Big mistakes, big questionsjma/BigMistake.pdf · Big mistakes, big questions: or how to think like...

Big mistakes, big questions:or how to think like a mathematician

Justin AllmanDuke University

Governor’s School WestMathematics

27 June 2015

Justin Allman Duke University Big mistakes, big questions:

The worst mistake ever made

Picture yourself in high-school algebra class.

You see the expression1

x ` y.

You write1

THIS IS FALSE AND ALSO THE WORST.

x ` y.

You write1

x ` y.

You write1

x ` y.

You write1

x ` y“

x ` y.

You write1

x ` y“

What does false mean in mathematics?

On the previous slide, false means that 1x`y is not always equal to

For example, set x “ 2 and y “ 3. Then

x ` y“

The above is called a counterexample.

If there exists even one counterexample to a statement, then it isconsidered false.

x ` y“

So now we can ask...

Lesson (Think Like a Mathematician Lesson #1)

Turn mistakes into interesting questions!

Question

Are there values of x and y for which the equation

x ` y“

actually holds?

Let’s investigate!

For the moment let’s assume that x and y must be real numbers (asis the case in most high-school algebra and calculus classes).

Question

x ` y“

actually holds?

Question

x ` y“

actually holds?

Question

x ` y“

actually holds?

Algebra time

Start with our equation:

x ` y“

Algebra time

Get a common denominator!

x ` y“

Algebra time

x ` y“

taking reciprocals gives the equation

x ` y “xy

Algebra time

x ` y“

x ` y “xy

and this implies the polynomial equation

px ` yq2 “ xy

Algebra time

x ` y“

x ` y “xy

and this implies the polynomial equation

px ` yq2 “ xy

where we must remember to exclude both x “ 0 and y “ 0.

More algebra

The equation

px ` yq2 “ xy

implies that the product of x and y is positive!

On the other hand the box above also implies that

xy “ px ` yq2 “ x2 ` 2xy ` y2.

Therefore

´xy “ x2 ` y2

and so the product of x and y must also be negative!

There is only one real number which can be both positive andnegative,

and zero must be excluded since neither x nor y areallowed to be zero.

More algebra

The equation

px ` yq2 “ xy

xy “ px ` yq2 “ x2 ` 2xy ` y2.

Therefore

´xy “ x2 ` y2

More algebra

The equation

px ` yq2 “ xy

xy “ px ` yq2 “ x2 ` 2xy ` y2.

Therefore

´xy “ x2 ` y2

More algebra

The equation

px ` yq2 “ xy

xy “ px ` yq2 “ x2 ` 2xy ` y2.

Therefore

´xy “ x2 ` y2

More algebra

The equation

px ` yq2 “ xy

xy “ px ` yq2 “ x2 ` 2xy ` y2.

Therefore

´xy “ x2 ` y2

There is only one real number which can be both positive andnegative, and zero must be excluded since neither x nor y areallowed to be zero.

Conclusion

There are exactly NO pairs of real numbers px , yq for which the equation

x ` y“

holds.

Allow negative results to generate more interesting questions.

What if we allow more numbers?

In particular, what if x and y can be complex numbers?

Conclusion

x ` y“

holds.

Conclusion

x ` y“

holds.

Conclusion

x ` y“

holds.

Primer on complex numbers

Let i denote the “imaginary number”?´1. Notice that i2 “ ´1.

As a set, the complex numbers are

C “ ta` bi : a P R, b P Ru

where R denotes the set of real numbers.

The complex numbers form a plane.

Example

Figure : The complex number 1`?3i , i.e. a “ 1, b “

Example

Figure : What is the length of the blue line?

Example

length of blue line is r “ 2;

length of angle between positive a-axis and r is θ “ π{3

Example

length of blue line is r “ 2;

length of angle between positive a-axis and r is θ “ π{3

The number r is called the modulus

The angle θ is called the argument

Using kindergarten trigonometry, we can write that 1 “ 2 cospπ{3qand

?3 “ 2 sinpπ{3q

Therefore 1`?

3i “ 2pcospπ{3q ` i sinpπ{3qq

And now an amazing formula:

Theorem (Euler’s Identity)

re iθ “ rpcos θ ` i sin θq

So the number we’ve been considering can be succinctly written as

2e iπ{3

?3 “ 2 sinpπ{3q

Therefore 1`?

2e iπ{3

?3 “ 2 sinpπ{3q

Therefore 1`?

2e iπ{3

?3 “ 2 sinpπ{3q

Therefore 1`?

2e iπ{3

?3 “ 2 sinpπ{3q

Therefore 1`?

2e iπ{3

?3 “ 2 sinpπ{3q

Therefore 1`?

2e iπ{3

?3 “ 2 sinpπ{3q

Therefore 1`?

2e iπ{3

Our problem

x ` y“

This implied that

px ` yq2 “ xy

Which can be rewritten as

x2 ` xy ` y2 “ 0

Using an algebraic trick, we have that

0 “ px ´ yqpx2 ` xy ` y2q .

Our problem

x ` y“

This implied that

px ` yq2 “ xy

x2 ` xy ` y2 “ 0

0 “ px ´ yqpx2 ` xy ` y2q .

Our problem

x ` y“

This implied that

px ` yq2 “ xy

x2 ` xy ` y2 “ 0

0 “ px ´ yqpx2 ` xy ` y2q .

Our problem

x ` y“

This implied that

px ` yq2 “ xy

x2 ` xy ` y2 “ 0

0 “ px ´ yqpx2 ` xy ` y2q .

Our problem

x ` y“

This implied that

px ` yq2 “ xy

x2 ` xy ` y2 “ 0

0 “ px ´ yqpx2 ` xy ` y2q“ x3 ´ y3.

Our problem

From the equation

x3 ´ y3 “ 0

we see that we must have x3 “ y3.

However, we are not permitted to have x “ y . Why not?

If so, the equation 0 “ x2 ` xy ` y2 becomes

0 “ x2 ` xpxq ` x2 “ 3x2

and therefore x “ 0 (and so also y “ 0).

Our problem

From the equation

x3 ´ y3 “ 0

0 “ x2 ` xpxq ` x2 “ 3x2

Our problem

From the equation

x3 ´ y3 “ 0

0 “ x2 ` xpxq ` x2 “ 3x2

Our problem

From the equation

x3 ´ y3 “ 0

0 “ x2 ` xpxq ` x2 “ 3x2

Our problem

What does it mean that x3 “ y3 but x ‰ y?

In words: x and y must be cube roots of the same number, butmust be distinct from each other.

Now we ask:

Question

Given a complex number z how does one find the cube roots of z?

Our problem

Now we ask:

Question

Our problem

Now we ask:

Question

Our problem

Now we ask:

Question

Cube roots

Let’s find the cube roots of 1.

Any cube root of 1 must be a root of the polynomial

t3 ´ 1

Recall that t3 ´ 1 “ pt ´ 1qpt2 ` t ` 1q

So t “ 1 is a cube root of 1.

There are two others! The complex roots of the quadratic t2` t ` 1.

The quadratic formula gives the answers:

t “´1`

2t “

´1´?´3

Cube roots

t3 ´ 1

t “´1`

2t “

´1´?´3

Cube roots

t3 ´ 1

t “´1`

2t “

´1´?´3

Cube roots

t3 ´ 1

t “´1`

2t “

´1´?´3

Cube roots

t3 ´ 1

t “´1`

2t “

´1´?´3

Cube roots

t3 ´ 1

t “´1`

2t “

´1´?´3

Cube roots

t3 ´ 1

t “ ´1

2i t “ ´

Cube roots

t3 ´ 1

t “ e2πi{3 t “ e4πi{3

Picture

Figure : The three cube roots of 1 in the complex plane. Picturedcounter-clockwise from the real axis: 1, e2πi{3, e4πi{3.

Picture

Figure : The three cube roots of 1 in the complex plane. Picturedcounter-clockwise from the real axis: 1, e2πi{3, e4πi{3.

Complex Arithmetic ðñ Plane Geometry

Multiplication by the number e iθ corresponds to rotation about theorigin by an angle of θ.

Multiplication by the re iθ corresponds to

dilating the distance from the origin by a factor of r ANDrotating about the origin by an angle of θ.

Notice that 1 “ 1 ¨ e i ¨0 so multiplication by 1 neitherstretches/shrinks nor rotates.

dilating the distance from the origin by a factor of r AND

rotating about the origin by an angle of θ.

Another picture

Figure : The three cube roots of 4` 3i in the complex plane. Notice that4` 3i “ 5e iθ where θ “ arctanp3{4q. What is the radius of the circle picturedabove?

Conclusion

By investigating our epic mistake we discovered...

Euler’s identity

how to factor a difference of two cubes

that the complex numbers link algebra and arithmetic to thegeometry of the plane

there is apparent symmetry in the roots of polynomial equations

There is beautiful mathematics in all kinds of places. One must simplyask the right questions.

Try to squeeze more juice out of the turnip. That is, change the problemslightly and see what else you can learn.

Conclusion

Euler’s identity

Conclusion

Euler’s identity

Conclusion

Euler’s identity

Conclusion

Euler’s identity

Conclusion

Euler’s identity

Conclusion

Euler’s identity

Another application of complex numbers

Consider the equationx2 “ y3

where x and y are complex numbers.

Here is a “real” picture:

Figure : The real curve x2“ y 3

Describe the singularity

The curve x2 “ y3 has a singularity at the origin.

Now write x “ re iθ and y “ ρe iφ where r and ρ are both positivereal numbers.

Then we have r2e2iθ “ ρ3e3iφ

Since equal complex numbers must have equal moduli, this impliesr2 “ ρ3.

Hence r “ ρ “ 1 or r “ ρ “ 0. (We already know about the “zero”case)

This means that x and y both live on circles of radius one in thecomplex plane.

This implies that any solution to x2 “ y3 actually lives on ageometric object which has the form of a Cartesian product:

Circleˆ Circle

Can you identify the boxed object?

Circleˆ Circle

Torus = Circle ˆ Circle

Our solution set lives on a torus!

Let’s finish the “argument”

Recall the condition: r2e2iθ “ ρ3e3iφ

Since equal complex numbers must have equal arguments (modulo2π) we see that 2θ ” 3φ

So our solution has the property that as it winds around the torus(say) in the “red” direction twice, it winds around in the “blue”direction three times.

Mathematica break...

Torus Knots

We just discovered the p2, 3q´torus knot aka the trefoil.

One can consider the pp, qq´torus knot by looking at solutions toxp “ yq

It turns out that to every singularity one can associate a knot (ormore generally a link). This is one way to measure the topologicalcomplexity of a singularity and is still an area of active mathematicalresearch.

Torus Knots

The end

Thank you!

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