Category: mathematics

This post is the second example in an ongoing list of various sequences of functions which converge to different things in different ways.

‍Example 2

A sequence of functions {fn:R→R}{fn:R→R} which converges to 0 uniformly but does not converge to 0 in L1L1.

This works because:  The sequence tends to 0 as n→∞n→∞ since the height of each function tends to 0 and the the region where fnfn is taking on this decreasing height is tending towards all of R+R+ ((0,n)(0,n) as n→∞n→∞) (and it’s already 0 on R−∪{0}R−∪{0}). The convergence is uniform because the number of times we have to keep “squishing” the rectangles until their height is less than ϵϵ does not depend on xx.

The details: Let ϵ>0ϵ>0 and choose N∈NN∈N so that N>1ϵN>1ϵ and let n>Nn>N. Fix x∈Rx∈R.

Case 1 (x≤0x≤0 or x≥nx≥n) Then fn(x)=0fn(x)=0 and so |fn(x)−0|=0<ϵ|fn(x)−0|=0<ϵ.

• Case 2 (0<x<n0<x<n ) Then fn(x)=1nfn(x)=1n and so |fn(x)−0|=1n<1N<ϵ|fn(x)−0|=1n<1N<ϵ

Finally, fn↛0fn↛0 in L1L1 since∫R|fn|=∫(0,n)1n=1nλ((0,n))=1.∫R|fn|=∫(0,n)1n=1nλ((0,n))=1.

‍Remark: Here’s a question you could ask: wouldn’t fn=nχ(0,1n)fn=nχ(0,1n) work here too? Both are tending to 0 everywhere and both involve rectangles of area 1. The answer is “kinda.” The problem is that the convergence of nχ(0,1n)nχ(0,1n) is pointwise. BUT Egoroff’s Theorem gives us a way to actually “make” it uniform!.

‍On the notation above:   For a measurable set X⊂RX⊂R, denote the set of all Lebesgue integrable functions f:X→Rf:X→R by L1(X)L1(X). Then a sequence of functions {fn}{fn} is said to converge in L1L1  to a function ff if limn→∞∫|fn−f|=0limn→∞∫|fn−f|=0.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

Given a sequence of real-valued functions {fn}{fn}, the phrase, “fnfn converges to a function ff” can mean a few things:

• fnfn converges uniformly
• fnfn converges pointwise
• fnfn converges almost everywhere (a.e.)
• fnfn converges in L1L1 (set of Lebesgue integrable functions)
• and so on…

Other factors come into play if the fnfn are required to be continuous, defined on a compact set, integrable, etc.. So since I do not have the memory of an elephant (whatever that phrase means…), I’ve decided to keep a list of different sequences that converge (or don’t converge) to different functions in different ways. With each example I’ll also include a little (and hopefully) intuitive explanation for why. Having these sequences close at hand is  especially useful when analysing the behavior of certain functions or constructing counterexamples.

The first sequence we’ll look at is one which converges almost everywhere, but does not converge in L1L1 (the set of Lebesgue integrable functions).

‍Example 1

A sequence of functions {fn:R→R}{fn:R→R} which converges to 0 almost everywhere but does not converge to 0 in L1L1.       

This works because: Recall that to say fn→0fn→0 almost everywhere means fn→0fn→0 pointwise on RR except for a set of measure 0. Here, the set of measure zero is the singleton set {0}{0} (at x=0x=0, fn(x)=nfn(x)=n and we can’t make this less than ϵϵ for any ϵ>0ϵ>0). So fnfn converges to 0 pointwise on (0,1](0,1]. This holds because if x<0x<0 or x>1x>1 then fn(x)=0fn(x)=0 for all nn. Otherwise, if x∈(0,1]x∈(0,1], we can choose nn appropriately:

The details:  Let ϵ>0ϵ>0 and x∈(0,1]x∈(0,1] and choose N∈NN∈N so that N>1xN>1x. Then whenever n>Nn>N, we have n>1xn>1x which implies x>1nx>1n and so fn(x)=0fn(x)=0. Hence |fnx−0|=0<ϵ|fnx−0|=0<ϵ.

Further*, fn↛0fn↛0 in L1L1 since∫R|fn|=∫[0,1n]n=nλ([0,1n])=1.∫R|fn|=∫[0,1n]n=nλ([0,1n])=1.

Remark: Notice that Egoroff’s theorem applies here! We just proved that fn→0fn→0 pointwise a.e. on RR, but Egoroff says that we can actually get uniform convergence a.e. on a bounded subset of RR, say (0,1](0,1].

In particular for each ϵ>0ϵ>0 we are guaranteed the existence of a subset E⊂(0,1]E⊂(0,1] such that fn→0fn→0 uniformly and λ((0,1]∖E)<ϵλ((0,1]∖E)<ϵ. In fact, it should be clear that that subset must be something like (ϵ2,1](ϵ2,1] (the “zero region” in the graph above). Then no matter where xx is in (0,1](0,1], we can always find nn large enough – namely all nn which satisfy 1n<ϵ21n<ϵ2 – so that fn(x)=0fn(x)=0, i.e. fn→ffn→f uniformly. And indeed, λ((0,1]∖(ϵ2,1]=ϵ/2<ϵλ((0,1]∖(ϵ2,1]=ϵ/2<ϵ as claimed.

‍On the notation above:   For a measurable set X⊂RX⊂R, denote the set of all Lebesgue integrable functions f:X→Rf:X→R by L1(X)L1(X). Then a sequence of functions {fn}{fn} is said to converge in L1L1  to a function ff if limn→∞∫|fn−f|=0limn→∞∫|fn−f|=0.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

In any area of math, it’s always good idea to keep a few counterexamples in your back pocket. Examples/non-examples from some of the different subsets of integral domains.

Z[i√5]Z[i5] is an integral domain which is not a UFD

That Z[i√5]Z[i5] an integral domain is easy to check (just computation).

• It’s not a UFD since we can write 6=2⋅3=(1+i√5)(1−i√5)6=2⋅3=(1+i5)(1−i5) as two distinct facorizations into irreducibles*

‍Z[x]Z[x] is a UFD which is not a PID

We know Z[x]Z[x] is a UFD because ZZ is a UFD (recall, a commutative ring RR is a UFD iff R[x]R[x] is a UFD).

• The ideal (2,x)={2f(x)+xg(x):f(x),g(x)∈Z[x]}(2,x)={2f(x)+xg(x):f(x),g(x)∈Z[x]} (polynomials with even constant term) is not principal**

‍Z[12+i√192]Z[12+i192] is a PID which is not a Euclidean domain

• This is a PID since it has a Dedekind-Hasse norm (see Dummit and Foote, 3rd ed., §8.2§8.2).
• It is not a Euclidean domain since it has no universal side divisors (ibid.).

ZZ is a Euclidean domain which is not a field

ZZ is a Euclidean domain via the absolute value norm (which gives the familiar division algorithm).

• It is not a field since the only elements which are units are 11 and −1−1.

‍  (*) Check 2,3,1+i√52,3,1+i5, and 1−i√51−i5 are indeed irreducible in Z[i√5]Z[i5]:

Write 2=αβ2=αβ for α,β∈Z[i√5]α,β∈Z[i5]. Then α=a+ib√5α=a+ib5 and N(α)=a2+5b2N(α)=a2+5b2 for some integers a,ba,b. Since 4=N(2)=N(α)N(β)4=N(2)=N(α)N(β), we must have a2+5b2=1,2a2+5b2=1,2 or 44. Notice b=0b=0 must be true (since a2+5b2∉{1,2,4}a2+5b2∉{1,2,4} for b≥1b≥1 and for any aa). Hence either α=a=1α=a=1 or 22. If α=1α=1 then αα is a unit. If α=2α=2, then we must have β=1β=1 and so ββ is a unit.

• Showing 3 is irreducible follows a similar argument.

‍Write 1+i√5=αβ1+i5=αβ with α=a+ib√5α=a+ib5 so that N(α)=a2+5b2∈{1,2,3,6}N(α)=a2+5b2∈{1,2,3,6} since 6=N(α)N(β)6=N(α)N(β). Consider two cases:  (case 1) If b=0b=0, then a2∈{1,2,3,6}a2∈{1,2,3,6} which is only true if a2=1a2=1 and so α=a=±1α=a=±1 is a unit. (case 2) If b>0b>0, we can only have b2=1b2=1 (since b2>1b2>1 gives a contradiction), and so a2+5∈{1,2,3,6}a2+5∈{1,2,3,6}, which implies a2=1a2=1. Hence α=±1±i√5α=±1±i5 and so N(α)=6N(α)=6. This implies N(β)=1N(β)=1 and so β=±1β=±1, which is a unit.

‍Showing 1−i√51−i5 is irreducible follows a similar argument.

principal in Z[x]Z[x]:

• Suppose to the contrary (2,x)=(f(x))(2,x)=(f(x)) for some polynomial f(x)∈Z[x]f(x)∈Z[x]. Since 2∈(f(x))2∈(f(x)), we must have 2=f(x)p(x)2=f(x)p(x) for some p(x)∈Z[x]p(x)∈Z[x]. Hence 0=degf(x)+degp(x)0=deg⁡f(x)+deg⁡p(x) which implies both f(x)f(x) and p(x)p(x) are constants. In particular, since 2=±1⋅±22=±1⋅±2, we need f(x),p(x)∈{±1,±2}f(x),p(x)∈{±1,±2}. If f(x)=±1f(x)=±1, then (f(x))=Z[x](f(x))=Z[x] which is a contradiction since (f(x))=(2,x)(f(x))=(2,x) mustbe a proper ideal (not every polynomial over Z[x]Z[x] has even constant term). It follows that f(x)=±2f(x)=±2. But since x∈(f(x))x∈(f(x)) as well, x=2r(x)x=2r(x) for some r(x)∈Z[x]r(x)∈Z[x]. But of course this is impossible for any polynomial with integer coefficients, r(x)r(x). Thus (2,x)(2,x) is not principal.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

Here is a list of some of the subsets of integral domains, along with the reasoning (a.k.a proofs) of why the bullseye below looks the way it does. Part 2 of this post will include back-pocket examples/non-examples of each.

Integral Domain: a commutative ring with 1 where the product of any two nonzero elements is always nonzero

Unique Factorization Domain (UFD): an integral domain where every nonzero element (which is not a unit) has a unique factorization into irreducibles

Principal Ideal Domain (PID): an integral domain where every ideal is generated by exactly one element

Euclidean Domain: an integral domain RR with a norm NN and a division algorithm (i.e. there is a norm NN so that for every a,b∈Ra,b∈R with b≠0b≠0, there are q,r∈Rq,r∈R so that a=bq+ra=bq+r with r=0r=0 or N(r)<N(b)N(r)<N(b))

Field: a commutative ring where every nonzero element has an inverse

Because… We can just choose the zero norm: N(r)=0N(r)=0 for all r∈Fr∈F.

Proof: Let FF be a field and define a norm NN so that N(r)=0N(r)=0 for all r∈Fr∈F. Then for any a,b∈Fa,b∈F with b≠0b≠0, we can writea=b(b−1a)+0.a=b(b−1a)+0.

Because… If I◃RI◃R is an arbitrary nonzero ideal in the Euclidean domain RR, then I=(d)I=(d), where d∈Id∈I such that dd has the smallest norm among all elements in II. Prove this using the division algorithm on dd and some a∈Ia∈I.

Proof: Let RR be a Euclidean domain with respect to the norm NN and let I◃RI◃R be an ideal. If I=(0)I=(0), then II is principle. Otherwise let d∈Id∈I be a nonzero element such that dd has the smallest norm among all elements in II. We claim I=(d)I=(d). That (d)⊂I(d)⊂I is clear so let a∈Ia∈I. Then by the division algorithm, there exist q,r∈Rq,r∈R so that a=dq+ra=dq+r with r=0r=0 or N(r)<N(d)N(r)<N(d). Then r=a−dq∈Ir=a−dq∈I since a,d∈Ia,d∈I. But my minimality of dd, this implies r=0r=0. Hence a=dq∈(d)a=dq∈(d) and so I⊂(d)I⊂(d).

Because…Every PID has the ascending chain condition (acc) on its ideals!* So to prove PID ⇒⇒ UFD, just recall that an integral domain RR is a UFD if and only if 1) it has the acc on principal ideals** and 2) every irreducible element is also prime.

Proof: Let RR be a PID. Then 1) RR has the ascending chain condition on principal ideals and 2) every irreducible element is also a prime element. Hence RR is a UFD.

Because… By definition.

Proof: By definition.

‍*Def: In general, an integral domain RR has the acc on its principal ideals if these two equivalent conditions are satisfied:

1. Every sequence I1⊂I2⊂⋯⊂⋯I1⊂I2⊂⋯⊂⋯ of principal ideals is stationary (i.e. there is an integer n0≥1n0≥1 such that In=In0In=In0 for all n≥n0n≥n0).
2. For every nonempty subset X⊂RX⊂R, there is an element m∈Xm∈X such that whenever a∈Xa∈X and (m)⊂(a)(m)⊂(a), then (m)=(a)(m)=(a).

**To see this, use part 1 of the definition above. If I1⊂I2⊂⋯I1⊂I2⊂⋯ is an acsending chain, consider their union I=⋃∞n=1InI=⋃n=1∞In. That guy must be a principal ideal (check!), say I=(m)I=(m). This implies that mm must live in some In0In0  for some n0≥1n0≥1 and so I=(m)⊂In0I=(m)⊂In0. But since II is the union, we have for all n≥n0n≥n0(m)=I⊃In⊃In0=(m).(m)=I⊃In⊃In0=(m).Voila!

Every field FF is a PID

because the only ideals in a field are (0)(0) and F=(1)F=(1)! And every field is vacuously a UFD since all elements are units. (Recall, RR is a UFD if every non-zero, non-invertible element (an element which is not a unit) has a unique factorzation into irreducibles).

‍In an integral domain, every maximal ideal is also a prime ideal. 

(Proof: Let RR be an integral domain and M◃RM◃R a maximal ideal. Then R/MR/M is a field and hence an integral domain, which implies M◃RM◃R is a prime ideal.)

Butut the converse is not true (see counterexample below). However, the converse is true in a PID because of the added structure!

(Proof: Let RR be a PID and (p)◃R(p)◃R a prime ideal for some p∈Rp∈R. Then pp is a prime – and hence an irreducible – element (prime ⇔⇔ irreducible in PIDs). Since in an integral domain a principal ideal is maximal whenever it is generated by an irreducible element, we conclude (p)(p) is maximal.)

This suggests that if you want to find a counterexample – an integral domain with a prime ideal which is not maximal – try to think of a ring which is not a PID:   In Z[x]Z[x], consider the ideal (p)(p) for a prime integer pp. Then (p)(p) is a prime ideal, yet it is not maximal since(p)⊂(p,x)⊂Z[x].(p)⊂(p,x)⊂Z[x].

‍If FF is a field, then F[x]F[x] – the ring of polynomials in xx with coefficients in FF – is a Euclidean domain with the norm N(p(x))=degp(x)N(p(x))=deg⁡p(x) where p(x)∈F[x]p(x)∈F[x].

By the integral domain hierarchy above, this implies every ideal in F[x]F[x] is of the form (p(x))(p(x)) (i.e. F[x]F[x] is a PID) and every polynomial can be factored uniquely into a product of prime polynomials (just like the integers)! The next bullet gives an “almost converse” statement.

‍If R[x]R[x] is a PID, the RR must be a field.

To see this, simply observe that R⊂R[x]R⊂R[x] and so RR must be an integral domain (since a subset of a integral domain inherets commutativity and the “no zero divisors” property). Since R[x]/(x)≅RR[x]/(x)≅R, it follows that R[x]/(x)R[x]/(x) is also an integral domain. This proves that (x)(x) is a prime ideal. But prime implies maximal in a PID! So R[x]/(x)R[x]/(x) – and therefore RR – is actually a field.

• This is how we know, for example, that Z[x]Z[x] is not a PID (in the counterexample a few bullets up) – ZZ is not a field!

‍For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

Bob Moses, who helped register Black residents to vote in Mississippi during the Civil Rights Movement, believed civil rights went beyond the ballot box. To Moses, who was a teacher as well as an activist, math literacy is a civil right: a requirement to earning a living wage in modern society. In 1982, he founded the Algebra Project to ensure that “students at the bottom get the math literacy they need.”

As a researcher who studies ways to improve the math experiences of students, we believe a new approach that expands access to algebra may help more students get the math literacy Moses, who died in 2021, viewed as so important. It’s a goal districts have long been struggling to meet.

Efforts to improve student achievement in algebra have been taking place for decades. Unfortunately, the math pipeline in the United States is fraught with persistent opportunity gaps. According to the Nation’s Report Card—a congressionally mandated project administered by the Department of Education—in 2022 only 29% of U.S. fourth graders and 20% of U.S. eighth graders were proficient in math. Low-income students, students of colour and multilingual learners, who tend to have lower scores on math assessments, often do not have the same access as others to qualified teachers, high-quality curriculum and well-resourced classrooms.

A new approach

The Dallas Independent School District—or Dallas ISD—is gaining national attention for increasing opportunities to learn by raising expectations for all students. Following in the footsteps of more than 60 districts in the state of Washington, in 2019 the Dallas ISD implemented an innovative approach of having students be automatically enrolled rather than opt in to honours math in middle school.

Under an opt-in policy, students need a parent or teacher recommendation to take honours math in middle school and Algebra 1 in eighth grade. That policy led both to low enrolment and very little diversity in honours math. Some parents, especially those who are Black or Latino, were not aware how to enroll their students in advanced classes due to a lack of communication in many districts.

In addition, implicit bias, which exists in all demographic groups, may influence teachers’ perceptions of the behaviour and academic potential of students, and therefore their subsequent recommendations. Public school teachers in the U.S. are far less racially and ethnically diverse than the students they serve.

Dallas ISD’s policy overhaul aimed to foster inclusivity and bridge educational gaps among students. Through this initiative, every middle school student, regardless of background, was enrolled in honours math, the pathway that leads to taking Algebra 1 in eighth grade, unless they opted out.

Flipping the switch from opt-in to opt-out led to a dramatic increase in the number of Black and Latino learners, who constitute the majority of Dallas students. And the district’s overall math scores remained steady. About 60% of Dallas ISD eighth graders are now taking Algebra 1, triple the prior level. Moreover, more than 90% are passing the state exam.

Efforts spread

Other cities are taking notice of the effects of Dallas ISD’s shifting policy. The San Francisco Unified School District, for example, announced plans in February 2024 to implement Algebra 1 in eighth grade in all schools by the 2026-27 school year.

In fall 2024, the district will pilot three programs to offer Algebra 1 in eighth grade. The pilots range from an opt-out program for all eighth graders—with extra support for students who are not proficient—to a program that automatically enrolls proficient students in Algebra 1, offered as an extra math class during the school day. Students who are not proficient can choose to opt in. Nationwide, however, districts that enroll all students in Algebra 1 and allow them to opt out are still in the minority. And some stopped offering eighth grade Algebra 1 entirely, leaving students with only pre-algebra classes. Cambridge, Massachusetts—the city in which Bob Moses founded the Algebra Project—is among them.

Equity concerns linger

Between 2017 and 2019, district leaders in the Cambridge Public Schools phased out the practice of placing middle school students into “accelerated” or “grade-level” math classes. Few middle schools in the district now offer Algebra 1 in eighth grade.

The policy shift, designed to improve overall educational outcomes, was driven by concerns over significant racial disparities in advanced math enrollment in high school. Completion of Algebra 1 in eighth grade allows students to climb the math ladder to more difficult classes, like calculus, in high school. In Cambridge, the students who took eighth grade Algebra 1 were primarily white and Asian; Black and Latino students enrolled, for the most part, in grade-level math.

Some families and educators contend that the district’s decision made access to advanced math classes even more inequitable. Now, advanced math in high school is more likely to be restricted to students whose parents can afford to help them prepare with private lessons, after-school programs or private schooling, they said.

While the district has tried to improve access to advanced math in high school by offering a free online summer program for incoming ninth graders, achievement gaps have remained persistently wide.

Perhaps striking a balance between top-down policy and bottom-up support will help schools across the U.S. realize the vision Moses dreamed of in 1982 when he founded the Algebra Project: “That in the 21st century every child has a civil right to secure math literacy—the ability to read, write and reason with the symbol systems of mathematics.”

For more insights like this, visit our website at www.international-maths-challenge.com.

Credit of the article given to Liza Bondurant, The Conversation

These are painful times for those hoping to see an international consensus and substantive action on global warming.

In the US, Republican presidential front-runner Mitt Romney said in June 2011: “The world is getting warmer” and “humans have contributed” but in October 2011 he backtracked to: “My view is that we don’t know what’s causing climate change on this planet.”

His Republican challenger Rick Santorum added: “We have learned to be sceptical of ‘scientific’ claims, particularly those at war with our common sense” and Rick Perry, who suspended his campaign to become the Republican presidential candidate last month, stated flatly: “It’s all one contrived phony mess that is falling apart under its own weight.”

Meanwhile, the scientific consensus has moved in the opposite direction. In a study published in October 2011, 97% of climate scientists surveyed agreed global temperatures have risen over the past 100 years. Only 5% disagreed that human activity is a significant cause of global warming.

The study concluded in the following way: “We found disagreement over the future effects of climate change, but not over the existence of anthropogenic global warming.

“Indeed, it is possible that the growing public perception of scientific disagreement over the existence of anthropocentric warming, which was stimulated by press accounts of [the UK’s] ”Climategate“ is actually a misperception of the normal range of disagreements that may persist within a broad scientific consensus.”

More progress has been made in Europe, where the EU has established targets to reduce emissions by 20% (from 1990 levels) by 2020. The UK, which has been beset by similar denial movements, was nonetheless able to establish, as a legally binding target, an 80% reduction by 2050 and is a world leader on abatement.

In Australia, any prospect for consensus was lost when Tony Abbott used opposition to the Labor government’s proposed carbon market to replace Malcolm Turnbull as leader of the Federal Opposition in late 2009.

It used to be possible to hear right-wing politicians in Australia or the USA echo the Democratic congressman Henry Waxman who said last year:

“If my doctor told me I had cancer, I wouldn’t scour the country to find someone to tell me that I don’t need to worry about it.”

But such rationality has largely left the debate in both the US and Oz. In Australia, a reformulated carbon tax policy was enacted in November only after a highly partisan debate.

In Canada, the debate is a tad more balanced. The centre-right Liberal government in British Columbia passed the first carbon tax in North America in 2008, but the governing Federal Conservative party now offers a reliable “anti-Kyoto” partnership with Washington.

Overviews of the evidence for global warming, together with responses to common questions, are available from various sources, including:

• Seven Answers to Climate Contrarian Nonsense, in Scientific American
• Climate change: A Guide for the Perplexed, in New Scientist
• Cooling the Warming Debate: Major New Analysis Confirms That Global Warming Is Real, in Science Daily
• Remind me again: how does climate change work? on The Conversation

It should be acknowledged in these analyses that all projections are based on mathematical models with a significant level of uncertainty regarding highly complex and only partially understood systems.

As 2011 Australian Nobel-Prize-winner Brian Schmidt explained while addressing a National Forum on Mathematical Education:

“Climate models have uncertainty and the earth has natural variation … which not only varies year to year, but correlates decade to decade and even century to century. It is really hard to design a figure that shows this in a fair way — our brain cannot deal with the correlations easily.

“But we do have mathematical ways of dealing with this problem. The Australian academy reports currently indicate that the models with the effects of CO₂ are with 90% statistical certainty better at explaining the data than those without.

“Most of us who work with uncertainty know that 90% statistical uncertainty cannot be easily shown within a figure — it is too hard to see …”

“ … Since predicting the exact effects of climate change is not yet possible, we have to live with uncertainty and take the consensus view that warming can cover a wide range of possibilities, and that the view might change as we learn more.”

But uncertainty is no excuse for inaction. The proposed counter-measures (e.g. infrastructure renewal and modernisation, large-scale solar and wind power, better soil remediation and water management, not to mention carbon taxation) are affordable and most can be justified on their own merits, while the worst-case scenario — do nothing while the oceans rise and the climate changes wildly — is unthinkable.

Some in the first world protest that any green energy efforts are dwarfed by expanding energy consumption in China and elsewhere. Sure, China’s future energy needs are prodigious, but China also now leads the world in green energy investment.

By blaiming others and focusing the debate on the level of human responsibility for warming and about the accuracy of predictions, the deniers have managed to derail long-term action in favour of short-term economic policies.

Who in the scientific community is promoting the denial of global warming? As it turns out, the leading figures in this movement have ties to conservative research institutes funded mostly by large corporations, and have a history of opposing the scientific consensus on issues such as tobacco and acid rain.

What’s more, those who lead the global warming denial movement – along with creationists, intelligent design writers and the “mathematicians” who flood our email inboxes with claims that pi is rational or other similar nonsense – are operating well outside the established boundaries of peer-reviewed science.

Austrian-born American physicist Fred Singer, arguably the leading figure of the denial movement, has only six peer-reviewed publications in the climate science field, and none since 1997.

After all, when issues such as these are “debated” in any setting other than a peer-reviewed journal or conference, one must ask: “If the author really has a solid argument, why isn’t he or she back in the office furiously writing up this material for submission to a leading journal, thereby assuring worldwide fame and glory, not to mention influence?”

In most cases, those who attempt to grab public attention through other means are themselves aware they are short-circuiting the normal process, and that they do not yet have the sort of solid data and airtight arguments that could withstand the withering scrutiny of scientific peer review.

When they press their views in public to a populace that does not understand how the scientific enterprise operates, they are being disingenuous.

With regards to claims scientists are engaged in a “conspiracy” to hide the “truth” on an issue such as global warming or evolution, one should ask how a secret “conspiracy” could be maintained in a worldwide, multicultural community of hundreds of thousands of competitive researchers.

As Benjamin Franklin wrote in his Poor Richard’s Almanac: “Three can keep a secret, provided two of them are dead.” Or as one of your present authors quipped, tongue-in-cheek, in response to a state legislator who was skeptical of evolution: “You have no idea how humiliating this is to me — there is a secret conspiracy among leading scientists, but no-one deemed me important enough to be included!”

There’s another way to think about such claims: we have tens-of-thousands of senior scientists in their late-fifties or early-sixties who have seen their retirement savings decimated by the recent stock market plunge. These are scientists who now wonder if the day will ever come when they are financially well-off-enough to do their research without the constant stress and distraction of applying for grants (the majority of which are never funded).

All one of these scientists has to do to garner both worldwide fame and considerable fortune (through book contracts, the lecture circuit and TV deals) is to call a news conference and expose “the truth”. So why isn’t this happening?

The system of peer-reviewed journals and conferences sponsored by major professional societies is the only proper forum for the presentation and debate of new ideas, in any field of science or mathematics.

It has been stunningly successful: errors have been uncovered, fraud has been rooted out and bogus scientific claims (such as the 1903 N-ray claim, the 1989 cold fusion claim, and the more-recent assertion of an autism-vaccination link) have been debunked.

This all occurs with a level of reliability and at a speed that is hard to imagine in other human endeavours. Those who attempt to short-circuit this system are doing potentially irreparable harm to the integrity of the system.

They may enrich themselves or their friends, but they are doing grievous damage to society at large.

For more such insights, log into our website https://international-maths-challenge.com

Credit of the article given to Jonathan Borwein (Jon), University of Newcastle and David H. Bailey, University of California, Davis

Does the thought of p-values and regressions make you break out in a cold sweat? Never fear – read on for answers to some of those burning statistical questions that keep you up 87.9% of the night.

• What are my hypotheses?

There are two types of hypothesis you need to get your head around: null and alternative. The null hypothesis always states the status quo: there is no difference between two populations, there is no effect of adding fertiliser, there is no relationship between weather and growth rates.

Basically, nothing interesting is happening. Generally, scientists conduct an experiment seeking to disprove the null hypothesis. We build up evidence, through data collection, against the null, and if the evidence is sufficient we can say with a degree of probability that the null hypothesis is not true.

We then accept the alternative hypothesis. This hypothesis states the opposite of the null: there is a difference, there is an effect, there is a relationship.

• What’s so special about 5%?

One of the most common numbers you stumble across in statistics is alpha = 0.05 (or in some fields 0.01 or 0.10). Alpha denotes the fixed significance level for a given hypothesis test. Before starting any statistical analyses, along with stating hypotheses, you choose a significance level you’re testing at.

This states the threshold at which you are prepared to accept the possibility of a Type I Error – otherwise known as a false positive – rejecting a null hypothesis that is actually true.

• Type what error?

Most often we are concerned primarily with reducing the chance of a Type I Error over its counterpart (Type II Error – accepting a false null hypothesis). It all depends on what the impact of either error will be.

Take a pharmaceutical company testing a new drug; if the drug actually doesn’t work (a true null hypothesis) then rejecting this null and asserting that the drug does work could have huge repercussions – particularly if patients are given this drug over one that actually does work. The pharmaceutical company would be concerned primarily with reducing the likelihood of a Type I Error.

Sometimes, a Type II Error could be more important. Environmental testing is one such example; if the effect of toxins on water quality is examined, and in truth the null hypothesis is false (that is, the presence of toxins does affect water quality) a Type II Error would mean accepting a false null hypothesis, and concluding there is no effect of toxins.

The down-stream issues could be dire, if toxin levels are allowed to remain high and there is some health effect on people using that water.

Do you know the difference between continuous and categorical variables?

• What is a p-value, really?

Because p-values are thrown about in science like confetti, it’s important to understand what they do and don’t mean. A p-value expresses the probability of getting a given result from a hypothesis test, or a more extreme result, if the null hypothesis were true.

Given we are trying to reject the null hypothesis, what this tells us is the odds of getting our experimental data if the null hypothesis is correct. If the odds are sufficiently low we feel confident in rejecting the null and accepting the alternative hypothesis.

What is sufficiently low? As mentioned above, the typical fixed significance level is 0.05. So if the probability portrayed by the p-value is less than 5% you reject the null hypothesis. But a fixed significance level can be deceiving: if 5% is significant, why is 6% not?

It pays to remember that such probabilities are continuous, and any given significance level is arbitrary. In other words, don’t throw your data away simply because you get a p-value of 6-10%.

• How much replication do I have?

This is probably the biggest issue when it comes to experimental design, in which the focus is on ensuring the right type of data, in large enough quantities, is available to answer given questions as clearly and efficiently as possible.

Pseudoreplication refers to the over-inflation of degrees of freedom (a mathematical restriction put in place when we calculate a parameter – e.g. a mean – from a sample). How would this work in practice?

Say you’re researching cholesterol levels by taking blood from 20 male participants.

Each male is tested twice, giving 40 test results. But the level of replication is not 40, it’s actually only 20 – a requisite for replication is that each replicate is independent of all others. In this case, two blood tests from the same person are intricately linked.

If you were to analyse the data with a sample size of 40, you would be committing the sin of pseudoreplication: inflating your degrees of freedom (which incidentally helps to create a significant test result). Thus, if you start an experiment understanding the concept of independent replication, you can avoid this pitfall.

• How do I know what analysis to do?

There is a key piece of prior knowledge that will help you determine how to analyse your data. What kind of variable are you dealing with? There are two most common types of variable:

1) Continuous variables. These can take any value. Were you to you measure the time until a reaction was complete, the results might be 30 seconds, two minutes and 13 seconds, or three minutes and 50 seconds.

2) Categorical variables. These fit into – you guessed it – categories. For instance, you might have three different field sites, or four brands of fertiliser. All continuous variables can be converted into categorical variables.

With the above example we could categorise the results into less than one minute, one to three minutes, and greater than three minutes. Categorical variables cannot be converted back to continuous variables, so it’s generally best to record data as “continuous” where possible to give yourself more options for analysis.

Deciding which to use between the two main types of analysis is easy once you know what variables you have:

ANOVA (Analysis of Variance) is used to compare a categorical variable with a continuous variable – for instance, fertiliser treatment versus plant growth in centimetres.

Linear Regression is used when comparing two continuous variables – for instance, time versus growth in centimetres.

Though there are many analysis tools available, ANOVA and linear regression will get you a long way in looking at your data. So if you can start by working out what variables you have, it’s an easy second step to choose the relevant analysis.

Ok, so perhaps that’s not everything you need to know about statistics, but it’s a start. Go forth and analyse!

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Sarah-Jane O’Connor*

Elliptic curves have a long and distinguished history that can be traced back to antiquity. They are prevalent in many branches of modern mathematics, foremost of which is number theory.

In simplest terms, one can describe these curves by using a cubic equation of the form

where A and B are fixed rational numbers (to ensure the curve E is nice and smooth everywhere, one also needs to assume that its discriminant 4A³ + 27B² is non-zero).

To illustrate, let’s consider an example: choosing A=-1 and B=0, we obtain the following picture:

At this point it becomes clear that, despite their name, elliptic curves have nothing whatsoever to do with ellipses! The reason for this historical confusion is that these curves have a strong connection to elliptic integrals, which arise when describing the motion of planetary bodies in space.

The ancient Greek mathematician Diophantus is considered by many to be the father of algebra. His major mathematical work was written up in the tome Arithmetica which was essentially a school textbook for geniuses. Within it, he outlined many tools for studying solutions to polynomial equations with several variables, termed Diophantine Equations in his honour.

One of the main problems Diophantus considered was to find all solutions to a particular polynomial equation that lie in the field of rational numbers Q. For equations of “degree two” (circles, ellipses, parabolas, hyperbolas) we now have a complete answer to this problem. This answer is thanks to the late German mathematician Helmut Hasse, and allows one to find all such points, should they exist at all.

Returning to our elliptic curve E, the analogous problem is to find all the rational solutions (x,y) which satisfy the equation defining E. If we call this set of points E(Q), then we are asking if there exists an algorithm that allows us to obtain all points (x,y) belonging to E(Q).

At this juncture we need to introduce a group law on E, which gives an eccentric way of fusing together two points (p₁ and p₂) on the curve, to obtain a brand new point (p₄). This mimics the addition law for numbers we learn from childhood (i.e. the sum or difference of any two numbers is still a number). There’s an illustration of this rule below:

Under this geometric model, the point p₄ is defined to be the sum of p₁ and p₂ (it’s easy to see that the addition law does not depend on the order of the points p₁, p₂). Moreover the set of rational points is preserved by this notion of addition; in other words, the sum of two rational points is again a rational point.

Louis Mordell, who was Sadleirian Professor of Pure Mathematics at Cambridge University from 1945 to 1953, was the first to determine the structure of this group of rational points. In 1922 he proved

where the number of copies of the integers Z above is called the “rank r(E) of the elliptic curve E”. The finite group Τ_E(Q) on the end is uninteresting, as it never has more than 16 elements.

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*Credit for article given to Daniel Delbourgo*

I work on the mathematics of sharing resources, which has led me to consider emotions such as envy, behaviour such as risk-taking and the best way to cut a cake.

Like, I suspect, many women, my wife enjoys eating dessert but not ordering it. I therefore dutifully order what I think she’ll like, cut it in half and invite her to choose a piece.

This is a sure-fire recipe for marital accord. Indeed, many mathematicians, economists, political scientists and others have studied this protocol and would agree. The protocol is known as the “cut-and-choose” procedure. I cut. You choose.

Cut-and-choose

Cut-and-choose is not limited to the dining table – it dates back to antiquity. It appears nearly 3,000 years ago in Hesiod’s poem Theogeny where Prometheus divides a cow and Zeus selects the part he prefers.

In more recent times, cut-and-choose has been enshrined in the UN’s 1982 Convention of the Law of the Sea where it was proposed as a mechanism to resolve disputes when dividing the seabed for mining.

To study the division of cake, cows and the seabed in a more formal way, various mathematical models have been developed. As with all models, these need to make a number of simplifying assumptions.

One typical assumption is that the people employing the cut-and-choose method are risk-averse. They won’t adopt a risky strategy that may give them less cake than a more conservative strategy.

With such assumptions in place, we can then prove what properties cake cutting procedures have and don’t have. For instance, cut-and-choose is envy free.

You won’t envy the cake I have, otherwise you would have taken this piece. And I won’t envy the piece you have, as the only risk-averse strategy is for me to cut the cake into two parts that I value equally.

On the other hand, the cutting of the cake is not totally equitable since the player who chooses can get cake that has more than half the total value for them.

With two players, it’s hard to do better than cut-and-choose. But I should record that my wife argues with me about this.

She believes it favours the second player since the first player inevitably can’t divide the cake perfectly and the second player can capitalise on this. This is the sort of assumption ignored in our mathematical models.

My wife might prefer the moving-knife procedure which doesn’t favour either player. A knife is moved over the cake, and either player calls “cut” when they are happy with the slice.

Again, this will divide the cake in such a way that neither player will envy the other (else they would have called “cut” themselves).

Three’s a crowd

Unfortunately, moving beyond two players increases the complexity of cutting cake significantly.

With two players, we needed just one cut to get to an envy free state. With three players, a complex series of five cuts of the cake might be needed. Of course, only two cuts are needed to get three slices.

The other three cuts are needed to remove any envy. And with four players, the problem explodes in our face.

An infinite number of cuts may be required to get to a situation where no one envies another’s cake. I’m sure there’s some moral here about too many cake cutters spoiling the dessert.

There are many interesting extensions of the problem. One such extension is to indivisible goods.

Suppose you have a bag of toys to divide between two children. How do you divide them fairly? As a twin myself, I know that the best solution is to ensure you buy two of everything.

It’s much more difficult when your great aunt gives you one Zhu Zhu pet, one Bratz doll and three Silly Bandz bracelets to share.

Online

More recently, I have been studying a version of the problem applicable to online settings. In such problems, not all players may be available all of the time. Consider, for instance, allocating time on a large telescope.

Astronomers will have different preferences for when to use the telescope depending on what objects are visible, the position of the sun, etcetera. How do we design a web-based reservation system so that astronomers can choose observation times that is fair to all?

We don’t want to insist all astronomers log in at the same time to decide an allocation. And we might have to start allocating time on the telescope now, before everyone has expressed their preferences. We can view this as a cake-cutting problem where the cake is made up of the time slots for observations.

The online nature of such cake-cutting problems poses some interesting new challenges.

How can we ensure that late-arriving players don’t envy cake already given to earlier players? The bad news is that we cannot now achieve even a simple property like envy freeness.

No procedure can guarantee situations where players don’t envy one another. But more relaxed properties are possible, such as not envying cake allocated whilst you are participating in the cutting of the cake.

Ham sandwich

There’s a brilliantly named piece of mathematics due to Arthur H. Stone and [John Tukey](http://www.morris.umn.edu/~sungurea/introstat/history/w98/Tukey.html, the Ham Sandwich Theorem which proves we can always cut a three-layered cake perfectly with a single cut.

Suppose we have three objects. Let’s call them “the top slice of bread”, “the ham filling” and “the bottom slice of bread”. Or if you prefer “the top layer” of the cake, “the middle layer” and “the bottom layer”.

The ham sandwich theorem proves a single slice can always perfectly bisect the three objects. Actually, the ham sandwich theorem works in any number of dimensions: any n objects in n-dimensional space can be simultaneously bisected by a single (n − 1) dimensional hyperplane.

So, in the case of the three-layered cake, n = 3, and the three-layered cake can be bisected (or cut) using a single, two-dimensional “hyperplane”. Such as, say, a knife.

Who would have thought that cutting cake would lead to higher dimensions of mathematics by way of a ham sandwich?

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*Credit for article given to Toby Walsh*

The objects pictured above are interesting structures – they are derived from the prime factorization of a given number n. They can be described in a number of ways – for example, as directed graphs. Because they are nicely structured, they actually form something more special – a lattice. Accordingly, these structures are called factor lattices.
It’s easy to start drawing these by hand following the instructions below.

1. The first node is 1
2. Draw arrows out of this node for each of the prime factors of n.
3. The arrows that you just drew should connect to nodes labled with the prime factors of n.

Now, for each of the new nodes that you drew do the following:

4. Start from a node x that is not equal to n.
5. Draw arrows out of this node for each of the prime factors of n/x.
6. The arrows that you just drew (one for each p = n/x) should connect to nodes labled with the numbers p*x.

7. Now repeat 4,5, and 6 for each new node that you have drawn that is not equal to n.

This process is recursive, and ends when you have the complete lattice. The process is well suited for implementation as a computer program – the images above were created using SAGE using output from a Java program based on the algorithm above.

Manually trying out the steps out for a number like n = 24 goes something like this: First write out the prime factorization of 24, 24=(2*2*2)*3 = (2^3)*3. Starting with 1, draw arrows out to 2 and 3. Now looking at each node and following the algorithm, from the 2 you will get arrows out to 4 and 6. From the 3 you will get an arrow out to 6 as well. From 4 you will get arrows out to 8 or 12. From 6 you will get an arrow out to 12 as well. From 8 and from 12 you get arrows out to 24, and you are done.

In general, the algorithm produces a lattice that can be described as follows. Each node is a factor of the given number n. Two nodes are connected by an edge if their prime factorization differs by a single prime number. In other words, if a and b are nodes, and p = b/a, then there is an arrow p:a–>b.

It’s a good exercise to make the connections between the lattice structure and the prime factorization of a number n.

1. What does the factor lattice of a prime number look like?
2. If a number is just a power of a prime, what does its lattice look like?
3. If you know the factorization, can you find the number of nodes without drawing the lattice.

The answer to the last question (3) can be expressed as:

For example, if n = 24= 2^3*3, then the number of nodes will be (3+1)(1+1) = 8

That these structures can be thought of as “lattices”comes from the fact that you can think of the arrows as an ordering of the nodes, ab. The number 1 is always the least node in the factor lattice for n, while n itself is the greatest node. The property that actually makes these structures a “lattice” is that for any two nodes there is always a lower-bound for any pair of nodes in the lattice, and always an upper-bound for the pair (these are often referred to as meets and joins).

The Wolfram Demonstrations Project has a nice factor lattice demo that will draw factor lattices for a large number of integers for you. There is also a good Wikipedia entry for lattices in general.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*