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Symmetry Anyone?

Willy Hereman

After Dinner Talk

SANUM 2008 Conference

Thursday, March 27, 2007, 9:00p.m.

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Tribute to Organizers

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Karin Hunter

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Karin Hunter

Andre Weideman

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Karin Hunter

Andre Weideman

Ben Herbst

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Karin Hunter

Andre Weideman

Ben Herbst

Dirk Laurie

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Karin Hunter

Andre Weideman

Ben Herbst

Dirk Laurie

Stefan van der Walt

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Karin Hunter

Andre Weideman

Ben Herbst

Dirk Laurie

Stefan van der Walt

Neil Muller

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Karin Hunter

Andre Weideman

Ben Herbst

Dirk Laurie

Stefan van der Walt

Neil Muller

and the Support Staff

Behind the Scenes

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A Big THANK YOU !

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Outline

• Symmetry Surrounding Us

• What is Symmetry?

• Father – Daughter Puzzle

• Steiner Trees – Puzzle

• Solving Quadratic, Cubic, Quartic Equations

• The Quintic and the French Revolutionary

• The Seven-Eleven Puzzle

• Modern Applications

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Symmetry Surrounding Us

• Ask people on US campus

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Symmetry Surrounding Us

• Ask people on US campus

• Ask an architect or artist

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Symmetry Surrounding Us

• Ask people on US campus

• Ask an architect or artist

• “Madam, I’m Adam”

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Symmetry Surrounding Us

• Ask people on US campus

• Ask an architect or artist

• “Madam, I’m Adam”

• Ask a mathematician

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What is Symmetry?

• It is all about transformations

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What is Symmetry?

• It is all about transformations

• Wigner: “... the unreasonable effectiveness of

mathematics in the natural sciences”

• “... the unreasonable effectiveness of using

symmetries in mathematics...”

• A simple example: father-daughter puzzle

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Father – Daughter Puzzle

Today, the ages of a father and his daughter add up

to 40 years. Five years from now, the father will be 4

times older than his daughter. How old is each today?

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Father – Daughter Puzzle

Today, the ages of a father and his daughter add up

to 40 years. Five years from now, the father will be 4

times older than his daughter. How old is each today?

Solution: F : age of the father (today)

D : age of the daughter (today)

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Father – Daughter Puzzle

Today, the ages of a father and his daughter add up

to 40 years. Five years from now, the father will be 4

times older than his daughter. How old is each today?

Solution: F : age of the father (today)

D : age of the daughter (today)

Then, F + D = 40

F + 5 = 4 (D + 5)

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Father – Daughter Puzzle

Today, the ages of a father and his daughter add up

to 40 years. Five years from now, the father will be 4

times older than his daughter. How old is each today?

Solution: F : age of the father (today)

D : age of the daughter (today)

Then, F + D = 40

F + 5 = 4 (D + 5)

Eliminate F = 40−D and solve

45−D = 4 (D + 5) or 5D = 25

Hence, D = 5 and F = 40−D = 35.

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Symmetry Reduces Complexity

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Symmetry Reduces Complexity

Solution: 20 + x : age of the father (today)

20− x : age of the daughter (today)

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Symmetry Reduces Complexity

Solution: 20 + x : age of the father (today)

20− x : age of the daughter (today)

Then,

25 + x = 4 (25− x) or 5x = 75

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Symmetry Reduces Complexity

Solution: 20 + x : age of the father (today)

20− x : age of the daughter (today)

Then,

25 + x = 4 (25− x) or 5x = 75

Hence, x = 15.

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Symmetry Reduces Complexity

Solution: 20 + x : age of the father (today)

20− x : age of the daughter (today)

Then,

25 + x = 4 (25− x) or 5x = 75

Hence, x = 15.

So, father is 20 + x = 35, daughter is 20− x = 5.

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Steiner Trees

Connecting Cities with Shortest Road System

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D

Three cities in equilateral triangle

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L = 1.87 D

One choice of a road system

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L = 1.73 D

Shortest road system connecting three cities

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D

Four cities in a square

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L = 2.83 D

One choice of a road system

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L = 2.73 D

Shortest road system connecting 4 cities

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Solving Quadratic, Cubic, Quartic Equations

• Quadratic: ax2 + bx + c = 0

Babylonians (400 BC)

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Solving Quadratic, Cubic, Quartic Equations

• Quadratic: ax2 + bx + c = 0

Babylonians (400 BC)

• Cubic: ax3 + bx2 + cx + d = 0

Italians (1525-1545): dal Ferro & Fior, Tartaglia

& Cardano

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Solving Quadratic, Cubic, Quartic Equations

• Quadratic: ax2 + bx + c = 0

Babylonians (400 BC)

• Cubic: ax3 + bx2 + cx + d = 0

Italians (1525-1545): dal Ferro & Fior, Tartaglia

& Cardano

• Quartic: ax4 + bx3 + cx2 + dx + e = 0

Cardano & Ferrari

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Solving Quadratic, Cubic, Quartic Equations

• Quadratic: ax2 + bx + c = 0

Babylonians (400 BC)

• Cubic: ax3 + bx2 + cx + d = 0

Italians (1525-1545): dal Ferro & Fior, Tartaglia

& Cardano

• Quartic: ax4 + bx3 + cx2 + dx + e = 0

Cardano & Ferrari

• Quintic: ax5 + bx4 + cx3 + dx2 + ex + f = 0

The equation that couldn’t be solved!

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Challenge Mathematica to Solve the

Quadratic, Cubic, Quartic, Quintic Equations

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The Quintic and the French Revolutionary

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The Quintic and the French Revolutionary

• Evariste Galois (1830): inventor of group theory,

which quintic equations can (cannot) be solved

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The Quintic and the French Revolutionary

• Evariste Galois (1830): inventor of group theory,

which quintic equations can (cannot) be solved

• Niels Hendrik Abel (1821): general quintic

equation can not be solved analytically

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The Quintic and the French Revolutionary

• Evariste Galois (1830): inventor of group theory,

which quintic equations can (cannot) be solved

• Niels Hendrik Abel (1821): general quintic

equation can not be solved analytically

• Joseph Liouville, Camille Jordan, Felix Klein,

Sophus Lie, ....

The Seven Eleven Puzzle

The sum and product of the prices of four items

is R 7.11

The Seven Eleven Puzzle

The sum and product of the prices of four items

is R 7.11

Hence,

x + y + z + w = 7.11

x y z w = 7.11

The Seven Eleven Puzzle

The sum and product of the prices of four items

is R 7.11

Hence,

x + y + z + w = 7.11

x y z w = 7.11

Solution:

The Seven Eleven Puzzle

The sum and product of the prices of four items

is R 7.11

Hence,

x + y + z + w = 7.11

x y z w = 7.11

Solution:

Prices: 1.20 1.25 1.50 3.16

One Solution Strategy :

• Convert to integer problem

x + y + z + w = 711

x y z w = 711 ∗ 106

One Solution Strategy :

• Convert to integer problem

x + y + z + w = 711

x y z w = 711 ∗ 106

• Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79

One Solution Strategy :

• Convert to integer problem

x + y + z + w = 711

x y z w = 711 ∗ 106

• Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79

• Thus, x = 79 ∗ 2a1 ∗ 3b1 ∗ 5c1 y = 2a2 ∗ 5c2

z = 2a3 ∗ 3b3 ∗ 5c3 w = 2a4 ∗ 3b4 ∗ 5c4

One Solution Strategy :

• Convert to integer problem

x + y + z + w = 711

x y z w = 711 ∗ 106

• Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79

• Thus, x = 79 ∗ 2a1 ∗ 3b1 ∗ 5c1 y = 2a2 ∗ 5c2

z = 2a3 ∗ 3b3 ∗ 5c3 w = 2a4 ∗ 3b4 ∗ 5c4

• With a1 + a2 + a3 + a4 = 6, b1 + b3 + b4 = 2,

c1 + c2 + c3 + c4 = 6.

One Solution Strategy :

• Convert to integer problem

x + y + z + w = 711

x y z w = 711 ∗ 106

• Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79

• Thus, x = 79 ∗ 2a1 ∗ 3b1 ∗ 5c1 y = 2a2 ∗ 5c2

z = 2a3 ∗ 3b3 ∗ 5c3 w = 2a4 ∗ 3b4 ∗ 5c4

• With a1 + a2 + a3 + a4 = 6, b1 + b3 + b4 = 2,

c1 + c2 + c3 + c4 = 6.

• Actually, x = n ∗ 79 with n = 1, 2, 3, 4, 5, 6, 8.

One Solution Strategy :

• Convert to integer problem

x + y + z + w = 711

x y z w = 711 ∗ 106

• Integer factors: 711 ∗ 106 = 26 ∗ 32 ∗ 56 ∗ 79

• Thus, x = 79 ∗ 2a1 ∗ 3b1 ∗ 5c1 y = 2a2 ∗ 5c2

z = 2a3 ∗ 3b3 ∗ 5c3 w = 2a4 ∗ 3b4 ∗ 5c4

• With a1 + a2 + a3 + a4 = 6, b1 + b3 + b4 = 2,

c1 + c2 + c3 + c4 = 6.

• Actually, x = n ∗ 79 with n = 1, 2, 3, 4, 5, 6, 8.

• Using y z w ≤ (y+z+w)3

27eliminates n = 5, 6, 8.

• Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237

because 711− 79 = 632, 711− 158 = 553, and

711− 237 = 474 are not five-folds (+ argument)

• Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237

because 711− 79 = 632, 711− 158 = 553, and

711− 237 = 474 are not five-folds (+ argument)

• Bingo! x = 4*79 = 316 = 79 ∗ 22 ∗ 30 ∗ 50

and y + z + w = 395

• Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237

because 711− 79 = 632, 711− 158 = 553, and

711− 237 = 474 are not five-folds (+ argument)

• Bingo! x = 4*79 = 316 = 79 ∗ 22 ∗ 30 ∗ 50

and y + z + w = 395

• So, at least one number is a five-fold: either only

one number is (excluded), or all three are

• Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237

because 711− 79 = 632, 711− 158 = 553, and

711− 237 = 474 are not five-folds (+ argument)

• Bingo! x = 4*79 = 316 = 79 ∗ 22 ∗ 30 ∗ 50

and y + z + w = 395

• So, at least one number is a five-fold: either only

one number is (excluded), or all three are

• Then, y = 5y′, z = 5z′, w = 5w′ and y′ + z′ + w′ = 79

y′ z′ w′ = 18000

• Reject x = 79, x = 2 ∗ 79 = 158, and x = 3 ∗ 79 = 237

because 711− 79 = 632, 711− 158 = 553, and

711− 237 = 474 are not five-folds (+ argument)

• Bingo! x = 4*79 = 316 = 79 ∗ 22 ∗ 30 ∗ 50

and y + z + w = 395

• So, at least one number is a five-fold: either only

one number is (excluded), or all three are

• Then, y = 5y′, z = 5z′, w = 5w′ and y′ + z′ + w′ = 79

y′ z′ w′ = 18000

• Not all three are five folds.

A single one cannot be a five fold (125 > 79)

• So, one must be a multiple of 25; the other

multiple of 5: y′ = 25y′′, z′ = 5z′′.

• Then, 25y′′ + 5z′′ + w′ = 79 and y′′ z′′ w′ = 144

• So, one must be a multiple of 25; the other

multiple of 5: y′ = 25y′′, z′ = 5z′′.

• Then, 25y′′ + 5z′′ + w′ = 79 and y′′ z′′ w′ = 144

• So, y′′ = 1 or 2

• So, one must be a multiple of 25; the other

multiple of 5: y′ = 25y′′, z′ = 5z′′.

• Then, 25y′′ + 5z′′ + w′ = 79 and y′′ z′′ w′ = 144

• So, y′′ = 1 or 2

• Test either case... conclude that y′′ = 2 is

impossible.

• So, one must be a multiple of 25; the other

multiple of 5: y′ = 25y′′, z′ = 5z′′.

• Then, 25y′′ + 5z′′ + w′ = 79 and y′′ z′′ w′ = 144

• So, y′′ = 1 or 2

• Test either case... conclude that y′′ = 2 is

impossible.

• Then, y′′ = 1. Bingo! y = 125

• So, one must be a multiple of 25; the other

multiple of 5: y′ = 25y′′, z′ = 5z′′.

• Then, 25y′′ + 5z′′ + w′ = 79 and y′′ z′′ w′ = 144

• So, y′′ = 1 or 2

• Test either case... conclude that y′′ = 2 is

impossible.

• Then, y′′ = 1. Bingo! y = 125

• Finally, we must solve 5z′′ + w′ = 54 and z′′w′ = 144

• So, one must be a multiple of 25; the other

multiple of 5: y′ = 25y′′, z′ = 5z′′.

• Then, 25y′′ + 5z′′ + w′ = 79 and y′′ z′′ w′ = 144

• So, y′′ = 1 or 2

• Test either case... conclude that y′′ = 2 is

impossible.

• Then, y′′ = 1. Bingo! y = 125

• Finally, we must solve 5z′′ + w′ = 54 and z′′w′ = 144

• Solve a quadratic equation: z′′ = 6, w′ = 24

• So, one must be a multiple of 25; the other

multiple of 5: y′ = 25y′′, z′ = 5z′′.

• Then, 25y′′ + 5z′′ + w′ = 79 and y′′ z′′ w′ = 144

• So, y′′ = 1 or 2

• Test either case... conclude that y′′ = 2 is

impossible.

• Then, y′′ = 1. Bingo! y = 125

• Finally, we must solve 5z′′ + w′ = 54 and z′′w′ = 144

• Solve a quadratic equation: z′′ = 6, w′ = 24

• Summary:

x = 316, y = 125, z = 25 ∗ 6 = 150, w = 5 ∗ 24 = 120

• Solution:

x = 316 = 79 ∗ 22 y = 125 = 53

z = 150 = 2 ∗ 3 ∗ 52 w = 120 = 23 ∗ 3 ∗ 5

• Solution:

x = 316 = 79 ∗ 22 y = 125 = 53

z = 150 = 2 ∗ 3 ∗ 52 w = 120 = 23 ∗ 3 ∗ 5

• Prices: 1.20 1.25 1.50 3.16

• Solution:

x = 316 = 79 ∗ 22 y = 125 = 53

z = 150 = 2 ∗ 3 ∗ 52 w = 120 = 23 ∗ 3 ∗ 5

• Prices: 1.20 1.25 1.50 3.16

• Challenge: Can the 7-11 puzzle be solved using

symmetries (group theory)?

Modern Applications

• Maxwell’s equations: merging electricity with

magnetism

Modern Applications

• Maxwell’s equations: merging electricity with

magnetism

• Einstein: general relativity

Modern Applications

• Maxwell’s equations: merging electricity with

magnetism

• Einstein: general relativity

• Merging general relativity and quantum mechanics

Modern Applications

• Maxwell’s equations: merging electricity with

magnetism

• Einstein: general relativity

• Merging general relativity and quantum mechanics

• String theory, super string theory

Modern Applications

• Maxwell’s equations: merging electricity with

magnetism

• Einstein: general relativity

• Merging general relativity and quantum mechanics

• String theory, super string theory

• A theory for everything

Literature

• Ian Stewart, Why Beauty is Truth:

The History of Symmetry, Basic Books, The

Perseus Books Group, April 2007, 290 pages.

Podcast series, University of Warwick, 2007

(7 episodes, ∼ 90 minutes total)

• Mario Livio, The Equation That Couldn’t Be

Solved, Simon & Schuster, 2005, 368 pages.

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Thank You!