π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Theπ story of and related puzzles

Narrator: Niraj Khare Carnegie Mellon University Qatar

Being with math is being with the truth and eternity!

Nov, 15, 2017 1 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Time line III (a): series expressions for π

• Ludolph Van Ceulen using archimedean method with 500 million sides calculated π calculated π to an accuracy of 20 decimal digits by 1596. By the time he died in 1610, he accurately found 35 digits! The digits were carved into his tombstone.

2 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Time line III (a): series expressions for π

• Ludolph Van Ceulen using archimedean method with 500 million sides calculated π calculated π to an accuracy of 20 decimal digits by 1596. By the time he died in 1610, he accurately found 35 digits! The digits were carved into his tombstone.

2 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Ludolph Van Ceulen: 3.14159265358979323846264338327950288...

Ludolph van Ceulen Dutch-German mathematician Ludolph van Ceulen was a German-Dutch mathematician from Hildesheim. He emigrated to the Netherlands. Wikipedia: Born: January 28, 1540, Hildesheim, Germany Died: December 31, 1610, Leiden, Netherlands Known for: pi Institution: Leiden University Notable student: Willebrord Snellius

3 / 33 • Around 250 BC, Archimedes proves that 10 223 220 1 3.1408 < 3 71 = 71 < π < 70 = 3 7 ≈ 3.1428.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Time line I: Ancient period

• The story starts in ancient Egypt and Babylon about 4000 years ago! • Around 450 BCE, Anaxagoras proposes ‘squaring the circle’ from a prison! The puzzle was finally ‘settled’ in 1882 AD.

4 / 33 • The story starts in ancient Egypt and Babylon about 4000 years ago!

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Time line I: Ancient period

• Around 450 BCE, Anaxagoras proposes ‘squaring the circle’ from a prison! The puzzle was finally ‘settled’ in 1882 AD. • Around 250 BC, Archimedes proves that 10 223 220 1 3.1408 < 3 71 = 71 < π < 70 = 3 7 ≈ 3.1428.

4 / 33 • The story starts in ancient Egypt and Babylon about 4000 years ago! • Around 450 BCE, Anaxagoras proposes ‘squaring the circle’ from a prison! The puzzle was finally ‘settled’ in 1882 AD.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Time line I: Ancient period

• Around 250 BC, Archimedes proves that 10 223 220 1 3.1408 < 3 71 = 71 < π < 70 = 3 7 ≈ 3.1428.

4 / 33 Definition A rational number is a ratio of two integers. In other words, a number q is a rational if there are integers a and b 6= 0 such a that q = b .

31 3 −5 Examples: 3.1 = 10 , 3 = 1 , −5 = 1 , 3.127127127 ··· 127

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A rational number

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples:{· · · , −3, −2, −1, 0, 1, 2, 3, ···}.

5 / 33 In other words, a number q is a rational if there are integers a and b 6= 0 such a that q = b .

31 3 −5 Examples: 3.1 = 10 , 3 = 1 , −5 = 1 , 3.127127127 ··· 127

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A rational number

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples:{· · · , −3, −2, −1, 0, 1, 2, 3, ···}. Definition A rational number is a ratio of two integers.

5 / 33 3 −5 3 = 1 , −5 = 1 , 3.127127127 ··· 127

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A rational number

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples:{· · · , −3, −2, −1, 0, 1, 2, 3, ···}. Definition A rational number is a ratio of two integers. In other words, a number q is a rational if there are integers a and b 6= 0 such a that q = b .

31 Examples: 3.1 = 10 ,

5 / 33 −5 −5 = 1 , 3.127127127 ··· 127

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A rational number

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples:{· · · , −3, −2, −1, 0, 1, 2, 3, ···}. Definition A rational number is a ratio of two integers. In other words, a number q is a rational if there are integers a and b 6= 0 such a that q = b .

31 3 Examples: 3.1 = 10 , 3 = 1 ,

5 / 33 3.127127127 ··· 127

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A rational number

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples:{· · · , −3, −2, −1, 0, 1, 2, 3, ···}. Definition A rational number is a ratio of two integers. In other words, a number q is a rational if there are integers a and b 6= 0 such a that q = b .

31 3 −5 Examples: 3.1 = 10 , 3 = 1 , −5 = 1 ,

5 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A rational number

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples:{· · · , −3, −2, −1, 0, 1, 2, 3, ···}. Definition A rational number is a ratio of two integers. In other words, a number q is a rational if there are integers a and b 6= 0 such a that q = b .

31 3 −5 Examples: 3.1 = 10 , 3 = 1 , −5 = 1 , 3.127127127 ··· 127

5 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A rational number

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples:{· · · , −3, −2, −1, 0, 1, 2, 3, ···}. Definition A rational number is a ratio of two integers. In other words, a number q is a rational if there are integers a and b 6= 0 such a that q = b .

31 3 −5 Examples: 3.1 = 10 , 3 = 1 , −5 = 1 , 3.127127127 ··· 127

5 / 33 ⇒ 1000x − x = 3127 − 3 ⇒ 999x = 3124 3124 ⇒ x = . 999

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof 3.127127127 ··· 127 is rational!

Proof. Let x = 3.127127127 ··· 127.

1000x = 3127.127127 ··· 127 x = 3.127127127 ··· 127

6 / 33 3124 ⇒ x = . 999

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof 3.127127127 ··· 127 is rational!

Proof. Let x = 3.127127127 ··· 127.

1000x = 3127.127127 ··· 127 x = 3.127127127 ··· 127 ⇒ 1000x − x = 3127 − 3 ⇒ 999x = 3124

6 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof 3.127127127 ··· 127 is rational!

Proof. Let x = 3.127127127 ··· 127.

1000x = 3127.127127 ··· 127 x = 3.127127127 ··· 127 ⇒ 1000x − x = 3127 − 3 ⇒ 999x = 3124 3124 ⇒ x = . 999

6 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof 3.127127127 ··· 127 is rational!

Proof. Let x = 3.127127127 ··· 127.

1000x = 3127.127127 ··· 127 x = 3.127127127 ··· 127 ⇒ 1000x − x = 3127 − 3 ⇒ 999x = 3124 3124 ⇒ x = . 999

6 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Not everything is rational!

Hippasus of Metapontum (/hpss/; Greek: , Hppasos; fl. 3rd century BC), was a Pythagorean philosopher. He was the first one to claim that there were irrational numbers!

7 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Not everything is rational!

Hippasus of Metapontum (/hpss/; Greek: , Hppasos; fl. 3rd century BC), was a Pythagorean philosopher. He was the first one to claim that there were irrational numbers!

7 / 33 but not both.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Quotes

• The oldest, shortest words “yes” and “no” are those which require the most thought. - Pythagoras • A statement or a proposition in mathematics is a sentence that is either true or false

8 / 33 • The oldest, shortest words “yes” and “no” are those which require the most thought. - Pythagoras

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Quotes

• A statement or a proposition in mathematics is a sentence that is either true or false but not both.

8 / 33 • The oldest, shortest words “yes” and “no” are those which require the most thought. - Pythagoras

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Quotes

• A statement or a proposition in mathematics is a sentence that is either true or false but not both.

8 / 33 Who is the manager?

Tom: “Mary is an engineer.” Alice: “Tom is the manager.”

Mary: “Alice is the manager.” Alice: “Mary is the manager.”

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof An engineer or a manager?

Tom, Mary and Alice work for ‘Logic is Fun’. Two of them are engineers and exactly one of them is a manager. The manager always lies and the engineers always speak the truth.

9 / 33 Tom: “Mary is an engineer.” Alice: “Tom is the manager.”

Mary: “Alice is the manager.” Alice: “Mary is the manager.”

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof An engineer or a manager?

Tom, Mary and Alice work for ‘Logic is Fun’. Two of them are engineers and exactly one of them is a manager. The manager always lies and the engineers always speak the truth. Who is the manager?

9 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof An engineer or a manager?

Tom, Mary and Alice work for ‘Logic is Fun’. Two of them are engineers and exactly one of them is a manager. The manager always lies and the engineers always speak the truth. Who is the manager?

Tom: “Mary is an engineer.” Alice: “Tom is the manager.”

Mary: “Alice is the manager.” Alice: “Mary is the manager.”

9 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof An engineer or a manager?

Tom, Mary and Alice work for ‘Logic is Fun’. Two of them are engineers and exactly one of them is a manager. The manager always lies and the engineers always speak the truth. Who is the manager?

Tom: “Mary is an engineer.” Alice: “Tom is the manager.”

Mary: “Alice is the manager.” Alice: “Mary is the manager.”

9 / 33 Therefore, Alice is an engineer. Thus, she always speak the truth. Hence, Tom and Mary are both managers. A contradiction! Therefore, our assumption must be false. So the negation of our assumption is true.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A proof by contradiction!

To prove: Alice is the manager. Proof. On the contrary assume that Alice is not the manager.

10 / 33 Hence, Tom and Mary are both managers. A contradiction! Therefore, our assumption must be false. So the negation of our assumption is true.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A proof by contradiction!

To prove: Alice is the manager. Proof. On the contrary assume that Alice is not the manager. Therefore, Alice is an engineer. Thus, she always speak the truth.

10 / 33 A contradiction! Therefore, our assumption must be false. So the negation of our assumption is true.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A proof by contradiction!

To prove: Alice is the manager. Proof. On the contrary assume that Alice is not the manager. Therefore, Alice is an engineer. Thus, she always speak the truth. Hence, Tom and Mary are both managers.

10 / 33 Therefore, our assumption must be false. So the negation of our assumption is true.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A proof by contradiction!

To prove: Alice is the manager. Proof. On the contrary assume that Alice is not the manager. Therefore, Alice is an engineer. Thus, she always speak the truth. Hence, Tom and Mary are both managers. A contradiction!

10 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A proof by contradiction!

To prove: Alice is the manager. Proof. On the contrary assume that Alice is not the manager. Therefore, Alice is an engineer. Thus, she always speak the truth. Hence, Tom and Mary are both managers. A contradiction! Therefore, our assumption must be false. So the negation of our assumption is true.

10 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A proof by contradiction!

To prove: Alice is the manager. Proof. On the contrary assume that Alice is not the manager. Therefore, Alice is an engineer. Thus, she always speak the truth. Hence, Tom and Mary are both managers. A contradiction! Therefore, our assumption must be false. So the negation of our assumption is true.

10 / 33 If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird. If Tom is not a bird then Tom cannot be a parrot.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Contrapositive

If president, then at least 35 years old.

11 / 33 If Tom is a parrot, then Tom is a bird. If Tom is not a bird then Tom cannot be a parrot.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Contrapositive

If president, then at least 35 years old. If not yet 35, then cannot be the president.

11 / 33 If Tom is a parrot, then Tom is a bird. If Tom is not a bird then Tom cannot be a parrot.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Contrapositive

If president, then at least 35 years old. If not yet 35, then cannot be the president.

11 / 33 then Tom is a bird. If Tom is not a bird then Tom cannot be a parrot.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Contrapositive

If president, then at least 35 years old. If not yet 35, then cannot be the president.

If Tom is a parrot,

11 / 33 If Tom is not a bird then Tom cannot be a parrot.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Contrapositive

If president, then at least 35 years old. If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird.

11 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Contrapositive

If president, then at least 35 years old. If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird. If Tom is not a bird then Tom cannot be a parrot.

11 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Contrapositive

If president, then at least 35 years old. If not yet 35, then cannot be the president.

If Tom is a parrot, then Tom is a bird. If Tom is not a bird then Tom cannot be a parrot.

11 / 33 • In 1761 to 1776, Lambert and Legendre proved that π is not a ratio of two integers.[Cajori, page 246] • In 1882, Ferdinand von Lindemann proved transcendence of π (i.e., squaring the circle is impossible). [Berggren, page 407]

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof History of π is irrational.

• There are conflicting claims who first ‘guessed’ that π is not a rational.But was believed by many by 5-th century AD.

12 / 33 • In 1882, Ferdinand von Lindemann proved transcendence of π (i.e., squaring the circle is impossible). [Berggren, page 407]

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof History of π is irrational.

• There are conflicting claims who first ‘guessed’ that π is not a rational.But was believed by many by 5-th century AD. • In 1761 to 1776, Lambert and Legendre proved that π is not a ratio of two integers.[Cajori, page 246]

12 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof History of π is irrational.

• There are conflicting claims who first ‘guessed’ that π is not a rational.But was believed by many by 5-th century AD. • In 1761 to 1776, Lambert and Legendre proved that π is not a ratio of two integers.[Cajori, page 246] • In 1882, Ferdinand von Lindemann proved transcendence of π (i.e., squaring the circle is impossible). [Berggren, page 407]

12 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof History of π is irrational.

• There are conflicting claims who first ‘guessed’ that π is not a rational.But was believed by many by 5-th century AD. • In 1761 to 1776, Lambert and Legendre proved that π is not a ratio of two integers.[Cajori, page 246] • In 1882, Ferdinand von Lindemann proved transcendence of π (i.e., squaring the circle is impossible). [Berggren, page 407]

12 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Lambart’s idea: I

Lemma Let x be a real number. If x is rational, then tan(x) is irrational.

13 / 33 π By previous lemma, tan( 4 ) must be an irrational. But π 1 tan( 4 ) = 1 = 1 . A contradiction!

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Lambart’s idea: II

Theorem π is irrational.

Proof. a Assume that π = b where a and b are integers. Thus, π a 4 = 4b .

14 / 33 But π 1 tan( 4 ) = 1 = 1 . A contradiction!

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Lambart’s idea: II

Theorem π is irrational.

Proof. a Assume that π = b where a and b are integers. Thus, π a π 4 = 4b .By previous lemma, tan( 4 ) must be an irrational.

14 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Lambart’s idea: II

Theorem π is irrational.

Proof. a Assume that π = b where a and b are integers. Thus, π a π 4 = 4b .By previous lemma, tan( 4 ) must be an irrational. But π 1 tan( 4 ) = 1 = 1 . A contradiction!

14 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Lambart’s idea: II

Theorem π is irrational.

Proof. a Assume that π = b where a and b are integers. Thus, π a π 4 = 4b .By previous lemma, tan( 4 ) must be an irrational. But π 1 tan( 4 ) = 1 = 1 . A contradiction!

14 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof

Theorem π is irrational.

‘Irrationality is not limited to numbers!’

15 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof On the contrary assume that π is a rational number. a Thus, π = b for integers a and b 6= 0. Without loss of generality, let a and b be both positive.

16 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof On the contrary assume that π is a rational number. a Thus, π = b for integers a and b 6= 0. Without loss of generality, let a and b be both positive.

16 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof On the contrary assume that π is a rational number. a Thus, π = b for integers a and b 6= 0. Without loss of generality, let a and b be both positive.

16 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof On the contrary assume that π is a rational number. a Thus, π = b for integers a and b 6= 0. Without loss of generality, let a and b be both positive.

16 / 33 For any positive integer n, define

bnxn(π − x)n xn(a − bx)n f(x) = = n! n!

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A bunny found!

17 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A bunny found!

For any positive integer n, define

bnxn(π − x)n xn(a − bx)n f(x) = = n! n!

17 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof A bunny found!

For any positive integer n, define

bnxn(π − x)n xn(a − bx)n f(x) = = n! n!

17 / 33 • For all non-negative integer k, f (k)(0) is an integer. Therefore, for all non-negative integer k, f (k)(π) is an integer too.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Some properties of f(x): I

• For all real x, f(x) = f(π − x). • For any non-negative integer k, f (k)(π − x) = −1kf (k)(x) where f (k)(x) denotes the k-th derivative of f with respect to x.

18 / 33 Therefore, for all non-negative integer k, f (k)(π) is an integer too.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Some properties of f(x): I

• For all real x, f(x) = f(π − x). • For any non-negative integer k, f (k)(π − x) = −1kf (k)(x) where f (k)(x) denotes the k-th derivative of f with respect to x. • For all non-negative integer k, f (k)(0) is an integer.

18 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Some properties of f(x): I

• For all real x, f(x) = f(π − x). • For any non-negative integer k, f (k)(π − x) = −1kf (k)(x) where f (k)(x) denotes the k-th derivative of f with respect to x. • For all non-negative integer k, f (k)(0) is an integer. Therefore, for all non-negative integer k, f (k)(π) is an integer too.

18 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Some properties of f(x): I

• For all real x, f(x) = f(π − x). • For any non-negative integer k, f (k)(π − x) = −1kf (k)(x) where f (k)(x) denotes the k-th derivative of f with respect to x. • For all non-negative integer k, f (k)(0) is an integer. Therefore, for all non-negative integer k, f (k)(π) is an integer too.

18 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Some properties of f(x): I

• For all real x, f(x) = f(π − x). • For any non-negative integer k, f (k)(π − x) = −1kf (k)(x) where f (k)(x) denotes the k-th derivative of f with respect to x. • For all non-negative integer k, f (k)(0) is an integer. Therefore, for all non-negative integer k, f (k)(π) is an integer too.

18 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof f (k)(0) is an integer: I

xn(a − bx)n f(x) = n! n xn X n  = a(n−i)(−1)ibixi n! i i=0   n n
a(n−i)(−1)ibixn+i X i = n! i=0

As f(x) is a polynomial of degree 2n, for all k > 2n f (k)(x) = 0.

19 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof f (k)(0) is an integer: II

f (k)(x) n i X c a(n−i)(−1) bi(n + i)(n + i − 1) ··· (n + i − {k − 1})xn+i−k = n! i=0 n
where c = i and when n + i ≥ k. When x = 0, only the term with n + i = k contributes. In particular f (k)(0) = 0 for all k < n. For k = n + i, f (k)(0) c a(n−i)(−1)ibi(k)(k − 1) ··· (k − {k − 1}) = n! c a(n−i)(−1)ibi(n + i)(n + i − 1) ··· (n) ··· (1) = n!

20 / 33 Note that g(x) + g(2)(x) = f(x) where g(2)(x) = g00(x).

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Some properties of f(x): II

Define g(x) = f(x) − f (2)(x) + f (4)(x) − f (6)(x) + ··· + (−1)kf (2k)(x) + ··· + (−1)(n−1)f (2(n−1))(x) + (−1)nf (2n)(x). Thus, g(0) and g(π) are integers.

21 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Some properties of f(x): II

Define g(x) = f(x) − f (2)(x) + f (4)(x) − f (6)(x) + ··· + (−1)kf (2k)(x) + ··· + (−1)(n−1)f (2(n−1))(x) + (−1)nf (2n)(x). Thus, g(0) and g(π) are integers. Note that g(x) + g(2)(x) = f(x) where g(2)(x) = g00(x).

21 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Some properties of f(x): II

Define g(x) = f(x) − f (2)(x) + f (4)(x) − f (6)(x) + ··· + (−1)kf (2k)(x) + ··· + (−1)(n−1)f (2(n−1))(x) + (−1)nf (2n)(x). Thus, g(0) and g(π) are integers. Note that g(x) + g(2)(x) = f(x) where g(2)(x) = g00(x).

21 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Some properties of f(x): II

Define g(x) = f(x) − f (2)(x) + f (4)(x) − f (6)(x) + ··· + (−1)kf (2k)(x) + ··· + (−1)(n−1)f (2(n−1))(x) + (−1)nf (2n)(x). Thus, g(0) and g(π) are integers. Note that g(x) + g(2)(x) = f(x) where g(2)(x) = g00(x).

21 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Anti-derivative of f(x)sin(x)

d {g0(x)sin(x) − g(x)cos(x))} dx = g00(x)sin(x) + g0(x)cos(x) − g0(x)cos(x) + g(x)sin(x) = g00(x)sin(x) + g(x)sin(x) = {g00(x) + g(x)}sin(x) = f(x)sin(x)

22 / 33 = g(π) + g(0)

R π Hence, 0 f(x)sin(x)dx is an integer for all positive integers n.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is an integer for all positive integers n.

Z π f(x)sin(x)dx 0 0 π = {g (x)sin(x) − g(x)cos(x)}|0 = {g0(π)sin(π) − g(π)cos(π)} − {g0(0)sin(0) − g(0)cos(0)}

23 / 33 R π Hence, 0 f(x)sin(x)dx is an integer for all positive integers n.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is an integer for all positive integers n.

Z π f(x)sin(x)dx 0 0 π = {g (x)sin(x) − g(x)cos(x)}|0 = {g0(π)sin(π) − g(π)cos(π)} − {g0(0)sin(0) − g(0)cos(0)} = g(π) + g(0)

23 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is an integer for all positive integers n.

Z π f(x)sin(x)dx 0 0 π = {g (x)sin(x) − g(x)cos(x)}|0 = {g0(π)sin(π) − g(π)cos(π)} − {g0(0)sin(0) − g(0)cos(0)} = g(π) + g(0)

R π Hence, 0 f(x)sin(x)dx is an integer for all positive integers n.

23 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is an integer for all positive integers n.

Z π f(x)sin(x)dx 0 0 π = {g (x)sin(x) − g(x)cos(x)}|0 = {g0(π)sin(π) − g(π)cos(π)} − {g0(0)sin(0) − g(0)cos(0)} = g(π) + g(0)

R π Hence, 0 f(x)sin(x)dx is an integer for all positive integers n.

23 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is not an integer for LARGE n.

Z π f(x)sin(x)dx 0 Z π bnxn(π − x)n = sin(x)dx 0 n! bn Z π = xn(π − x)nsin(x)dx n! 0

24 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is not an integer for LARGE n.

bn Z π 0 < xn(π − x)nsin(x)dx n! 0

25 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is not an integer for LARGE n.

bn Z π 0 < xn(π − x)nsin(x)dx n! 0

25 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof War between exponent and factorial!

bn π 2n 0 < (π) n! 2 6n y = n!

26 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof War between exponent and factorial!

bn π 2n 0 < (π) n! 2 6n y = n!

26 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof War between exponent and factorial!

bn π 2n 0 < (π) n! 2 6n y = n!

26 / 33 bn Z π 0 < xn(π − x)nsin(x)dx n! 0 bn π 2n < (π) n! 2 < 1.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is not an integer for LARGE n:III

As n → ∞, bn π 2n lim (π) = 0. n→∞ n! 2

27 / 33 bn π 2n < (π) n! 2 < 1.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is not an integer for LARGE n:III

As n → ∞, bn π 2n lim (π) = 0. n→∞ n! 2

bn Z π 0 < xn(π − x)nsin(x)dx n! 0

27 / 33 < 1.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is not an integer for LARGE n:III

As n → ∞, bn π 2n lim (π) = 0. n→∞ n! 2

bn Z π 0 < xn(π − x)nsin(x)dx n! 0 bn π 2n < (π) n! 2

27 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is not an integer for LARGE n:III

As n → ∞, bn π 2n lim (π) = 0. n→∞ n! 2

bn Z π 0 < xn(π − x)nsin(x)dx n! 0 bn π 2n < (π) n! 2 < 1.

27 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof R π 0 f(x)sin(x)dx is not an integer for LARGE n:III

As n → ∞, bn π 2n lim (π) = 0. n→∞ n! 2

bn Z π 0 < xn(π − x)nsin(x)dx n! 0 bn π 2n < (π) n! 2 < 1.

27 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Bunny or a pigeon?

For large integer n, bn R π n n • n! 0 x (π − x) sin(x)dx is an integer. • bn Z π 0 < xn(π − x)nsin(x)dx < 1. n! 0

28 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Formal and informal references: Informal References

• Documentaries: • Math and rise of civilizations: http://motionpic.com/catalogue/ math-rise-of-civilization-science-docs-documentaries-2/ • BBC: Story of Mathematics: http: //www.bbc.co.uk/programmes/b00dxjls/episodes/guide • Websites: • The history of pi by David Wilson http://sites.math.rutgers.edu/~cherlin/History/ Papers2000/wilson.html • Archimedes’ Approximation of Pi http://itech.fgcu.edu/faculty/clindsey/mhf4404/ archimedes/archimedes.html • Euclid’s Elements https://mathcs.clarku.edu/~djoyce/java/elements/ • Wekipedia: Ludolph Van Ceulen’s biography https://en.wikipedia.org/wiki/Ludolph_van_Ceulen 29 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Formal and informal references: Formal References(I)

George E. Andrews, Peter Paule. Some questions concerning computer-generated proofs of a binomial double-sum identity, J. Symbolic Comput. 16(1993), 147–151. P. Backman, The history of Pi, The Golem Press. Boulder Colorado, 1971. J.L.Berggren, J. , Borwein, P. Borwein, Pi: A Source Book, Springer, 2004. D. Blatner, The joy of Pi, Walker Publishing Company, Inc Newyork, 1997. F. Cajori, A history of Mathematics, MacMillan and Co. London, 1926. Sir T. Heath,A History of Greek Mathematics: From Thales 30 / 33 to Euclid, Volume 1, Dover Publications, inc. Newyork, 1981. J. J. O’Connor, E.F. Robertson, The MacTutor History of Mathematics Archive,World Wide Web., 1996. π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Formal and informal references: Formal References (II)

Bourbaki, N Fonctions d’une variable relle, chap. IIIIII, Actualits Scientifiques et Industrielles (in French), 1074, Hermann, pp. 137138, 1949. Jeffreys, Harold, Scientific Inference (3rd ed.), Cambridge University Press, p. 268, 1973. Niven, Ivan, ”A simple proof that is irrational” (PDF), Bulletin of the American Mathematical Society, 53 (6), p. 509, 1947.

31 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Acknowledgments

I would specially like to thank: i) Prof. Marion Oliver for his support, interest in history of mathematics and encouragement for the concept of ‘Explore Math’. ii) Prof. H. Demirkoparan and Prof. Z. Yilma for their help and suggestions. iii) Kara, Angela, Catalina and Geetha for promoting the event and taking care of logistics. iv) Ghost of Ludolph Van Ceulen for haunting me and pushing me to explore more about Pi :).

32 / 33 A: A half of an egg! Because nothing is better than eternal bliss, and a half of an egg is better than nothing.

π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Something to carry home!

Q: What will a logician choose: a half of an egg or eternal bliss?

33 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Something to carry home!

Q: What will a logician choose: a half of an egg or eternal bliss?

A: A half of an egg! Because nothing is better than eternal bliss, and a half of an egg is better than nothing.

33 / 33 π is irrational Philosophy of Mathematics Niven and Bourbaki’s proof Something to carry home!

Q: What will a logician choose: a half of an egg or eternal bliss?

A: A half of an egg! Because nothing is better than eternal bliss, and a half of an egg is better than nothing.

33 / 33