# Challenge Math Challenge - October 2018

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#### fresh_42

Mentor
2018 Award
The conjugacy classes are defined the usual way in group theory: Conjugacy class - Wikipedia. For a group G and element x in it, a conjugacy class is the set of all values of g.x.g-1 for all g in G.
I know what conjugacy classes are. However, you deal with two groups $A_5$ and $S_5$ and neither said where $g$ is from nor what $x$ is. This is needed to unambiguously speak of conjugacy classes. Sorry, but my crystal ball is out of order. My guess is that you talk about $S_5$, for otherwise you won't get $24$ elements in a class, but why?
The cycle content is the set of lengths of cycles in it.
A cycle is $(n_1 \,\ldots \,n_k)$ with one length $k$. Now what are cycles in a cycle? And how can there be more than one length?
The number of elements is for the conjugacy class.
You wrote e.g. "5:even 24" and in the legend "(cycle content): (parity) (number of elements)"
Now does this mean we have $24$ $5-$cycles, or that each conjugacy class of one $5-$cycle contains $24$ elements? I assume the latter, as you just said the number of elements in a class.
Sharing a set of cycle lengths means all the elements having the same set of cycle lengths.
So you mean if $\pi = \sigma_1 \ldots \sigma_k$ is a presentation of $\pi$ with disjoint cycles $\sigma_i$, then you inspect the set $\{\,\operatorname{ord}(\sigma_i)\,|\,1 \leq i \leq k\,\}$? And the set, not the ordered vector $(\operatorname{ord}(\sigma_i))_i$, is that correct?
Now $1⋅2^2:even 15$ means: All conjugacy classes of a product of two transpositions have $15$ elements? And you could have omitted all the $1^k$ as they are the identity anyway? Why don't you say this instead of a cryptic "$1⋅2^2:even 15$" code.
There is a theorem about permutations that conjugation of a permutation does not change its cycle lengths.]Symmetric group - Wikipedia and Alternating group - Wikipedia both discuss the contents of the groups' conjugacy classes. For the symmetric group, "two elements of Sn are conjugate in Sn if and only if they consist of the same number of disjoint cycles of the same lengths."

I brought in S5 because I wanted to derive the sizes of the classes of A5.
Does this mean you considered $|gA_5g^{-1}|$ for $g\in S_5$? What for? Both classes of $A_5$ have sixty elements.
An is Simple contains my argument about class sizes and normal subgroups near the bottom of the second page.
Thanks, I have a proof. I only try to understand yours. You do not need to explain standard vocables, you need to use them correctly (conjugacy classes without the source of conjugation and element), and explain your inventions: set of cycle lengths, cycles contain cycles.

#### fresh_42

Mentor
2018 Award
I reviewed your proof and there are still some open questions.
Solution of 10d
The group A5 is the group of all even permutations of 5 symbols.

I will show that A5 is simple by using some results from my solution for 10b. In particular, every normal subgroup of a group must include complete conjugacy classes of that group, including the identity-element class.

We start with the classes of S5, the group of all permutations of 5 symbols. With their parities and sizes, the are:
$${1^5: even\ 1, 1^3 \cdot 2: odd\ 10, 1 \cdot 2^2: even\ 15, 1^2 \cdot 3: even\ 20, 2 \cdot 3: odd\ 20, 1 \cdot 4: odd\ 30, 5: even\ 24}$$
E.g. you say that there are $15$ elements in $g(n_1n_2)(n_3n_4)g^{-1}\; , \;(g \in S_5)$. How do we know? There are $15$ pairs of disjoint transpositions, but why does conjugation operate transitive on each class? I see that it is clear for single cycles as they generate a subgroup of $S_5$ and Lagrange tells us the size, which in my opinion you should have mentioned, but why is $|g(n_1n_2)(n_3n_4)g^{-1}|=15$?
Selecting out the even ones for A4 ...
$A_5$ [for possible readers]
... and splitting the ones that get split, I find
$${1^5: 1, 1 \cdot 2^2: 15, 1^2 \cdot 3: 20, 5_1: 12, 5_2: 12}$$
What is $5_i$ and what does split how?
I then used Mathematica ...
... which I find difficult to accept as a proof but it also can be seen by inspection ...
... to find the sums of all selections of the class sizes that include the identity class. They are {1, 13, 16, 21, 25, 28, 33, 36, 40, 45, 48, 60}.
Of these, only 1 and 60 evenly divide the order of A5, 60, and that means that A5 has no nontrivial normal subgroups. Thus, A5 is simple.
O.k. I can accept that this is a proof, although the way it is presented is a bit sloppy and strange.

An alternative proof and which I find is much better to read is the following:
1. $A_n$ are generated by permutations of the form $(12k)$.
2. A normal subgroup of $A_n$ which contains a $3-$ cycle also contains $(12k)$ and thus the entire group.
3. Choose a permutation of a normal subgroup of $A_n$ which permutes the least possible number of elements.
4. Show that this permutation is necessarily a $3-$cycle.

#### nuuskur

1. Since $\infty\notin \emptyset$ and $\emptyset\subset\mathbb C$ is open, then $\emptyset\in \overline{\tau}$. Also, as $\infty\in\overline{\mathbb C}$ and $\overline{\mathbb C}^c=\emptyset\subset \mathbb C$ is compact, then $\overline{\mathbb C}\in\overline{\tau}$
2. Pick $U,V\in\overline{\tau}\neq\emptyset$. For $U\cap V$ there are three possibilities (four, but two are symmetrical).
1. Neither $U$ nor $V$ contain $\infty$ and both are open in $\mathbb C$, then $U\cap V\subseteq\mathbb C$is open and $U\cap V\in\overline{\tau}$.
2. It holds that $\infty\in U\cap V$ and both $U^c$ and $V^c$ are compact in $\mathbb C$. As compactness is preserved under finite unions, then $(U\cap V)^c\subset\mathbb C$ is compact (de Morgan laws) and $U\cap V\in\overline{\tau}$.
3. It holds that $\infty\in V$ and $V^c\subset\mathbb C$ is compact and $\infty\notin U$ and $U\subseteq\mathbb C$ is open. Then certainly $\infty\notin U\cap V$, $V\cap\mathbb C$ is open in $\mathbb C$ and thus $U\cap V\in\overline{\tau}$.
3. Let $I\neq\emptyset$ and $V_i\in\overline{\tau}, i\in I$. Check two mutually exclusive cases.
1. For every $i\in I$ it holds that $\infty\notin V_i$ and $V_i\subseteq\mathbb C$ is open. Then $V:=\bigcup_{i\in I}V_i\subseteq\mathbb C$ is open and $\infty\notin V$. Thus $V\in\overline{\tau}$.
2. Exists $j\in I$ with $\infty\in V_j$ and $V_j^c\subset\mathbb C$ compact. For every $i\in I$ it holds that $V_i^c\subseteq\mathbb C$ is closed. Closedness is preserved under arbitrary intersections, therefore (by de Morgan laws) the subset
$$V^c=\left (\bigcup _{i\in I} V_i\right )^c = \bigcap _{i\in I} V_i^c \subseteq V_j^c$$
is closed. Since $V_j^c$ is compact (also closed, because we are in $T_2$), the subset $V^c$ is compact. We also have $\infty\in V$, therefore $V\in \overline{\tau}$.​

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#### nuuskur

In $T_2$ spaces every compact subset is closed. At 2.3, since $V^c\subset \mathbb C$ is compact, it is also closed, thus its complement (in $\mathbb C$) is necessarely open. I don't think this works if we didn't have compact $\Rightarrow$ closed.

Clearly the standard euclidean topology $\tau\subset\overline{\tau}$. We need only check the case if $x\in\mathbb C$ and $y=\infty$.

Wo.l.o.g assume $x=0$ and take $U$ to be the open unit ball. Denote by $\overline{U}$ the euclidean closed unit ball. Take $V = \overline{\mathbb C}\setminus {\overline{U}}$. The sets $U,V\in\overline{\tau}$.
Then $\infty\in V$, $0\in U$ and $U\cap V =\emptyset$ (the disconnection occurs on the sphere in between).
Given an open cover of $\overline{\mathbb C}$ we must have an open set $V$ in the open cover with $\infty\in V$. By definition $V^c$ is compact in $\mathbb C$, thus we have a finite subcover $V^c\subseteq \bigcup _{j=1}^n U_j$, therefore $\overline{\mathbb C} = V\cup \bigcup_{j=1}^n U_j$ is a finite subcover.

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#### fresh_42

Mentor
2018 Award
In $T_2$ spaces every compact subset is closed. At 2.3, since $V^c\subset \mathbb C$ is compact, it is also closed, thus its complement (in $\mathbb C$) is necessarely open. I don't think this works if we didn't have compact $\Rightarrow$ closed.

Clearly the standard euclidean topology $\tau\subset\overline{\tau}$. We need only check the case if $x\in\mathbb C$ and $y=\infty$.

Wo.l.o.g assume $x=0$ and take $U$ to be the open unit ball. Denote by $\overline{U}$ the euclidean closed unit ball. Take $V = \overline{\mathbb C}\setminus {\overline{U}}$. The sets $U,V\in\overline{\tau}$.
Then $\infty\in V$, $0\in U$ and $U\cap V =\emptyset$ (the disconnection occurs on the sphere in between).
Given an open cover of $\overline{\mathbb C}$ we must have an open set $V$ in the open cover with $\infty\in V$. By definition $V^c$ is compact in $\mathbb C$, thus we have a finite subcover $V^c\subseteq \bigcup _{j=1}^n U_j$, therefore $\overline{\mathbb C} = V\cup \bigcup_{j=1}^n U_j$ is a finite subcover.
A bit short, nevertheless correct. Instead of w.l.o.g. $0\in U$ one could simply take the shifted balls $|x-z|<1$, and the finite subcover $U_1,\ldots ,U_n$ in the last part could have had the mention that first there has been a cover $\cup_{\alpha \in I}U_\alpha$ of $\hat{\mathbb{C}}$ but o.k.

The procedure in the above is called Alexandroff extension or 1-point-compactification. It is an important tool in topology since the original space, here $\mathbb{C}$ is openly embedded in a compact Hausdorff space. It can be done with any topological space, at least as long as closed subsets of compact sets are compact again.

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#### nuuskur

@fresh_42
I haven't looked too much into it, yet, but at first glance it seems trivialising. The extended topology gains virtually every set in $2^{\overline{\mathbb C}}$. I'm probably crazy for saying that :D

#### fresh_42

Mentor
2018 Award
@fresh_42
I haven't looked too much into it, yet, but at first glance it seems trivialising. The extended topology gains virtually every set in $2^{\overline{\mathbb C}}$. I'm probably crazy for saying that :D
I'm not sure I understood your remark. What is it that's trivial? And what do you mean by gain sets? Since the original topology is still intact, yourself noted $\tau \subseteq \bar{\tau}$, we still have the original sets in $\tau$ resp. $\tau^C$.

#### Infrared

Gold Member
It can be done with any topological space, at least as long as closed subsets of compact sets are compact again.
Fortunately, closed subsets of compact sets are always compact (in any topological space).

#### fresh_42

Mentor
2018 Award
Fortunately, closed subsets of compact sets are always compact (in any topological space).
Thanks, I was too lazy to check. I thought at least some separation property might be necessary.

#### Math_QED

Homework Helper
Solution for (2)

(a) We claim that $$\sigma (\mathcal{F}) = \{A \subseteq X \mid |A| \leq |\mathbb{N}| \lor |A^c| \leq |\mathbb{N}|\}$$

Clearly the right side is contained in the left side, as we can write countable sets as countable union of singeltons. It is also clear that the right set contains $\mathcal{F}$, since singeltons are finite. It remains to check that the right side is a sigma-algebra, which is evident. Just note that the countable union of countable sets remains countable. The other inclusion then follows by minimality.

(b) Define $$\mathcal{S}:=\{B \in \sigma(A) \mid \exists S_0: |S_0| \leq |\mathbb{N|: B \in \sigma(S_0)}\}$$ If we believe the claim the exercise asks for, we have to prove that $\sigma(A)= \mathcal{S}$. Clearly, the right side is contained in the left side. The other inclusion follows again because the right side is a sigma algebra containing $A$ (because $A \in \sigma(\{A\})$).

#### fresh_42

Mentor
2018 Award
Solution for (2)

(a) We claim that $$\sigma (\mathcal{F}) = \{A \subseteq X \mid |A| \leq |\mathbb{N}| \lor |A^c| \leq |\mathbb{N}|\}$$

Clearly the right side is contained in the left side, as we can write countable sets as countable union of singeltons. It is also clear that the right set contains $\mathcal{F}$, since singeltons are finite. It remains to check that the right side is a sigma-algebra, which is evident. Just note that the countable union of countable sets remains countable. The other inclusion then follows by minimality.

(b) Define $$\mathcal{S}:=\{B \in \sigma(A) \mid \exists S_0: |S_0| \leq |\mathbb{N|: B \in \sigma(S_0)}\}$$ If we believe the claim the exercise asks for, we have to prove that $\sigma(A)= \mathcal{S}$. Clearly, the right side is contained in the left side. The other inclusion follows again because the right side is a sigma algebra containing $A$ (because $A \in \sigma(\{A\})$).
The first part is correct, although very short. I'll give a more complete solution for the sake of readability.

A $\sigma-$algebra $\sigma(\mathcal{B})$ over a set $\mathcal{B}$ is a subset of the power set $P(\mathcal{B})$, which contains $\mathcal{B}$ as an element or equivalently $\emptyset$, is closed under complements to $\mathcal{B}$, and closed under the union of countable many sets. For an arbitrary set $X$ and a family of subsets $B \subseteq \mathcal{P}(X)$, $\sigma(B)$ denotes the intersection of all $\sigma-$algebras of subsets of $X$ that contain $B$, i.e. $\sigma(B)=\cap \{\,\sigma(C)\,|\,C\subseteq B\,\}$ with complements are taken to $X$.

We define $\mathcal{A}:=\{\,A\subseteq X\,|\,A \text{ is countable or }X-A \text{ is countable }\,\}$ and show $\mathcal{A}=\sigma(\mathcal{F})$.
Every countable set $A=\{\,x_i\,|\,i\in \mathbb{N}\,\}=\bigcup_{i\in\mathbb{N}}\{\,x_i\,\}\in \sigma(\mathcal{F})$ and also each set with a countable complement. Thus $\mathcal{A}\subseteq \sigma(\mathcal{F})\,.$ In order to show $\sigma(\mathcal{F})\subseteq \mathcal{A}$, we show that $\mathcal{A}$ is a $\sigma-$algebra which contains $\mathcal{F}\,.$
1. $\emptyset \in \mathcal{A}$ since it is countable.
2. If $A\in\mathcal{A}$ then $X-A\in \mathcal{A}$ since $A=X-(X-A)\,.$
3. Let $A_i\in \mathcal{A}\; , \;i\in \mathbb{N}.$ Then either all $A_i$ are countable, and then $\cup_{i\in \mathbb{N}}A_i$ is countable, too, and in $\mathcal{A}$, or there is an index $j$ with an uncountable set $A_j$. In this case is $X-A_j$ countable. Then we have that $X-\cup_{i\in \mathbb{N}}A_i = \cap_{i\in \mathbb{N}} X-A_i \subseteq X-A_j$ is countable, and again $\cup_{i\in \mathbb{N}}A_i \in \mathcal{A}\,.$
4. If $A=\{\,x\,\}\in \mathcal{F}$, then it is countable and so is $A \in \mathcal{A}$, i.e. $\mathcal{F}\subseteq \mathcal{A}\,. \square$

Your solution to the second part confuses me. $S \subseteq \mathcal{P}(X)$ is a given set of sets, but you define an (other?) $S$ in the first line? Regarding your question about what has to be shown: Given any set $A$ from the $\sigma-$algebra of $S$, which are both given, prove that there is a countable subset $S_0$ of the given set $S$, such that $A \in \sigma(S_0)$.

Your definition looks a bit right, as one should consider $\mathcal{A}:=\bigcup\{\,\sigma(C)\,|\,C\subseteq S \text{ countable }\,\}$. But this is at prior simply a set of sets. It should be shown that this is already a $\sigma-$algebra, and that it contains $S$. If this has been shown, the proof is a final small conclusion from that.

#### Math_QED

Homework Helper
The first part is correct, although very short. I'll give a more complete solution for the sake of readability.

A $\sigma-$algebra $\sigma(\mathcal{B})$ over a set $\mathcal{B}$ is a subset of the power set $P(\mathcal{B})$, which contains $\mathcal{B}$ as an element or equivalently $\emptyset$, is closed under complements to $\mathcal{B}$, and closed under the union of countable many sets. For an arbitrary set $X$ and a family of subsets $B \subseteq \mathcal{P}(X)$, $\sigma(B)$ denotes the intersection of all $\sigma-$algebras of subsets of $X$ that contain $B$, i.e. $\sigma(B)=\cap \{\,\sigma(C)\,|\,C\subseteq B\,\}$ with complements are taken to $X$.

We define $\mathcal{A}:=\{\,A\subseteq X\,|\,A \text{ is countable or }X-A \text{ is countable }\,\}$ and show $\mathcal{A}=\sigma(\mathcal{F})$.
Every countable set $A=\{\,x_i\,|\,i\in \mathbb{N}\,\}=\bigcup_{i\in\mathbb{N}}\{\,x_i\,\}\in \sigma(\mathcal{F})$ and also each set with a countable complement. Thus $\mathcal{A}\subseteq \sigma(\mathcal{F})\,.$ In order to show $\sigma(\mathcal{F})\subseteq \mathcal{A}$, we show that $\mathcal{A}$ is a $\sigma-$algebra which contains $\mathcal{F}\,.$
1. $\emptyset \in \mathcal{A}$ since it is countable.
2. If $A\in\mathcal{A}$ then $X-A\in \mathcal{A}$ since $A=X-(X-A)\,.$
3. Let $A_i\in \mathcal{A}\; , \;i\in \mathbb{N}.$ Then either all $A_i$ are countable, and then $\cup_{i\in \mathbb{N}}A_i$ is countable, too, and in $\mathcal{A}$, or there is an index $j$ with an uncountable set $A_j$. In this case is $X-A_j$ countable. Then we have that $X-\cup_{i\in \mathbb{N}}A_i = \cap_{i\in \mathbb{N}} X-A_i \subseteq X-A_j$ is countable, and again $\cup_{i\in \mathbb{N}}A_i \in \mathcal{A}\,.$
4. If $A=\{\,x\,\}\in \mathcal{F}$, then it is countable and so is $A \in \mathcal{A}$, i.e. $\mathcal{F}\subseteq \mathcal{A}\,. \square$

Your solution to the second part confuses me. $S \subseteq \mathcal{P}(X)$ is a given set of sets, but you define an (other?) $S$ in the first line? Regarding your question about what has to be shown: Given any set $A$ from the $\sigma-$algebra of $S$, which are both given, prove that there is a countable subset $S_0$ of the given set $S$, such that $A \in \sigma(S_0)$.

Your definition looks a bit right, as one should consider $\mathcal{A}:=\bigcup\{\,\sigma(C)\,|\,C\subseteq S \text{ countable }\,\}$. But this is at prior simply a set of sets. It should be shown that this is already a $\sigma-$algebra, and that it contains $S$. If this has been shown, the proof is a final small conclusion from that.
Sorry for the second part. I'll try to be more explicit.

Define $$\mathcal{B} =\{C \in \sigma(S)\mid \exists S_0 \subseteq S \mathrm{\ at \ most \ countable \ with\ } C \in \sigma(S_0)\}$$

One then shows that $\mathcal{B}$ is a sigma-algebra containing $S$ and we are done.

This $\mathcal{B}$ looks different than your $\mathcal{A}$, but works . I just called it curled S but that was confusing as something was already called S. For some strange reason I must have thought that our sigma algebra was generated by a certain set A. Apologies for that. Maybe too much mathematics for today...

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#### fresh_42

Mentor
2018 Award
Sorry for the second part. I'll try to be more explicit.

Define $$\mathcal{B} =\{C \in \sigma(S)\mid \exists S_0 \subseteq S \mathrm{\ at \ most \ countable \ with\ } C \in \sigma(S_0)\}$$

One then shows that $\mathcal{B}$ is a sigma-algebra containing $S$ and we are done.

This $\mathcal{B}$ looks different than your $\mathcal{A}$, but works . I just called it curled S but that was confusing as something was already called S. For some strange reason I must have thought that our sigma algebra was generated by a certain set A. Apologies for that. Maybe too much mathematics for today...
O.k., but still very short. Let me post my proof for all who are less familiar with the constructions.

Let $\mathcal{A}=\bigcup \{\,\sigma(C)\,|\,C\subseteq S \text{ countable}\,\}\,.$ We show that $\mathcal{A}$ is a $\sigma-$algebra over $X$ which contains $S.$
1. $\emptyset \in \sigma(\emptyset)=\{\emptyset,X\}$ and $\emptyset$ is countable with $\emptyset \subseteq S$, so $\emptyset \in \mathcal{A}\,.$
2. With $A\in \mathcal{A}$ we have a countable set $C\subseteq S$ with $A\in \sigma(C)$, and so is $X-A \in \mathcal{A}$.
3. Let $A_i \in \mathcal{A}$ for $i \in \mathbb{N}$. Then there are countable sets $C_i \subseteq S$ with $A_i \in \sigma(C_i)$. The set $C=\bigcup_{i\in \mathbb{N}} C_i \subseteq S$ is also countable. Since $C_i \subseteq C$ and $\sigma$ is monotone, we have $A_i\in \sigma(C)$ and so $\bigcup_{i\in \mathbb{N}} A_i \in \sigma(C) \in \mathcal{A}$.
If $A \in S$, then $\{\,A\,\}\subseteq S$ is countable and $A \in \sigma(\{A\})=\{\,\emptyset, A,X-A,X\,\}$, i.e. $A\in \mathcal{A}$ and thus $S\subseteq \mathcal{A}$.

Since $\mathcal{A}$ is a $\sigma-$algebra which contains $S$, we have $\sigma(S)\subseteq \mathcal{A}$. This means that for all $A\in \sigma(S)$ there is a countable set $S_0\subseteq S$ with $A\in \sigma(S_0)$.

#### YoungPhysicist

It’s November 9th now.Will there be a November math challenge?

#### Math_QED

Homework Helper
It’s November 9th now.Will there be a November math challenge?
I think there are still too much open problems of the previous editions.

#### YoungPhysicist

I think there are still too much open problems of the previous editions.
Oh~~all right.It is just that since I joined PF, there is a new math challenge every month.

#### fresh_42

Mentor
2018 Award
Oh~~all right.It is just that since I joined PF, there is a new math challenge every month.
We currently have 25 open problems of all possible degrees of difficulty:

As nobody even has asked about hints or at least attempted to solve them, we concluded a lack of interest. And to provide a collection of questions à la 'Solve this integral / series!' didn't appear worthwhile to us.

#### Math_QED

Homework Helper
Here is another solution for (11), using complex analysis. I hope this is the elegant solution you were looking for. The method I use is the same as in this video:

Note that the roots of $x^2 - x +1$ over $\mathbb{C}$ are given by $\frac{1\pm \sqrt{3}i}{2}$. Only the one with the positive square root sign lies in the upper half plane.

These are poles of order 1. Using the technique in the video, one obtains:

$$\int_{-\infty}^{+\infty} \frac{4}{x^2 - x +1} = 2 \pi i Res\left(\frac{4}{x^2-x+1}, \frac{1+\sqrt{3}i}{2}\right) = 2\pi i \lim_{x \to \frac{1+\sqrt{3}i}{2}} \frac{4}{x-1/2 + \sqrt{3}/2i} = \frac{8 \pi}{\sqrt{3}}$$

#### Fred Wright

Dear Dr. @wrobel ,
The quadrature solution of problem #1 results in an intractable integral:
$$\int \frac {x^2dx}{\sqrt {x^4-\frac {1}{4}} }=\sqrt {C_2}(\frac{2^{\frac {7}{2}}}{3}t^{\frac {3}{2}} +C_1)$$
I note that the integrand is complex when $\left | x \right | < 1$ and real otherwise. How am I wrong?

#### wrobel

The quadrature solution of problem #1 results in an intractable integral
There is no need to compute such integrals explicitly. For example, if you obtained an equation $\dot x=\frac{1}{\sin(x^2)}$ then you
can write
$$\int\sin(x^2)dx=t+const$$ and that is all. Integration in quadrature does not imply that you will present all the integrals and inverse functions as combination of elementary functions.

#### nuuskur

Not lack of interest, rather lack of time

#### Math_QED

Homework Helper
Not lack of interest, rather lack of time
Indeed, same for me. Over time, I think the questions will get solved though.

#### Greg Bernhardt

We currently have 25 open problems of all possible degrees of difficulty:
Maybe compile these into a new competition for better organization?

#### lpetrich

Good idea. I'd like to see that.

#### fresh_42

Mentor
2018 Award

"Math Challenge - October 2018"

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