mei powerpoint template · 2019. 7. 3. · maths, she noted, isn't that different from novel-...

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How do you solve a

problem like Maryam?

Proof and problem

solving

Jo Sibley

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In case you hadn’t heard already… In 2014, Dr. Maryam Mirzakhani became the first woman to win Maths’

highest honour, the Fields Medal.

The celebrated scholar hadn't always been a maths whizz. Growing

up in Iran, her dream was to be a writer. It wasn't until high

school that Mirzakhani's passion for numbers took hold. By 17, she

was winning international competitions. At Harvard, her thesis solved a

problem (calculating the volume of hyperbolic surfaces) that had

stumped mathematicians for years.

As a professor at Stanford, Mirzakhani was respected for her ability to

creatively mix maths theories with a willingness to pursue seemingly

unsolvable problems. Maths, she noted, isn't that different from novel-

writing since "your problem evolves like a live character."

Mirzakhani passed away of breast cancer at just 40 years old.

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In her words… “The beauty of mathematics only shows itself to

more patient followers.” annual report | 2008

“You have to spend some

energy and effort to see the

beauty in math.”

A Tenacious Explorer of

Abstract Surfaces | august

2014

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Cubes Conjecture:

Every cube number is either a multiple of 9, or 1 more or 1 less than a multiple of 9.

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Cubes Conjecture:

Every cube number is either a multiple of 9, or 1 more or 1 less than a multiple of 9.

Proof: Each cube number is the cube of some integer n. This integer n is either a multiple of 3, or 1 more or 1 less than a multiple of 3. So these 3 cases are exhaustive.

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Cubes Proof: Each cube number is the cube of some integer n. This integer n is either a multiple of 3, or 1 more or 1 less than a multiple of 3. So these 3 cases are exhaustive:

Case 1: If n = 3p, then n³ = 27p³, which is a multiple of 9.

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Case 2: If n = 3p+1, then n³ = 27p³+27p²+9p+1, which is 1 more than a multiple of 9.

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Case 2: If n = 3p+1, then n³ = 27p³+27p²+9p+1, which is 1 more than a multiple of 9.

Case 3: If n = 3p−1, then n³ = 27p³−27p²+9p−1, which is 1 less than a multiple of 9.

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Cubes – pretty!

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Cubes – add marzipan

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Cubes

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Cubes – explode!

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Cubes – organise

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In her words… “It's invaluable to have a friend who shares your

interests and helps you stay motivated.” Interview

with Research Fellow Maryam Mirzakhani | 2008

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Team task Prove that n3+2n is a multiple of 3 for all integer

values of n.

Can you prove it a different way?

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In her words…

“I like crossing the imaginary boundaries people

set up between different fields—it's very

refreshing. There are lots of tools, and you don't

know which one would work. It's about being

optimistic and trying to connect things.”

A Tenacious

Explorer of Abstract

Surfaces | august

2014

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KS3 or A level? Prove that 0. 3 45 =

115

333

Do you:

Just ‘know’ it’s 345

999 and cancel down because it’s

a 3-digit recurring decimal?

Make 𝑥 = 0. 3 45 and multiply by 1000?

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115

333

Can you use this as an example of an A level

technique?

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115

333

0. 3 45 = 0.345 + 0.000345 + 0.000000345 +…

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115

333

0. 3 45 = 0.345 + 0.000345 + 0.000000345 +…

= 0.345 + 0. 345 ×1

1000+ 0. 345 ×

1

1000

2

+…

𝑎 = 0.345, 𝑟 =1

1000

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115

333

0. 3 45 = 0.345 + 0.000345 + 0.000000345 +…

= 0.345 + 0. 345 ×1

1000+ 0. 345 ×

1

1000

2

+…

𝑎 = 0.345, 𝑟 =1

1000

𝑆∞ =𝑎

1 − 𝑟=

0.345

1 −1

1000

=0.345

9991000

=345

999=115

333

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Side note… 𝑆∞ =

115

333

𝑆∞ =115

333 Q.E.D.

𝑆∞ =115

333 WWWWW

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Side note… 𝑆∞ =

115

333 innit

𝑆∞ =115

333 Boo-yah

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In her words… “There are different characters, and you are

getting to know them better. Things evolve, and

then you look back at a character, and it’s

completely different from your first impression.”

Meet the First Woman to Win

Math’s Most Prestigious Prize

August 2014

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Martian coinage Only two coins permitted, 3 marv and 5 marv.

Starbucks can only charge integer prices.

Prove that all bills above 7 marv can be paid

using combinations of only these two coins.

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Martian coinage Step 1. It is easy to see how to make 8 from

3 and 5 coins.

Step 2. Assume k can be made where k ≥ 8.

There are two cases:

Case 1: k uses at least one 5 coin. Then

replace this with two 3 coins to make a new

total of (k+1).

Case 2:

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Martian coinage Step 1. It is easy to see how to make 8 from

3 and 5 coins.

Step 2. Assume k can be made where k ≥ 8.

There are two cases:

Case 1: k uses at least one 5 coin. Then

replace this with two 3 coins to make a new

total of (k+1).

Case 2: k uses only 3 coins. Then replace

three 3 coins (possible since k>7) with two

5 coins to make a new total of (k+1).

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In her words… “There are times when I feel like I'm in a big forest and

don't know where I'm going. But then somehow I come to

the top of a hill and can see everything more clearly. When

that happens it's really exciting.”

Brilliant 10: Princeton University ~ Maryam Mirzakhani |

october 2005

“I don't have any particular recipe… Doing research is

challenging as well as attractive. It is like being lost in a

jungle and trying to use all the knowledge that you can

gather to come up with some new tricks, and with some

luck you might find a way out.”

Stanford's Maryam Mirzakhani Wins Field's Medal | august

2014

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Pythagorean triple threat Proposition:

If x, y, z are non-zero integers such that x2+ y2 =z2

then xyz is divisible by 60.

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…this is a BIG solve.

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Maryam’s Forest

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When 60 is factorised into primes, we have

60=2235. Thus, because prime factorisation is

unique, we need to prove that those factors appear

in the factors which make up the numbers x, y and

z.

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60=2235

So we have to show that

(a) at least one of x, y and z is divisible by 3; and

(b) at least one of x, y and z is divisible by 5; and…

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60=2235

So we have to show that

(a) at least one of x, y and z is divisible by 3; and

(b) at least one of x, y and z is divisible by 5; and…

(c) either one of x, y, z is divisible by 4 or at least

two of them are even.

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Pick a path to try If x, y, z are non-zero integers such that x2+ y2 =z2


(a) at least one of x, y and z is divisible by 3

Suppose none of x, y and z is divisible by 3.

An integer p is not divisible by 3 when it has one of

two forms, namely:

p = 3m+1 or p = 3m+2, where m is an integer.

That is another way of saying that when p is

divided by 3 it has remainder 1 or 2.

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Follow the path If x, y, z are non-zero integers such that x2+ y2 =z2


(a) at least one of x, y and z is divisible by 3

Now we can see that in turn

p2 = 9m2 + 6m + 1 = 3(3m2 + 2m) + 1

i.e. has remainder 1 on division by 3; or

p2 = 9m2 + 12m + 4 = 3(3m2 + 4m + 1) + 1

i.e. has remainder 1 on division by 3.

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Use efficient notation If x, y, z are non-zero integers such that x2+ y2 =z2 then xyz is divisible by 60. (a) at least one of x, y and z is divisible by 3

We can usefully use a notation [0], [1] and [2] to

stand for the collection of numbers which have

remainder 0, 1 or 2 when divided by 3 respectively:

So [0] = {…, -6, -3, 0, 3, 6,…..}

[1] = {…, -5, -2, 1, 4, 7,…..}

[2] = {…, -4, -1, 2, 5, 8,…..}

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Pause and check you’re still on the path

If x, y, z are non-zero integers such that x2+ y2 =z2 then xyz is divisible by 60. (a) at least one of x, y and z is divisible by 3

What we have just shown is that [1]2 = [1] and also that [2]2 = [1]. Similarly [0]2 = [0].

We can easily see that

[0] + [0] = [0] i.e. two multiples of 3 add to a multiple of 3;

[0] + [1] = [1] i.e. a multiple of 3 plus an integer with remainder 1 on division by 3 add to a another integer which has remainder 1.

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Ta-dah! Now if we return to x2 + y2 = z2, we can see by trial

that if none of x, y or z is divisible by 3 we are

implying that x, y and z is each [1] or [2], which

both square to [1] so we are saying that

[1] + [1] = [1],

which is simply not true – a contradiction.

Hence we deduce that at least one of x, y and z is

divisible by 3. This is the only conclusion possible

since it is the only other possibility if our

assumption that none was a multiple of 3 leads to

a contradiction.

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Maryam’s Forest

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In her words… “I think it's rarely about what you actually learn in

class . . . it's mostly about things that you stay

motivated to go and continue to do on your own.”

Maryam Mirzakhani press conference after

winning Field's Medal | august 2014

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Jo Sibley

07477 778533

[email protected]

amsp.org.uk

Advanced_Maths (AMSP)

@JusSumChick (me) More comics

than maths!

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