Motivational Quotes and Stories (for students)

Motivational Quotes and Stories (Life / Education / Math)

Just to share some motivational quotes and stories, including those that are relevant to life, education and math. Do also check out Motivational Books for the Student (Educational).

If you have any other quotes / stories that you like, do post it in the comments! 🙂

1) “Success consists of going from failure to failure without loss of enthusiasm.”

Winston Churchill

This applies especially to students in higher education (e.g. Junior Colleges in Singapore), where it is quite common to “fail” an exam by getting below 50%. Do not despair, and continue to study hard, and you will achieve success eventually.

“There is nothing noble in being superior to your fellow man; true nobility is being superior to your former self.”
― Ernest Hemingway

Do not compare yourself with your classmates, everyone is unique. Focus on improving yourself day by day.

Kirby tried his qualifying exam again, on the same two topics. “This time, they said, ‘You passed,’” he says. “They didn’t say it with any enthusiasm, but they said, ‘You passed.’” His committee recommended that Kirby move into some other field than topology.

But Kirby was not one to be deterred by discouragement from his teachers. He waited until their backs were turned, so to speak, and identified a topologist — Eldon Dyer — who had been away when Kirby took his qualifying exam. Kirby kept going to Dyer with questions, and “at some point it sort of became obvious that I was his student,” Kirby says. “And he told somebody later on that he realized at some point or other he was stuck with me.”

(https://www.simonsfoundation.org/science_lives_video/robion-kirby/)

Inspirational story from Rob Kirby (famous mathematician) on how to ignore discouragement, even from teachers. This is applicable to students in Singapore who are sometimes told by teachers / school to drop certain subjects (e.g. drop Higher Chinese / drop A Maths), where the motive may not be purely in the student’s interest. Sometimes the reason that the school wants the student to drop the subject is to protect the school’s ranking in the exams / boost principal’s KPI etc. In this case, the student should follow his own judgement on whether to drop the subject.

“Everyone is No. 1” Motivational Song by Andy Lau.

Beautiful Lyrics (Chinese):
我的路不是你的路
我的苦不是你的苦
每个人都有潜在的能力
把一切去征服

我的泪不是你的泪
我的痛不是你的痛
一样的天空不同的光荣
有一样的感动

不需要自怨自艾的惶恐
只需要沉着
只要向前冲
告诉自己天生我才必有用

Everyone is NO.1
只要你凡事不问能不能
用一口气交换你一生
要迎接未来不必等

Everyone is NO.1
成功的秘诀在你肯不肯
流最热的汗
拥最真的心
第一名属于每个人

我的手不是你的手
我的口不是你的口
只要一条心
暴风和暴雨
都变成好朋友

不需要自怨自艾的惶恐
只需要沉着
只要向前冲
告诉自己天生我才必有用

不害怕路上有多冷
直到还有一点余温
我也会努力狂奔

Everyone is NO.1
只要你凡事不问能不能
用一口气交换你一生
要迎接未来不必等

Everyone is NO.1
成功的秘诀在你肯不肯
流最热的汗
拥最真的心
第一名属于每个人
5)

Success is not final, failure is not fatal: it is the courage to continue that counts.

Winston Churchill

Galatians 6:4-6Easy-to-Read Version (ERV)

⁴Don’t compare yourself with others. Just look at your own work to see if you have done anything to be proud of. ⁵You must each accept the responsibilities that are yours.

Curious Inequality: 2^p(|a|^p+|b|^p)>=|a+b|^p

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This inequality often appears in Analysis: $2^p(|a|^p+|b|^p)\geq |a+b|^p$ , for $p\geq 1$ , and $a,b\in\mathbb{R}$ . It does seem quite tricky to prove, and using Binomial Theorem leads to a mess and doesn’t work!

It turns out that the key is to use convexity, and we can even prove a stronger version of the above, namely $2^{p-1}(|a|^p+|b|^p)\geq |a+b|^p$ .

Proof: Consider $f(x)=|x|^p$ which is convex on $\mathbb{R}$ . Let $a,b\in\mathbb{R}$ . By convexity, we have $\displaystyle\boxed{f(ta+(1-t)b)\leq tf(a)+(1-t)f(b)}$ for $0\leq t\leq 1$ .

Choose $t=1/2$ . Then we have $f(\frac{1}{2}a +\frac 12 b)\leq \frac 12 f(a)+\frac 12 f(b)$ , which implies $|\frac{a+b}{2}|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p$ .

Thus,

$\begin{aligned}|a+b|^p&\leq 2^{p-1}|a|^p+2^{p-1}|b|^p\\ &\leq 2^p|a|^p+2^p|b|^p\\ &=2^p(|a|^p+|b|^p) \end{aligned}$

Z(D_2n), Center of Dihedral Group D_2n

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Question: What is $Z(D_{2n})$ , the center of the dihedral group $D_{2n}$ ?

Algebraically, the dihedral group may be viewed as a group with two generators $a$ and $b$ , i.e. $\boxed{D_{2n}=\{1,a,a^2,\dots,a^{n-1},b,ab,a^2b,\dots,a^{n-1}b\}}$ with $a^n=b^2=1$ , $bab=a^{-1}$ .

Answer: $Z(D_2)=D_2$

$Z(D_4)=D_4$ .

For $n\geq 3$ , $Z(D_{2n})=\begin{cases}1&,\ n\ \text{is odd}\\ \{1,a^{n/2}\}&,\ n\ \text{is even} \end{cases}$

Proof: For $n=1$ , $D_2=\{1, b\}\cong\mathbb{Z}_2$ which is abelian. Thus, $Z(D_2)=D_2$ .

For $n=2$ , $D_4=\{1,a,b,ab\}\cong V$ , the Klein four-group, which is also abelian. Thus, $Z(D_4)=D_4$ .

Let $A=\{1,a,a^2,\dots,a^{n-1}\}$ , $B=\{b,ab,a^2b,\dots,a^{n-1}b\}$ . Clearly elements in $A$ commute with each other.

Let $a^k$ be an element in $A$ . ( $0\leq k\leq n-1$ ). Let $a^lb$ be an element in $B$ . ( $0\leq l\leq n-1$ )

$\begin{aligned}a^k(a^lb)=(a^lb)a^k&\iff a^kb=ba^k\\ &\iff a^kba^{-k}b^{-1}=1\\ &\iff a^kb(bab)^kb=1\ (\text{here we used}\ bab=a^{-1})\\ &\iff a^kb(ba^kb)b=1\\ &\iff a^{2k}=1\\ &\iff k=0\ \text{or}\ n/2 \end{aligned}$

I.e. the only element in $A$ (other than 1) that is in the center is $a^{n/2}$ , which is only possible if $n$ is even.

Let $a^kb$ , $a^lb$ be two distinct elements in $B$ . ( $0\leq k< l\leq n-1$ )

$\begin{aligned}(a^kb)(a^lb)=(a^lb)(a^kb)&\iff ba^l=a^{l-k}ba^k\\ &\iff ba^{l-k}=a^{l-k}b \end{aligned}$

By earlier analysis, this is true iff $l-k=n/2$ . Each $a^kb\ (0\leq k\leq n-2)$ is not in the center since we may consider $l=k+1$ , i.e. $a^{k+1}b$ . Then $l-k=1<n/2$ . (since $n\geq 3$ ). $a^{n-1}b$ also does not commute with $a^{n-2}b$ for the same reason.

Therefore,

For $n\geq 3$ , $Z(D_{2n})=\begin{cases}1&,\ n\ \text{is odd}\\ \{1,a^{n/2}\}&,\ n\ \text{is even} \end{cases}$

Hardest JC Maths Paper (Prelims/Promos)

Just heard from some sources that AJC (Anderson Junior College) Math papers are considered the most difficult of all JCs, beating RI/HCI in terms of difficulty.

Do check out this page on how to calculate JC Ranking points.

Quotes:

AJC might have the most challenging Math papers, but that doesn’t equate to having the best math results. Other schools do better (RI/HCI)

From: http://forums.sgclub.com/singapore/ajc_good_jc_386546.html

H2 Maths this year was quite easy for both paper 1 & 2. It is definitely no where of the standard of AJC Maths Exam Papers, which are famously known for very challenging questions.

From: http://doggy94-in-air.blogspot.sg/

Introducing WordAds 2.0

WordAds 2.0 is out. Hope it is good news for international WordPress blogs, who are getting much lower ad revenue compared to the US and Europe.

WordAds

Today, we’re excited to introduce you to a new WordAds. On the front end, it’s a simpler and more streamlined experience like never before. On the back-end we have launched a real-time bidding platform to maximize earnings and ad creative control. Say hello to WordAds 2.0!

WordAds 2.0 is now fully integrated where you control the rest of your blog, in WordPress.com’s main Settings interface. You can also view your Earnings reports here and manage your payout information.

Existing WordAds users aren’t the only ones to benefit from the changes in WordAds 2.0. For new users, we have done away with the separate application process. Any family friendly WordPress.com blog with minimal page views will be considered for immediate admission to WordAds.

Bigger changes are now live in our real time bidding environment. We have dozens of ad agencies and buyers bidding in real time on each of our global…

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Projective Space Explicit Homotopy (RP1 to RP2)

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The above video describes the real projective plane ( $\mathbb{R}P^2$ ).

The projective space $\mathbb{R}P^n$ can be defined as the quotient space of $S^n$ by the equivalence relation $x\sim -x$ for $x\in S^n$ .

Notation: For $x=(x_1,\dots, x_{n+1})\in S^n$ , we write $[x_1, x_2,\dots, x_{n+1}]$ for the corresponding point in $\mathbb{R}P^n$ . Let $f,g: \mathbb{R}P^1\to\mathbb{R}P^2$ be the maps defined by $f[x,y]=[x,y,0]$ and $g[x,y]=[x,-y,0]$ .

How do we construct an explicit homotopy between $f$ and $g$ ? A common mistake is to try the “straight-homotopy”, e.g. $F([x,y],t)=[x,(1-2t)y,0]$ . This is a mistake as it passes through the point [0,0,0] which is not part of the projective plane.

A better approach is to consider $F:\mathbb{R}P^1\times I\to\mathbb{R}P^2$ , defined by $\boxed{F([x,y],t)=[x,(\cos\pi t)y, (\sin\pi t)y]}$ .

Note that if $x^2+y^2=1$ , then $x^2+[(\cos\pi t)y]^2+[(\sin\pi t)y]^2=x^2+y^2=1$ .

$F([x,y],0)=[x,y,0]$

$F([x,y],1)=[x,-y,0]$

Story of Edison (Motivational)

Calculus World Cup

Just to share this news:

The National Taiwan University is holding the first ever Calculus World Cup (CWC) in February 2016. It’s the first time students from global top universities will be able to compete over Calculus in e-sports. The competition will be held on PaGamO – a social online gaming platform for education. The top 12 teams will be invited to Taiwan for the final round, and great prizes with a value of over $70,000 await the finalists!
Official website: http://cwc.pagamo.com.tw

Registration: https://pagamo.com.tw/calculus_cup

Facebook: https://www.facebook.com/PaGamo.glo

Contractible space as Codomain implies any two maps Homotopic

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Recall that a space Y is contractible if the identity map $\text{id}_Y$ is homotopic to a constant map. Let Y be contractible space and let X be any space. Then, for any maps $f,g: X\to Y$ , $f\simeq g$ .

Proof: Let Y be a contractible space and let X be any space. $\text{id}_Y\simeq c$ , where $c$ is a constant map. There exists a map $F: Y\times [0,1]\to Y$ such that $F(y,0)=\text{id}_Y(y)=y$ , for $y\in Y$ . $F(y,1)=c(y)=b$ for some point $b\in Y$ .

Let $f,g: X\to Y$ be any two maps. Consider $G:X\times [0,1]\to Y$ where $G(x,t)=\begin{cases} F(f(x),2t),&\ \ \ \text{for}\ 0\leq t\leq 1/2\\ F(g(x),-2t+2),&\ \ \ \text{for}\ \frac 12<t\leq 1 \end{cases}$

When $t=\frac 12$ , $F(f(x),1)=b$ , $F(g(x),1)=b$ . Therefore G is cts.

$G(x,0)=F(f(x),0)=f(x)$ ,

$G(x,1)=F(g(x),0)=g(x)$ .

Therefore $f\simeq g$ .

Motivational Video of the Weekend

Just chanced upon this video on YouTube, really made a lot of sense and is very uplifting. Dream big, and don’t set any limits on yourself and what you can do.

Some other books by Nick:

Outer measure of Symmetric Difference Zero implies Measurability

Free Career Quiz: Please help to do!

Just came across this neat beginner’s Lebesgue Theory question. As students of analysis know, just to show a set is measurable is no easy feat. The usual way is to use the Caratheodory definition, where a set E is said to be measurable if for any set A, $m^*(A)=m^*(A\cap E)+m^*(A\cap E^c)$ . This can be quite tedious.

Question: Suppose E is a Lebesgue measurable set and let F be any subset of $\mathbb{R}$ such that $m^*(E\Delta F)=0$ (Symmetric Difference is Zero). Show that F is measurable.

The short way to do this is to note that $m^*(E\Delta F)=0$ implies $m^*(E\setminus F)=0$ , and $m^*(F\setminus E)=0$ . This in turn (using a lemma that any set with outer measure zero is measurable) implies the measurability of $E\setminus F$ and $\setminus F\setminus E$ .

Next comes the critical observation: $\boxed{E\cap F=E\setminus (E\setminus F)=E\cap (E\setminus F)^c}$ . Using the fact that the collection of measurable sets is a $\sigma$ -algebra, we can conclude $E\cap F$ is measurable.

Thus $F=(E\cap F)\cup (F\setminus E)$ is the union of two measurable sets and thus is measurable.

Interesting indeed!

Math Joke (Fields Arranged by Purity)

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I came across this joke on another blog: http://phdlife.warwick.ac.uk/

Quite true! A Math student will understand this at the university level and beyond, where Math has no more numbers and is full of symbols and jargon! Although even the most abstract Math has applications, the applications are only discovered years later, hence Pure Math is indeed one of the most pure subjects around.

$math joke$

Fail H2 Maths Promos or Prelims

For H2 (or H1) Maths students who are getting low marks for internal school exams, do not be overly discouraged. The current trend for schools is to set very tough internal exams (i.e. Promos and Prelims) to spur students to study hard, and (hopefully) ace the eventual final A level exams. If you look at the actual A Level Ten Year Series, you will find that the standard of questions is much easier than Prelim level.

A rule of thumb is that the eventual A level grade is 2 grades above the internal school grade. E.g., in internal exams a student getting D for H2 Maths is most likely equivalent to a B in the final A levels, provided the student continues to study hard.

Jumping from E to A grade has been done by many seniors. Do not give up, continue to believe in yourself, and keep calm while constantly revising.

Do check out this highly condensed H2 Math Notes (comes with free exam papers). The key thing to do before exams is to remember Math formulas (many students forget the AP/GP formulae for instance, and lost some free marks). Constant practice and exposure to questions is also a must.

Here are some sources of true stories:

1) https://www.facebook.com/RJConfessions/posts/220752441406251

To all the J1 and J2 kids who are struggling with math, let me share with you my personal experience. I took H2 math by the way, and refused to drop to H1 when people started dropping.

J1 CT 1: Math: U
J1 promos: Math: S
J2 CT1: Math S
J2 CT2: Math S
J2 Prelims: Math E
A levels: Math A.

The moral of the story is simple: It can be done. My math teacher used to motivate us with stories of seniors who have also flunked their way through math in the 2 years and clinched an A at the end. I didnt really believed it could happen, but I guess I chose to believe it anyways.

2) https://www.reddit.com/r/singapore/comments/3nkq4t/jc_prelims_alevels_correlation/

H2 Math: E A

H2 Chem: D B

H2 Econs: D D

H1 Physics: U A

H1 GP: B A

The grades on the left were prelims and right were my actual results. Of course it depends on your school and how hard they set the prelim papers

Fermat’s Two Squares Theorem (Gaussian Integers approach)

Today we will discuss Fermat’s Two Squares Theorem using the approach of Gaussian Integers, the set of numbers of the form a+bi, where a, b are integers. This theorem is also called Fermat’s Christmas Theorem, presumably because it is proven during Christmas.

Have you ever wondered why $5=1^2+2^2$ , $13=2^2+3^2$ can be expressed as a sum of two squares, while not every prime can be? This is no coincidence, as we will learn from the theorem below.

Theorem: An odd prime p is the sum of two squares, i.e. $p=a^2+b^2$ where a, b are integers if and only if $p\equiv 1 \pmod 4$ .

(=>) The forward direction is the easier one. Note that $a^2\equiv 0\pmod 4$ if a is even, and $a^2\equiv 1\pmod 4$ if a is odd. Similar for b. Hence $p=a^2+b^2$ can only be congruent to 0, 1 or 2 (mod 4). Since p is odd, this means $p\equiv 1\pmod 4$ .

(<=) Conversely, assume $p\equiv 1\pmod 4$ , where p is a prime. p=4k+1 for some integer k.

First we prove a lemma called Lagrange’s Lemma: If $p\equiv 1\pmod 4$ is prime, then $p\mid (n^2+1)$ for some integer n.

Proof: By Wilson’s Theorem, $(p-1)!=(4k)!\equiv -1\pmod p$ . $(4k)!\equiv [(2k)!]^2\equiv -1\pmod p$ . We may see this by observing that $4k\equiv p-1\equiv -1\pmod p$ , $4k-1\equiv -2\pmod p$ , …, $4k-(2k-1)=2k+1\equiv -2k\pmod p$ . Thus $[(2k)!]^2+1\equiv 0\pmod p$ and hence $p\mid n^2+1$ , where $n=(2k)!$ .

Then $p\mid (n+i)(n-i)$ . However $p\nmid (n+i)$ since $p\nmid n=(2k)!$ . Similarly, $p\nmid (n-i)$ . Therefore $p$ is not a Gaussian prime, and it is thus not irreducible.

$p=\alpha\beta$ with $N(\alpha)>1$ and $N(\beta)>1$ . $N(p)=N(\alpha)N(\beta)$ , which means $p^2=N(\alpha)N(\beta)$ . Thus we may conclude $N(\alpha)=p$ , $N(\beta)=p$ .

Let $\alpha=a+bi$ . Then $p=a^2+b^2$ and we are done.

This proof is pretty amazing, and shows the connection between number theory and ring theory.

Holder’s Inequality Trick

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Holder’s Inequality is a very useful inequality in Functional Analysis, hence many results can be proved by applying Holder’s Inequality.

Suppose that $\mu(\Omega)<\infty$ and $1\leq p_1<p_2\leq\infty$ . Prove that if $f_n\to f$ in $L^{p_2}$ , then $f_n\to f$ in $L^{p_1}$ .

Proof: Assume $f_n\to f$ in $L^{p_2}$ . Then there exists $N$ such that if $n\geq N$ , then $\|f_n-f\|_{p_2}=(\int |f_n-f|^{p_2} d\mu)^{1/p_2}<\epsilon$ .

Then, $\begin{aligned} \|f_n-f\|_{p_1}^{p_1}&=\int |f_n-f|^{p_1} d\mu\\ &=\||f_n-f|^{p_1}\cdot 1\|_1\\ &\leq\||f_n-f|^{p_1}\|_{p_2/p_1}\|1\|_{(p_2/p_1)'},\ \ \ \text{where } (p_2/p_1)'=\frac{p_2}{p_2-p_1}\ \text{is the Holder conjugate}\\ &=(\int |f_n-f|^{p_2}d\mu)^{p_1/p_2}(\int 1 d\mu)^\frac{p_2-p_1}{p_2}\\ &<\epsilon^{p_1}\cdot \mu(\Omega)^\frac{p_2-p_1}{p_2} \end{aligned}$

Since $\epsilon>0$ is arbitrary, $\|f_n-f\|_{p_1}^{p_1}\to 0$ as $n\to\infty$ .

Therefore, $\|f_n-f\|_{p_1}\to 0$ .

245A, Notes 4: Modes of convergence

What's new

If one has a sequence $latex {x_1, x_2, x_3, ldots in {bf R}}&fg=000000$ of real numbers $latex {x_n}&fg=000000$, it is unambiguous what it means for that sequence to converge to a limit $latex {x in {bf R}}&fg=000000$: it means that for every $latex {epsilon > 0}&fg=000000$, there exists an $latex {N}&fg=000000$ such that $latex {|x_n-x| leq epsilon}&fg=000000$ for all $latex {n > N}&fg=000000$. Similarly for a sequence $latex {z_1, z_2, z_3, ldots in {bf C}}&fg=000000$ of complex numbers $latex {z_n}&fg=000000$ converging to a limit $latex {z in {bf C}}&fg=000000$.

More generally, if one has a sequence $latex {v_1, v_2, v_3, ldots}&fg=000000$ of $latex {d}&fg=000000$-dimensional vectors $latex {v_n}&fg=000000$ in a real vector space $latex {{bf R}^d}&fg=000000$ or complex vector space $latex {{bf C}^d}&fg=000000$, it is also unambiguous what it means for that sequence to converge to a limit $latex {v in {bf R}^d}&fg=000000$ or $latex {v in {bf C}^d}&fg=000000$; it means…

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mu is countably additive if and only if it satisfies the Axiom of Continuity

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Let $\mu$ be a finite, non-negative, finitely additive set function on a measurable space $(\Omega, \mathcal{A})$ . Show that $\mu$ is countably additive if and only if it satisfies the Axiom of Continuity: For $E_n\in\mathcal{A}, E_n\downarrow\emptyset \implies \mu(E_n)\to 0$ .

(=>) Assume $\mu$ is countably additive. Let $E_n\in\mathcal{A}$ , $E_n\downarrow\emptyset$ . Then,

$\displaystyle \lim_n \mu (E_n)=\mu (\cap_{n=1}^\infty E_n)=\mu (\emptyset)$ .

Suppose $\mu(\emptyset)=c$ . Then $\mu(\emptyset)=\mu(\cup_{n=1}^\infty \emptyset)=\sum_{n=1}^\infty c$ implies $c=0$ .

(<=) Assume $\mu$ satisfies Axiom of Continuity. Let $A_n\in\mathcal{A}$ be mutually disjoint sets. Define $E_n=\cup_{i=1}^\infty A_i\setminus \cup_{i=1}^n A_i$ .

Then $E_n\downarrow\emptyset$ . $\lim_n \mu(E_n)=0$ , $\lim_n \mu(\cup_{i=1}^\infty A_i)-\mu (\cup_{i=1}^n A_i)=0$ . $\lim_n \mu(\cup_{i=1}^n A_i)=\mu (\cup_{i=1}^\infty A_i)$ .

Therefore

$\begin{aligned}\mu(\cup_{i=1}^\infty A_i)&=\lim_n \mu(\cup_{i=1}^n A_i)\\ &=\lim_n \sum_{i=1}^n \mu (A_i)\\ &=\sum_{i=1}^\infty \mu(A_i) \end{aligned}$

Measure that is absolutely continuous with respect to mu

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Let $(X,\mathcal{M},\mu)$ be a measure space, and let $f:X\to [0,\infty]$ be a measurable function. Define the map $\lambda:\mathcal{M}\to[0,\infty]$ , $\lambda(E):=\int_X \chi_E f d\mu$ , where $\chi_E$ denotes the characteristic function of $E$ .

(a) Show that $\lambda$ is a measure and that it is absolutely continuous with respect to $\mu$ .

(b) Show that for any measurable function $g:X\to[0,\infty]$ , one has $\int_X g d\lambda=\int_X gf d\mu$ in $[0,\infty]$ .

Proof: For part (a), we routinely check that $\lambda$ is indeed a measure.

$\lambda(\emptyset)=\int_X \chi_\emptyset f d\mu=\int_X 0 d\mu=0$ . Let $E_i$ be mutually disjoiint measurable sets.

$\begin{aligned} \lambda(\cup_{i=1}^\infty E_i)&=\int_X \chi_{\cup_{i=1}^\infty E_i} f d\mu\\ &=\int_X (\sum_{i=1}^\infty \chi_{E_i}) f d\mu\\ &=\sum_{i=1}^\infty \int_X \chi_{E_i} f d\mu\\ &=\sum_{i=1}^\infty \lambda (E_i) \end{aligned}$

If $\mu (E)=0$ , then $\chi_{E} f=0$ a.e., thus $\lambda (E)=0$ . Therefore $\lambda\ll\mu$ .

(b) We note that when $g$ is a characteristic function, i.e. $g=\chi_E$ ,

$\begin{aligned} \int_X g d\lambda&=\int_X \chi_E d\lambda\\ &=\lambda (E)\\ &=\int_X \chi_E f d\mu\\ &=\int_X gf d\mu \end{aligned}$

Hence the equation holds. By linearity, we can see that the equation holds for all simple functions. Let $(\psi_n)$ be a sequence of simple functions such that $\psi_n\uparrow g$ . Then by the Monotone Convergence Theorem, $\lim_{n\to\infty} \int \psi_n d\lambda=\int g d\lambda$ .

Note that $\psi_n f\uparrow gf$ , thus by MCT, $\lim_{n\to\infty}\int\psi_n f d\mu=\int g f d\mu$ . Note that $\int \psi_n d\lambda=\int \psi_n f d\mu$ . Hence, $\int g d\lambda=\int gf d\mu$ , and we are done.

f integrable implies set where f is infinite is measure zero

Let $(X,\mathcal{M},\mu)$ be a measure space. Let $f:X\to [0,\infty]$ be a measurable function. Suppose that $\int_X f d\mu<\infty$ .

(a) Show that the set $\{x\in X:f(x)=\infty\}\subseteq X$ is of $\mu$ -measure 0. (Intuitively, this is quite obvious, but we need to prove it rigorously.)

(b) Show that the set $\{x\in X:f(x)\neq 0\}\subseteq X$ is $\sigma$ -finite with respect to $\mu$ . i.e. it is a countable union of measurable sets of finite $\mu$ -measure.

We may use Markov’s inequality, which turns out to be very useful in this question.

Proof: (a) Let $E_k=\{x\in X:f(x)\geq k\}$ , where $k\in\mathbb{N}$ . Denote $E_\infty=\{x\in X:f(x)=\infty\}$ .

$E_K \downarrow E_\infty$ , and $\mu (E_1)\leq\frac{1}{1}\int_X f d\mu<\infty$ . (Markov Inequality!)

Then

$\begin{aligned}\mu(E_\infty)&=\lim_{k\to\infty}\mu (E_k)\\ &\leq\lim_{k\to\infty}\frac{1}{k}\int f d\mu\ \ \ \text{(Markov Inequality)}\\ &=0 \end{aligned}$

Therefore, $\mu(E_\infty)=0$ .

(b) Let $S_k=\{x\in X:f(x)\geq\frac{1}{k}\}$ , $k\in\mathbb{N}$ .

$\{x\in X:f(x)\neq 0\}=\cup_{k=1}^\infty S_k$

Therefore, $\mu(S_k)\leq k\int f d\mu<\infty$ , and we have expressed the set as a countable union of measurable sets of finite measure.

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Markov Inequality + PSLE One Dollar Question

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Markov inequality is a useful inequality that gives a rough upper bound of the measure of a set in terms of an integral. The precise statement is: Let $f$ be a nonnegative measurable function on $\Omega$ . The Markov inequality states that for all $K>0$ , $\displaystyle \mu\{x\in\Omega:f(x)\geq K\}\leq\frac{1}{K}\int fd\mu$ .

The proof is rather neat and short. Let $E_K:=\{x\in\Omega: f(x)\geq K\}$ Then,

$\begin{aligned} \int f d\mu &\geq \int_{E_K} fd\mu\\ &\geq \int_{E_K}K d\mu\\ &=K \mu(E_K) \end{aligned}$

Therefore, $\mu(E_K)\leq\frac{1}{K}\int fd\mu$ .

Free Career Personality Quiz (Please help to do!)

URL: https://career-test.com/s/sgamb?reid=210

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Necessary and Sufficient Condition for Integrability in finite measure space

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The results of this Personality Test is quite surprisingly accurate, do give it a try to see if you are a Careerist, Entrepreneur, Harmonizer, Idealist, Hunter, Internationalist or Leader?

Do try out this Free Career Guidance Personality Test at https://career-test.com/s/sgamb?reid=210 while it is still available!

Benefits of doing the (Free) Career Test:

Get familiar with the top companies in Singapore
There are seven distinct career types based on career preferences, goals and personality. Get to know yours! (My result: Harmonizer)
Take part in the annual Universum survey and win prizes!

Let $(\Omega, \Sigma, \mu)$ be a finite measure space. Suppose that $f\geq 0$ is a measurable function on $\Omega$ . Let $E_n:=\{\omega\in\Omega:f(\omega)\geq n\}$ for each $n\in \mathbb{N}\cup\{0\}$ . Show that $f$ is integrable if and only if $\sum_{n=0}^\infty \mu(E_n)<\infty$ .

This proof has a cute solution that is potentially very short. We will elaborate more on this proof. Other approaches include using Markov’s Inequality / Chebyshev’s inequality.

Proof: Consider $g=\sum_{n=1}^\infty I_{E_n}$ .

Note that for each $n\geq 0, n\in\mathbb{N}$ on $E_n\setminus E_{n+1}$ , $g=n$ , while $n\leq f<n+1$ . Therefore $g\leq f<g+1$ on $\Omega$ .

Integrating with respect to $\mu$ , we get $\int_{\Omega} gd\mu\leq\int_{\Omega} fd\mu<\int_{\Omega} g+1d\mu$ .

$\displaystyle\boxed{\sum_{n=1}^\infty \mu (E_n)\leq \int_\Omega f d\mu<\sum_{n=1}^\infty\mu(E_n)+\mu(\Omega)}$

(=>) Now assuming f is integrable, i.e. $\int fd\mu<\infty$ , we have $\sum_{n=1}^\infty \mu(E_n)<\infty$ . $\mu(E_0)=\mu(\Omega)<\infty$ . Therefore $\sum_{n=0}^\infty\mu(E_n)<\infty$ .

(<=) Conversely, if $\sum_{n=0}^\infty\mu(E_n)=\sum_{n=1}^\infty \mu(E_n)+\mu(\Omega)<\infty$ , then $\int_\Omega f d\mu<\infty$ .

We are done.

Note: For a more rigorous proof of $\int gd\mu=\sum_{n=1}^\infty \mu (E_n)$ we can use MCT (Monotone Convergence Theorem).

Let $g_k=\sum_{n=1}^k I_{E_n}$ . Then $g_k\uparrow g$ . By MCT, $\int gd\mu=\lim_{k\to\infty} \int g_k d\mu=\lim_{k\to\infty} \sum_{n=1}^k \mu (E_n)=\sum_{n=1}^\infty \mu(E_n)$ .

Interesting Analysis Question (Measure Theory)

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Let $f\geq 0$ be a measurable function with $\int f d\mu<\infty$ . Show that for any $\epsilon>0$ , there exists a $\delta(\epsilon)>0$ such that for any measurable set $E\in\mathcal{A}$ with $\mu(E)<\delta(\epsilon)$ , we have $\int_E f d\mu<\epsilon$ .

Proof: For $M>0$ , we define $f_M(x)=\min (f(x),M)\leq M$ , for all $x\in \Omega$ .

Then $f=f_M+(f-f_M)$ .

Let $\delta=\epsilon/2M$ . Then for any $E\in\mathcal{A}$ with $\mu(E)<\delta$ ,

$\begin{aligned}\int_E f_M d\mu &\leq \int_E M d\mu\\ &=\mu (E)M\\ &<\delta M\\ &=(\epsilon/2M)M\\ &=\epsilon/2 \end{aligned}$

Note that $f_M\uparrow f$ . By Monotone Convergence Theorem,

$\int f d\mu=\lim_{M\to\infty}\int f_M d\mu$ .

Therefore $\lim_{M\to\infty}\int f-f_M d\mu=0$ .

We can choose $M$ sufficiently large such that

$\int_E f-f_M d\mu \leq \int f-f_M d\mu <\epsilon/2$ .

Then

$\begin{aligned} \int_E f d\mu&=\int_E f_M d\mu+\int_E f-f_M d\mu\\ &<\epsilon/2+\epsilon/2\\ &=\epsilon \end{aligned}$

We are done!

Conjugacy Classes of non-abelian group of order p^3

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Let p be a prime, and let G be a non-abelian group of order $p^3$ . We want to find the number of conjugacy classes of G.

First we prove a lemma: Z(G) has order p.

Proof: We know that since G is a non-trivial p-group, then $Z(G)\neq 1$ . Since $Z(G)\trianglelefteq G$ , by Lagrange’s Theorem, $|Z(G)|=p,p^2,\text{or }p^3$ .

Case 1) $|Z(G)|=p$ . We are done.

Case 2) $|Z(G)|=p^2$ . Then $|G/Z(G)|=p^3/p=p$ . Thus $G/Z(G)$ is cyclic which implies that G is abelian. (contradiction).

Case 3) $|Z(G)|=p^3$ . This means that the entire group G is abelian. (contradiction).

Next, let $O(x_1),\dots, O(x_n)$ be the distinct conjugacy classes of G.

$O(x_i)=\{gx_i g^{-1}:g\in G\}$ , where $C_G(x_i)=\{g\in G:gx_i=x_ig\}$ .

Then by the Class Equation, we have $\displaystyle p^3=|G|=\sum_{i=1}^n [G:C_G(x_i)]$ .

If $x_i\in Z(G)$ , then $C_G(x_i)=G$ , which means $[G:C_G(x_i)]=1$ .

If $x_i \notin Z(G)$ , then $C_G(x_i)\neq G$ . Since $x_i\in C_G(x_i)$ , thus $Z(G)\subsetneq C_G(x_i)$ . Thus we have $p=|Z(G)|<|C_G(x_i)|<|G|=p^3$ . Since $C_G(x_i)$ is a subgroup of $G$ , Lagrange’s Theorem forces $|C_G(x_i)|=p^2$ . Thus $[G:C_G(x_i)]=p^3/p^2=p$ .

By the Class Equation, we thus have $p^3=p+(n-p)p$ , which leads us to $\boxed{n=p^2+p-1}$ .

Group of order 56 is not simple + Affordable Air Purifier

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Let G be a group of order 56. Show that G is not simple.

Proof:

We will use Sylow’s Theorem to show that either the 2-Sylow subgroup or 7-Sylow subgroup is normal.

$|G|=2^3\cdot 7$

By Sylow’s Theorem $n_2\mid 7, n_2\equiv 1\pmod 2$ . Thus $n_2=1,7$ .

Also, $n_7\mid 8, n_7\equiv 1\pmod 7$ . Therefore $n_7=1, 8$ .

If $n_2=1$ or $n_7=1$ , we are done, as one of the Sylow subgroups is normal.

Suppose to the contrary $n_2=7$ and $n_7=8$ .

Number of elements of order 7 = 8 x (7-1)=48

Remaining elements = 56-48=8. This is just enough for one 2-Sylow subgroup, thus $n_2=1$ . This is a contradiction.

Therefore, a group of order 56 is simple.

Universal Property of Kernel Question

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In this blog post, we will discuss a category theory question, in the framework of homomorphisms of abelian groups.

Let $\phi:M'\to M$ be a homomorphism of abelian groups. Suppose that $\alpha:L\to M'$ is a homomorphism of abelian groups such that $\phi\circ\alpha$ is the zero map. (One example is the inclusion $\mu:\ker\phi\to M'$ )

Are the following true or false?

(i) There is a unique homomorphism $\alpha_0:\ker\phi\to L$ such that $\mu=\alpha\circ\alpha_0$ .

(ii) There is a unique homomorphism $\alpha_1:L\to\ker\phi$ such that $\alpha=\mu\circ\alpha_1$ .

It turns out that (i) is false. We may construct a trivial counterexample as follows. Consider $L=M=0$ , and $M'=\mathbb{Z}/2\mathbb{Z}$ . Let $\alpha$ , $\phi$ be both the zero maps. Then certainly $\phi\circ\alpha=0$ . $\ker\phi=\mathbb{Z}/2\mathbb{Z}$ . Then, for any $\alpha_0$ , $\alpha\circ\alpha_0(x)=0$ , and hence is not equals to the the inclusion map $\mu$ .

It turns out that (ii) is true, in fact it is the famous universal property of the kernel, that any homomorphism yielding zero when composed with $\phi$ has to factor through $\ker\phi$ .

First we will prove uniqueness. Let $\alpha=\mu\circ\alpha_1=\mu\circ\beta$ , where $\beta$ is another such map with the property (ii). Then for all $x\in L$ , $\mu\alpha_1(x)=\mu\beta(x)$ , which implies $\mu(\alpha_1(x)-\beta(x))=0$ . Since $\mu$ is the inclusion map, this means that $\alpha_1(x)-\beta(x)=0$ and thus $\alpha_1(x)=\beta (x)$ .

Next, we will prove existence. Consider $\alpha_1:L\to\ker\phi, \alpha_1(l)=\alpha(l)$ . Note that $\phi(\alpha(l))=0$ by definition thus $\alpha(l)\in\ker\phi$ .

Next we prove it is a homomorphism. $\alpha_1(l_1l_2)=\alpha(l_1l_2)=\alpha(l_1)\alpha(l_2)=\alpha_1(l_1)\alpha_1(l_2)$ .

Finally by construction it is easy to see that $\mu\alpha_1(l)=\mu\alpha(l)=\alpha(l)$ for all $l\in L$ .

Group of order 432 is not simple

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For this blog post, we shall show that a group of order $432=2^4\cdot 3^3$ is not simple. We will be using several previous posts as lemmas to prove this nontrivial result.

Suppose to the contrary G is simple. By Sylow’s Third Theorem, $n_3\equiv 1\pmod 3$ , $n_3\mid 16$ . This means that $n_3$ is 1, 4 or 16.

We recall that if $n_3=1$ , then the Sylow 3-subgroup is normal.

Let $Q_1$ and $Q_2$ be two distinct Sylow 3-subgroups of $G$ such that $|Q_1\cap Q_2|$ is maximum.

Using our previous lemma regarding index of intersection of Sylow subgroups, we split our analysis into three cases, the hardest of which is Case 3.

Case 1) If $|Q_1\cap Q_2|=1$ , $[Q_1:Q_1\cap Q_2]=3^3$ . Thus $n_3=1\pmod {27}$ , which allows us to conclude $n_3=1$ .

Case 2) If $|Q_1\cap Q_2|=3$ , $[Q_1:Q_1\cap Q_2]=3^2$ . Thus $n_3=1 \pmod 9$ . Similarly, we can conclude $n_3=1$ and we are done.

Case 3) If $|Q_1\cap Q_2|=9$ , then $[Q_1:Q_1\cap Q_2]=3$ .

By another previous lemma regarding index of least prime divisor, $Q_1\cap Q_2\trianglelefteq Q_1$ . Thus, $Q_1\leq N_G(Q_1\cap Q_2)$ . Thus $|N_G(Q_1\cap Q_2)|=3^3\cdot 2, 3^3\cdot 2^2, 3^3\cdot 2^3,\text{or }3^3\cdot 2^4$ .

If $N_G(Q_1\cap Q_2)=G$ , then $Q_1\cap Q_2\trianglelefteq G$ which is a contradiction. Hence we suppose $N_G(Q_1\cap Q_2)\neq G$ . Let $[G:N_G(Q_1\cap Q_2)]=k$ . The possible values of k are $k=2, 2^2, 2^3, 2^4$ .

Next, we use the fact that if $G$ is a simple group and $H$ is a subgroup of index $k$ , then $|G|$ divides $k!$ .

Thus, $432\mid k!$ , which forces k=16.

But then $Q_1=N_G(Q_1\cap Q_2)$ and similarly $Q_2=N_G(Q_1\cap Q_2)$ . Thus $Q_1=Q_2$ . This is a contradiction to $|Q_1\cap Q_2|=9$ .

Therefore all cases lead to contradiction and thus G is not simple.

Sylow subgroup intersection of a certain index + F1 Trespasser

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Back to our topic on Sylow theory…

Let $G$ be a finite group, where $q$ is a prime divisor of $G$ . Suppose that whenever $Q_1$ and $Q_2$ are two distinct Sylow q-subgroups of $G$ , $Q_1\cap Q_2$ is a subgroup of $Q_1$ of index at least $q^a$ . Prove that the number $n_q$ of Sylow q-subgroups of G satisfies $n_q\equiv 1\pmod {q^a}$ .

Proof: Let $\Omega=\{Q_1,Q_2\dots,Q_n\}$ be the set of all Sylow q-subgroups of G. Fix $P=Q_k\in \Omega$ . Consider the group action of P acting on $\Omega$ by conjugation.

$\phi:P\times\Omega\to\Omega$ , $\phi_x(Q_i)=xQ_i x^{-1}$

By Orbit-Stabilizer Theorem, $|O(Q_i)|=|P|/|N_p(Q_i)|$ .

We claim that $N_p(Q_i)=Q_i\cap P$ , since any element x outside of $Q_i$ cannot normalise $Q_i$ , since otherwise if $x \neq Q_i$ , $xQx^{-1}=Q_i$ , then $\langle Q_i, x\rangle$ will be a larger q-subgroup of G than $Q_i$ .

Thus, $|O(Q_i)|=|P|/|Q_i\cap P|\geq q^a$ , i.e. $q^a\mid |O(Q_i)|$ .

$|O(P)|=1$ .

The orbits form a partition of $\Omega$ , thus $|\Omega|=1+\sum{|O(Q_i)|}$ , where the sum runs over all orbits other than $O(P)$ .

Thus, $n_q\equiv 1\pmod {q^a}$ .

How to type LaTeX in WordPress without using “$latex”

Currently, LaTeX is well supported in WordPress, however there is one practical issue when typing LaTeX in WordPress, the need to type “$latex” for every single math expression! It gets pretty troublesome after a while.

For those familiar with LaTeX, one would know that for ordinary LaTeX typesetting, typing double $ will do, there is no need to type the word LaTeX. If I remember correctly, for Blogger there is no need to type “$latex”, hence it is a uniquely WordPress issue.

So far, I have not found any solution to this issue. (I am using WordPress.com hosted WordPress). Any readers who happen to know a solution, please enlighten me by dropping a comment! It will be greatly appreciated.

WordPress vs Blogger LaTeX:

https://mathtuition88.com/2014/12/25/math-formula-in-wordpress-vs-blogspot/

LaTeX Beginner’s Guide

S_n has Trivial Center for n greater than 3

We will prove that for $n\geq 3$ , $Z(S_n)=1$ .

Proof:

Clearly, $1\in Z(S_n)$ . Let $1\neq \sigma\in S_n$ . There exists $a, b$ distinct elements of $\{1, 2, \dots , n\}$ such that $\sigma(a)=b$ .

Consider the transposition $\tau =(b\ c)$ , where $c$ is distinct from $a, b$ . (Since $n\geq 3$ , such a $c$ exists.)

Then, $\tau\sigma (a)=\tau (b)=c$

$\sigma\tau (a)=\sigma (a)=b$

$\therefore \tau\sigma\neq\sigma\tau$

$\therefore \sigma \notin Z(S_n)$ .

Therefore, $Z(S_n)=1$

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Weierstrass M Test and Lebesgue’s Dominated Convergence Theorem

Previously, we wrote a blog post about Weierstrass M Test. It turns out Weierstrass M Test is a special case of Lebesgue’s Dominated Convergence Theorem, a very powerful theorem in Measure Theory, where the measure is taken to be the counting measure.

Lebesgue Dominated Convergence Theorem: Let $(f_n)$ be a sequence of integrable functions which converges a.e. to a real-valued measurable function $f$ . Suppose that there exists an integrable function $g$ such that $|f_n|\leq g$ for all $n$ . Then, $f$ is integrable and $\displaystyle \int f d\mu=\lim_n \int f_n d\mu$ .

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Singapore Haze & Subgroup of Smallest Prime Index

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Previously, we proved that any subgroup of index 2 is normal. It turns out that there is a generalisation of this theorem. Let $p$ be the smallest prime divisor of a group $G$ . Then, any subgroup $H\leq G$ of index $p$ is normal in $G$ .

Proof: Let $H$ be a subgroup of $G$ of index $p$ . Let $G$ act on the left cosets of $H$ by left multiplication: $\forall x\in G$ , $x\cdot gH=xgH$ .

This group action induces a group homomorphism $\phi:G\to S_p$ .

Let $K=\ker \phi$ . If $x\in K$ , then $xgH=gH$ for all $g\in G$ . In particular when g=1, xH=H, i.e. $x\in H$ .

Thus $K\subseteq H$ . In particular, $K\trianglelefteq H$ , since $\ker\phi$ is a normal subgroup of $G$ .

We have $G/K\cong \phi(G)\leq S_p$ . Thus $|G/K|\mid p!$ .

Also note that $|G|=|G/K||K|$ . Note that $|G/K|\neq 1$ since $|G/K|=[G:H][H:K]=p[H:K]\geq p$ .

Let $q$ be a prime divisor of $|G/K|$ . Then $q\leq p$ since $|G/K|\mid p!$ . Also, $q\mid |G|$ . Since $p$ is the smallest prime divisor of $|G|$ , $p\leq q$ . Therefore, $p=q$ , i.e. $|G/K|=p$ .

Then $p=p[H:K] \implies [H:K]=1$ , i.e. H=K. Thus, H is normal in G.

Proof of Wilson’s Theorem using Sylow’s Theorem

Wilson’s theorem $(p-1)!\equiv -1 \pmod p$ is a useful theorem in Number Theory, and may be proved in several different ways. One of the interesting proofs is to prove it using Sylow’s Third Theorem.

Let $G=S_p$ , the symmetric group on p elements, where p is a prime.

$|G|=p!=p(p-1)!$

By Sylow’s Third Theorem, we have $n_p\equiv 1\pmod p$ . The Sylow p-subgroups of $S_p$ have $p-1$ p-cycles each.

There are a total of $(p-1)!$ different p-cycles (cyclic permutations of p elements).

Thus, we have $n_p (p-1)=(p-1)!$ , which implies that $n_p=(p-2)!$

Thus $(p-2)!\equiv 1\pmod p$ , and multiplying by p-1 gives us $(p-1)!\equiv p-1\equiv -1\pmod p$ which is precisely Wilson’s Theorem. 🙂

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Tips to write the start of composition (argumentative essay) 议论文的开头如何入手（一）

Chinese Tuition Singapore

写议论文有个“万能公式”，我们称为“三段论”。就是把文章分为三个部分：开头（提出论点），中间（将论点分为几个部分加以概括并运用论据进行阐述），结尾（总结，进一步深化论点）。

都说“良好的开头是成功的一半”，所以一篇作文写出一个好的开头是十分有必要的。

有学生很苦恼，不知道该怎么写议论文的开头。如果套用“三段论”，他们往往在开头用一句话就结束了。其实议论文的开头有很多种写法。

我们以“怎样才算是一个幸福的家庭”这个题目来探讨议论文开头的写法。

这个题目的意思就是让学生阐述构成幸福家庭的条件。有一个学生认为，幸福家庭的必备条件是“父母爱护孩子，孩子孝顺父母，兄弟姐妹之间互相关心”。那我们就利用这个观点来写。

本文先介绍比较常用的几种开头方式：

1. 开门见山式。come straight to the point

这种方式的特点是在开头就将文章的论点摆出开，直截了当，一看便知文章的主旨。

例：人从一出生开始首先面对的“小社会”便是家庭。家庭的幸福与否关系到一个人的幸福与否。当然，一个幸福的家庭是需要每个家庭成员来共同努力维护的。父母爱护孩子，孩子孝顺父母，兄弟姐妹之间互相关心，这样才算是一个幸福的家庭。
2. 设问句式。（rhetorical question）

就议论的问题提出疑问，在回答问题的过程中提出自己的观点。

例：家庭与每个人的成长和生活是密不可分的。家庭环境甚至可以影响一个人的人生。人人都希望生活在幸福的家庭中，那么什么样的家庭才算是幸福的呢？幸福的家庭是需要家庭成员之间互相关怀照顾的。父母爱护孩子，孩子孝顺父母，兄弟姐妹之间互相关心，这样的家庭才会幸福。

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Zmn/Zm isomorphic to Zn

The following is a simple proof of why $\mathbb{Z}_{mn}/\mathbb{Z}_m\cong\mathbb{Z}_n$ .

For instance $\mathbb{Z}_6/\mathbb{Z}_2\cong\mathbb{Z}_3$ . Note that the tricky part is that $\mathbb{Z}_2$ is not actually the usual {0,1}, but rather {0,3} (considered as part of $\mathbb{Z}_6$ ). Hence the elements of $\mathbb{Z}_6/\mathbb{Z}_2$ are {0,3}, {1, 4}, {2, 5}, which can be seen to be isomorphic to $\mathbb{Z}_3$ .

A sketch of a proof is as follows. Consider $\phi:\mathbb{Z}_{mn}/\mathbb{Z}_m\to \mathbb{Z}_n$ , where $\mathbb{Z}_m=\{0,n,2n,\dots,(m-1)n\}$ , defined by $\phi (a+\mathbb{Z}_m)=a$ .

We may check that it is well-defined since if $a+\mathbb{Z}_m=a'+\mathbb{Z}_m$ , then $a\equiv a' \pmod n$ , and thus $\phi (a+\mathbb{Z}_m)=\phi (a'+\mathbb{Z}_m)$ .

It is a fairly straightforward to check it is a homomorphism, $\begin{aligned}\phi (a+\mathbb{Z}_m+a'+\mathbb{Z}_m)&=\phi (a+a'+\mathbb{Z}_m)\\ &=a+a'\\ &=\phi (a+\mathbb{Z}_m)+\phi(a'+\mathbb{Z}_m) \end{aligned}$

Injectivity is clear since $\ker \phi=0+\mathbb{Z}_m$ , and surjectivity is quite clear too.

Hence, this ends the proof. 🙂

Do check out some Recommended Books on Undergraduate Mathematics, and also download the free SG50 Scientific Pioneers Ebook, if you haven’t already.

Free Ebook: Singapore’s Scientific Pioneers

Source: http://www.asianscientist.com/pioneers/

The non-commercial book Singapore’s Scientific Pioneers, sponsored by grants from the SG50 Celebration Fund and Nanyang Technological University, is dedicated to all scientists in Singapore, past, present and, most of all, aspiring. Read more from Asian Scientist Magazine at: http://www.asianscientist.com/pioneers/

The free downloadable book also includes a Mathematician from National University of Singapore, Professor Louis Chen, one of the discoverers of the Chen-Stein method.

To download the ebook, click here to download (official mirror).

The Chen-Stein method is part of the branch of Mathematics known as Probability. Advanced probability requires a lot of Measure Theory, another type of Math classified under Analysis.

In the book, Professor Chen mentioned one of his favourite books is “One Two Three . . . Infinity: Facts and Speculations of Science (Dover Books on Mathematics)“.

It turns out that this book is very inspiring, and many reviewers on Amazon said that after they read this book during their childhood, they became inspired to become mathematicians/scientists!

Sample review (from Amazon): “It seems that almost all the reviewers had the same experience: we read this book at an early age, and it was so fascinating, so inspiring, and so magical that it directed us into math and science for the rest of our lives. In my case the book was loaned to me when I was about 12, by my best friend’s father.”

Hence, if you are looking for a Math/Science book for your child, this book may be one of the top choices. 🙂

Geometric n-simplex is convex

Given the definition of a geometric n-simplex:

$\displaystyle\sigma^n=\{x=\sum_{i=0}^{n}t_i a^i \mid t_i\geq 0\ \text{and }\sum_{i=0}^{n}=1\}\subseteq\mathbb{R}^n$

where $\{a^0,\dots, a^n\}$ are geometrically independent, we can show that the n-simplex is convex (i.e. given any two points, the line connecting them lies in the simplex).

Write $x=\sum_{i=0}^n t_i a^i$ , $y=\sum_{i=0}^n s_i a^i$ .

Consider the line from x to y: $\{ty+(1-t)x\mid 0\leq t\leq 1\}$ .

$\begin{aligned} ty+(1-t)x&=t\sum_{i=0}^n s_i a^i+(1-t)\sum_{i=0}^n t_i a^i\\ &=\sum_{i=0}^n (s_i t+t_i-tt_i)a_i\\ s_it+t_i-tt_i&=s_i t+t_i (1-t)\\ &\geq 0(0)+(0)(1-1)\\ &=0\\ \sum_{i=0}^n s_i t+t_i-tt_i &=t\sum_{i=0}^n s_i+\sum_{i=0}^n t_i -t\sum_{i=0}^n t_i\\ &=t(1)+(1)-t(1)\\ &=1 \end{aligned}$

Thus the line lies inside the simplex, and thus the simplex is convex.

Recommended Books for Math Majors

Tiger Mom Amy Chua Sets Up Tuition Center in Singapore

Source: http://www.cnbc.com/2015/08/20/queen-of-the-tiger-moms-takes-on-singapore.html

“Tiger Mom” Amy Chua has started a Tuition Center in Singapore. Amy Chua is the famous author of Battle Hymn of the Tiger Mother, which is an interesting book which has both supporters and critics.

The Tuition Centre is called Keys Academy, located in North Bridge Road, Singapore.

Do check it out, and more importantly do read the book Battle Hymn of the Tiger Mother to see if you agree with the author! Amy Chua does have some good points to be made, as many of the top students in Western countries are Asians. Do check out her book to read about her method.

Some other books written by Amy Chua are:

The Triple Package: How Three Unlikely Traits Explain the Rise and Fall of Cultural Groups in America

Day of Empire: How Hyperpowers Rise to Global Dominance–and Why They Fall

World on Fire: How Exporting Free Market Democracy Breeds Ethnic Hatred and Global Instability

Cycle Decomposition of Permutations is Unique

Cycle decomposition of Permutations into disjoint cycles is unique (up to reordering of cycles).

A proof can be found here.

The condition of disjoint is crucial. For example, the permutation (1 3 2) can be factored into (2 3)(1 2), where the two cycles are not disjoint. (1 3 2)=(1 2)(1 3) is also another decomposition, the two cycles are also not disjoint.

Wolframalpha can calculate permutations, a useful tool to replace manual calculations. Take note though that Wolframalpha’s convention is multiplying permutations from left to right, while most books follow the convention of multiplying right to left.

Recommended Math Books from Amazon

“偷得浮生半日闲”诗句分析—出现于中三课文《乌敏岛》

Chinese Tuition Singapore

在中三下华文课本中，有一篇题目为《乌敏岛》的课文。课文介绍了乌敏岛的自然风光和纯朴的人文环境。在文章的最后，作者写道“如果能’偷得浮生半日闲’，何不暂时摆脱现实的束缚，和三五好友到乌敏岛游玩呢？”。

“偷得浮生半日闲”出自于唐代诗人李涉的七言绝句《题鹤林寺僧舍》。全文如下：

终日昏昏醉梦间，忽闻春尽强登山。

因过竹院逢僧话，偷得浮生半日闲。

大意是：作者整日昏昏沉沉处于醉梦之中，消磨人生。忽然有一天才意识到春天就要过去了，于是勉强去爬山。在游览寺院的时候碰到一位高僧，便与其闲聊，难得在这纷纷扰扰的世事中获得片刻的清闲。

这首诗的创作背景是李涉官途不顺，被皇帝贬官后又流放到南方，所以其情绪消极终日昏昏沉沉。而在一次偶然机会，登山之时偶遇高僧，闲聊之中，不料解开了苦闷的心结，化解了世俗的烦扰，使得自己心情得以放松。

再回到课文《乌敏岛》，从课文的开始，作者就强调“踏上乌敏岛，映入眼帘的是一幅和繁忙市区截然不同的景象”。市区的人们熙熙攘攘，为生活而忙于奔走，有很多世事的烦扰。而乌敏岛却是一个别样的世界，这里没有喧嚣，人们的生活简单平静而又质朴。来到这里，看看美不胜收的风景，体验淳朴宁静的生活，相信你也会暂时忘记现实生活的烦恼和忧愁。

“偷得浮生半日闲”，人生漂浮不定，难得半日的清闲。

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Weierstrass M-test Proof and Special Case of Abel’s Theorem

First, let us recap what is Weierstrass M-test:

Weierstrass M-test:

Let $\{f_n\}$ be a sequence of real (or complex)-valued functions defined on a set A, and let $\{M_n\}$ be a sequence satisfying $\forall n\in\mathbb{N}, \forall x\in A$

$|f_n (x)|\leq M_n$ , and also $\sum_{n=1}^\infty M_n=M<\infty$ .

Then, $\sum_{n=1}^\infty f_n(x)$ converges uniformly on A (to a function f).

Proof:

Let $\epsilon >0$ . $\exists N\in\mathbb{N}$ such that $m\geq N$ implies $|M-\sum_{n=1}^m M_n|<\epsilon$ .

For $m\geq N, \forall x\in A$ ,

$\begin{aligned} |f(x)-\sum_{n=1}^m f_n(x)|&=|\sum_{n=m+1}^\infty f_n (x)|\\ &\leq\sum_{n=m+1}^\infty |f_n (x)|\\ &\leq \sum_{n=m+1}^\infty M_n\\ &=|M-\sum_{n=1}^m M_n|\\ &<\epsilon \end{aligned}$

Thus, $\sum_{n=1}^\infty f_n (x)$ converges uniformly.

Application to prove Abel’s Theorem (Special Case):

Consider the special case of Abel’s Theorem where all the coefficients $a_i$ are of the same sign (e.g. all positive or all negative).

Then, for $x\in [0,1]$ ,

$|a_n x^n|\leq |a_n|:=M_n$

Then by Weierstrass M-test, $\sum_{n=1}^\infty a_n x^n$ converges uniformly on [0,1] and thus $\lim_{x\to 1^-} \sum_{n=1}^\infty a_n x^n=\sum_{n=1}^\infty a_n$ .

Check out some books suitable for Math Majors here!

If Ratio Test Limit exists, then Root Test Limit exists, and both are equal

The limit for ratio test is $\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|}$ , while the limit for root test is $\lim_{n\to\infty}|a_n|^{1/n}$ . Something special about these two limits is that if the former exists, the latter also exists and they are equal!

Proof:

Let $\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=L$ . There exists $N\in\mathbb{N}$ such that $n\geq N \implies ||\frac{a_{n+1}}{a_n}|-L|<\epsilon$ .

i.e. $L-\epsilon<|\frac{a_{n+1}}{a_n}|<L+\epsilon$

For $n>N$ ,

$|a_n|=\frac{|a_n|}{|a_{n-1}|}\cdot \frac{|a_{n-1}|}{|a_{n-2}|}\cdots \frac{|a_{N+1}|}{|a_N|}\cdot |a_N| < (L+\epsilon)^{n-N}\cdot |a_N|$ .

Taking nth roots,

$|a_n|^{1/n}<(L+\epsilon)^\frac{n-N}{n}\cdot |a_N|^{\frac{1}{n}}$

Taking limits,

$\lim_{n\to\infty}|a_n|^{\frac{1}{n}}\leq (L+\epsilon)$

Since $\epsilon$ is arbitrary, $\lim_{n\to\infty}|a_n|^{\frac{1}{n}}\leq L$ .

Similarly, we can show $\lim_{n\to\infty}|a_n|^\frac{1}{n}\geq L$ .

Thus, $\lim_{n\to\infty}|a_n|^\frac{1}{n}=L$ .

This is considered a rather tricky (though not that difficult) proof, hope it helps whoever is searching for it!

Note that the converse is false, we can see that by considering the “rearranged” geometric series: 1/2,1, 1/8, 1/4, 1/32, … (source: https://www.maa.org/sites/default/files/0025570×33450.di021200.02p0190s.pdf)

where the ratio alternates from 2 to 1/8 and hence does not exist.

However, the root test limit of the first 2n terms is defined:

$\begin{aligned} |a_{2n}|&=\frac{|a_{2n}|}{a_{2n-1}}\cdot \frac{|a_{2n-1}|}{|a_{2n-2}|}\cdot \frac{|a_2|}{|a_1|}\cdot |a_1|\\ &=2\cdot \frac{1}{8}\cdot 2 \cdot \frac{1}{8} \cdots 2 \cdot \frac{1}{2}\\ &=2^n \cdot (\frac{1}{8})^{n-1}\cdot \frac{1}{2}\\ &=(\frac{1}{4})^{n-1} \end{aligned}$

Thus, $|a_{2n}|^\frac{1}{2n}=\frac{1}{4}^{\frac{n-1}{2n}}\to \frac{1}{2}$ .

To learn more about epsilon-delta proofs, check out one of the Recommended Analysis Books for Undergraduates.

Z[Sqrt(-2)] is a Principal Ideal Domain Proof

It turns out that to prove $\mathbb{Z}[\sqrt{-2}]$ is a Principal Ideal Domain, it is easier to prove that it is a Euclidean domain, and hence a PID.

(Any readers who have a direct proof that $\mathbb{Z}[\sqrt{-2}]$ is a PID, please comment below, as it would be very interesting to know such a proof. 🙂 )

Proof:

As mentioned above, we will prove that it is a Euclidean domain.

Let $a, b\in\mathbb{Z}[\sqrt{-2}], b\neq 0$ .

We need to show: $\exists q, r\in \mathbb{Z}[\sqrt{-2}]$ such that $a=bq+r$ , with $N(r)<N(b)$ .

Consider $\frac{a}{b}=c_1+c_2 \sqrt{-2} \in \mathbb{Q}[\sqrt{-2}]$ . Define $q=q_1+q_2 \sqrt{-2}$ where $q_1, q_2$ are the integers closest to $c_1, c_2$ respectively.

Then, $\frac{a}{b}=q+\alpha$ , where $\alpha=\alpha_1+\alpha_2 \sqrt{-2}$ .

$a=bq+b\alpha$ .

Take $r=b\alpha$ .

$\begin{aligned} N(r)&=N(b\alpha)\\ &=N(b)\cdot N(\alpha)\\ &=N(b)\cdot (\alpha_1^2+2\alpha_2^2)\\ &\leq N(b)\cdot ({\frac{1}{2}}^2+2(\frac{1}{2})^2)\\ &=N(b)\cdot (\frac{3}{4})\\ &<N(b) \end{aligned}$

Check out recommended Abstract Algebra books: Recommended Books for Math Undergraduates

Is Z[x] a Principal Ideal Domain?

In the previous post, we showed that a Euclidean domain is a Principal Ideal Domain (PID).

Consider the Polynomial Ring $\mathbb{Z}[x]$ . We can show that it is not a PID and hence also not a Euclidean domain.

Proof: Consider the ideal $<2,x>=\{ 2f(x)+xg(x)\vert f(x), g(x) \in \mathbb{Z} [x]\}$ .

Suppose to the contrary $<2,x>=<p(x)>=\{ f(x)p(x)\vert f(x)\in \mathbb{Z}[x]\}$ .

Note that $2\in <2,x>$ , hence $2\in <p(x)>$ .

2=f(x)p(x)

p(x)=2 or -2.

<p(x)>=<2>

However, $x\in <2,x>$ but $x\notin <2>$ . (contradiction!)

Check out this page for Recommended Singapore Math books!

Proof that a Euclidean Domain is a PID (Principal Ideal Domain)

Previously, we defined what is a Euclidean Domain and what is a PID. Now, we will prove that in fact a Euclidean Domain is always a PID (Principal Ideal Domain). This proof will be elaborated, it can be shortened if necessary.

Proof:

Let R be a Euclidean domain.

Let I be a nonzero ideal of R. (If I is a zero ideal, then I=(0) )

Choose $b\in I, b\neq 0$ such that $d(b)=\min \{ d(i): i\in I\}$ , where d is the Euclidean function. By the well-ordering principle, every non-empty set of positive integers contains a least element, hence b exists.

Let $a\in I$ be any element in I. $\exists q,r \in R$ such that $a=bq+r$ , with either r=0, or d(r)<d(b). (This is the property of Euclidean domain.)

We can’t have d(r)<d(b) as that will contradict minimality of d(b). Thus, r=0, and a=bq. Hence every element in the ideal is a multiple of b, i.e. I=(b). Thus R is a PID (Principal Ideal Domain).

Check out other pages on our blog:

What is Singapore Math + Recommended Singapore Math Books

Definition of Euclidean Domain and Principal Ideal Domain (PID)

A Euclidean domain is an integral domain $R$ with a function $d:R\setminus \{0\}\to \mathbb{N}$ satisfying the following:

(1) $d(a)\leq d(ab)$ for all nonzero $a,b$ in $R$ .

(2) for all $a,b \in R$ , $b\neq 0$ , $\exists q, r, \in R$ such that $a=bq+r$ , with either $r=0$ or $d(r)<d(b)$ .

(d is known as the Euclidean function)

On the other hand, a Principal ideal domain (PID) is an integral domain in which every ideal is principal (can be generated by a single element).

Video on Computing Homology Groups

This video follows after the previous video on Simplicial Complexes.

If you are looking for quality Math textbooks to study from (including Linear Algebra and Calculus, the two most popular Math courses), check out my page on Recommended Math Books for students!