varphi \mathbb Z_4 \qquad \mathbb Z_3 \rtimes _\varphi (\mathbb Z_2\oplus\mathbb Z_2).
$$
In the abelian case there are ##\mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3##.
Quick remark. Being a Sylow ##p-##subgroup is invariant with respect to conjugating, thus if ##H## was a unique Sylow ##p-##subgroup, we would have##gHg^{-1} = H, g\in G,## i.e ##H## would be normal.
Fact. Suppose a non-abelian group ##G## has order ##pq## (primes) with ##q\equiv 1 \pmod{p}##, then there are precisely ##q## Sylow##p-##subgroups in ##G##.
Proof of fact. Let n_p,n_q be the numbers of Sylow p,q-subgroups respectively. By assumption q>p. By theorem 3 n_q \equiv 1 \pmod{q} and n_q\mid p, which forces n_q = 1, thus we have a (normal) subgroup of order q, which makes it cyclic, therefore abelian.
For n_p we thus have two choices: n_p = 1,q. If n_p = 1, then G\cong \mathbb Z_p \oplus \mathbb Z_q would be abelian, thus it must be that n_p = q.
Our non-abelian group G therefore contains three Sylow 2-subgroups, call them H_1,H_2,H_3, which by theorem 2 are all conjugate to each other. We have \mbox{Sym}(H_1,H_2,H_3) \cong S_3. We show that G\cong S_3. Define
<br />
\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}<br />
(also called conjugating)
Firstly, if H is a Sylow p-subgroup, then conjugating it gives another Sylow p-subgroup, hence it must hold that gH_ig^{-1} = H_j for some j.
The map \varphi (g) is injective, because if gH_ig^{-1} = gH_jg^{-1}, then
<br />
h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.<br />
The argument is symmetrical, thus H_i=H_j. It is surjective due to finiteness.
Take g,h\in G, then
<br />
\varphi (gh) (H_j) = (gh)H_j(gh)^{-1} = g(hH_jh^{-1})g^{-1} = g (\varphi (h)(H_j)) g^{-1} = \varphi (g) (\varphi (h)(H_j)) = (\varphi (g) \varphi (h))(H_j).<br />
To show it's an isomorphism, it suffices to show it's injective (because |G| = |S_3| = 6). Suppose \varphi (g) = \mbox{id} i.e gH_jg^{-1}=H_j, then by definition g\in N(H_j) the normaliser of H_j. By theorem 3 all N(H_j) = H_j and g\in H_1\cap H_2\cap H_3 = \{e\} i.e g=e, which makes \varphi an isomorphism.
In the abelian case there are
<br />
\mathbb Z_8 \qquad \mathbb Z_4 \oplus \mathbb Z _2 \qquad \mathbb Z_2 \oplus\mathbb Z_2\oplus\mathbb Z_2<br />
Suppose G is non-abelian of order 8. If there is an element of order 8, then G is cyclic, thus abelian. Suppose the maximal order is 2 and pick g,h\in G. If gh=e, then they commute. Suppose (gh)^2 = e, then
<br />
gh = geh = g(ghgh)h = (gg)hg(hh) = ehge = hg<br />
and again G would be abelian. Thus we must have an element a of order 4. Its generated subgroup \langle a \rangle =: H is of index 2, therefore normal. Take b\notin H, then we have bab^{-1}\in H due to normality. Notice that
<br />
(bab^{-1})^4 = bab^{-1}bab^{-1}bab^{-1}bab^{-1} = baaaab^{-1} = e.<br />
If also (bab^{-1}) ^2 = e, then baab^{-1} = e would lead to aa=e, contradicting the order of a.
Next, we find the possible values of bab^{-1}.
If bab^{-1} = a^4 = e, then a=e, which is impossible.
If bab^{-1} = a, then ba = ab, but this will make G abelian. Indeed, pick g,h\in G. If they're both in H, then they commute. Suppose g\notin H and write g = ba^m and h = a^n, then
<br />
gh = ba^ma^n = ba^na^m = a^n ba^m = hg.<br />
and if both reside outside H, then writing g = ba^m and h = ba^n would similarly lead to gh=hg. But our group is not abelian.
If bab^{-1} = a^2, then (bab^{-1})^2 = e would contradict the order of bab^{-1}.
Thus, the only possibility is bab^{-1}=a^3 =a^{-1}.
There are two cases to consider.
b is of order 2, then we have G \cong \langle a,b \mid \mbox{ord}(a) = 4, \mbox{ord}(b) = 2, bab^{-1} = a^{-1} \rangle this is the dihedral group D_4.
b is of order 4. We show G is the dicyclic group \mbox{Dic}_2, where
<br />
\mbox{Dic}_2 \cong \langle x,y \mid x^{4} = e, x^2 = y^2, x^{-1} = y^{-1}xy \rangle.<br />
Notice that
<br />
bab^{-1} = a^{-1} \Rightarrow b^{-1}ab = b^{-1}(ba^{-1}b^{-1})b = a^{-1}.<br />
So we have left to show a^2 = b^2. Considering the natural projection G\to G/H\cong\mathbb Z_2, we have b^2 \in H. Consider possible values of b^2.
b^2 = a^4 = e contradicts order of b.
b^2 = a contradicts order of a
b^2 = a^2 is what we want.
b^2 = a^{-1} contradicts order of a^{-1} (hence order of a).
Thus b^2 = a^2 must hold.
In the abelian case there are
<br />
\mathbb Z_{12}\qquad \mathbb Z_6 \oplus \mathbb Z_2.<br />
Suppose G is non-abelian of order 12. Let H_p denote a Sylow p-subgroup and n_p the number of Sylow p-subgroups in G. Then, by theorem 1, we consider subgroups H_2 and H_3. The case n_2=n_3=1 yields normality and H_2 \cong \mathbb Z_4 or H_2 \cong \mathbb Z_2\oplus \mathbb Z_2 and H_3 \cong \mathbb Z_3 and that leads to one of the previously mentioned abelian structures.
Case 2 : n_3= 4. Call them H^1,H^2,H^3,H^4. Define
<br />
\varphi : G \to \mbox{Sym}(H^1,H^2,H^3,H^4) \cong S_4, \quad \varphi (g) := gH^jg^{-1}.<br />
Since conjugation preserves the property of being a Sylow p-subgroup and injectivity of each \varphi (g) is trivial, the map is well-defined. It's a routine task to check it's a morphism.
Notice, if, say, aH^ia^{-1} = H^j, then aN(H^{i})a^{-1} = N(H^j), so taking normalisers preserves the property of being conjugate.
Now, suppose \varphi (g) = \mbox{id}, then by definition g\in N(H^j). By theorem 3 N(H^j) = H^j, thus g=e is forced and \varphi is injective. Furthermore, each H^j contains two generators, therefore G has 8 elements of order 3.
Consider the alternating group A_4. Trivially \varphi (G) \cap A_4 \leqslant\varphi (G) and A_4 also contains 8 elements of order 3, thus G\cong A_4.
Case 3: n_3 = 1. Call it H and pick a Sylow 2-subgroup K (of order 4). As H is normal it holds HK is a subgroup of G. As H\cap K = \{e\} (by orders of elements), then
<br />
|HK| = \frac{|H||K|}{|H\cap K|} \Rightarrow HK = G.<br />
We show G\cong H\rtimes _\varphi K, the semidirect product under \varphi : K\to \mbox{Aut}(H), where
<br />
(h,k)\cdot (h',k') := (h\varphi (k)(h'),kk')<br />
This will yield two more groups, namely the following:
<br />
\mathbb Z_3 \rtimes_\varphi \mathbb Z_4 \qquad \mathbb Z_3 \rtimes _\varphi (\mathbb Z_2\oplus\mathbb Z_2).<br />
Define
<br />
\varphi : K\to \mbox{Aut}(H),\quad \varphi (k)(h) := khk^{-1}.<br />
I will omit the routine checks. We show
<br />
\tau : G\to H\rtimes _\varphi K, \quad g=hk \mapsto (h,k),<br />
is an isomorphism. Although, let's have some fun and do it the category theory way, for a change. First, we check \tau is a morphism. Take g=hk, g'=h'k'\in G, then due to normality kh'k^{-1} \in H and we get
<br />
\begin{align*}<br />
\tau (hkh'k') &= \tau (h(kh'k^{-1})kk') \\ &= (h(kh'k^{-1}),kk')\\ &= (h\varphi (k)(h'),kk')\\ &= (h,k)(h',k')\\ &= \tau (hk) \tau (h'k').<br />
\end{align*}<br />
Also \tau (e) = \tau (ee) = (e,e). Now define
<br />
\omega : H\rtimes _\varphi K \to G, \quad (h,k) \mapsto hk.<br />
Verify it, too, is a morphism. Take (h,k),(h',k')\in H\rtimes _\varphi K, then
<br />
\begin{align*}<br />
\omega ((h,k)(h',k')) &= \omega ((h\varphi (k)(h'), kk'))\\ &= h(kh'k^{-1})kk' \\ &= hkh'k' \\ &= \omega ((h,k)) \omega ((h',k')).<br />
\end{align*}<br />
Also \omega ((e,e)) = ee = e. The equalities \tau\omega = \mbox{id}_{H\rtimes _\varphi K} and \omega\tau = \mbox{id}_G are straightforward to verify.
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#100
member 587159
nuuskur said:
Right, I wrote (E) for S=X and T=S. Perhaps, "similarly to (E)" would have been a better choice of words. The proof is the same up to labelling, regardless. Since \Sigma _T \cup \Sigma _{S\cap T^c} \subseteq \Sigma _S we have
<br />
\sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right ) \subseteq \Sigma _S.<br />
Conversely, for every A\in\Sigma_S, we have
<br />
A = A \cap (T\cup S\cap T^c) = [A\cap T] \cup [A\cap (S\cap T^c)]\in \sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right )<br />
I went through your solution again and I think it is correct! Well done!
mhh, the following is a cheapshot
. But more thanks for the tough questions for which I couldn't not solve - I am keen to learn from the solutions posted by others and from the learning resources.
