π is irrational

James A. Swenson

University of Wisconsin – Platteville

swensonj@uwplatt.edu

April 29, 2011MAA-WI Annual Meeting

James A. Swenson (UWP) π is irrational 4/29/11 1 / 23

Thanks for coming!

I hope you’ll enjoy the talk; please feel free to get involved!

James A. Swenson (UWP) π is irrational 4/29/11 2 / 23

Outline

1 History

2 Plan & Setup

3 Proof

James A. Swenson (UWP) π is irrational 4/29/11 3 / 23

Our topic

π is irrational

The first proof that π cannot be written in the form a/b, where a and bare whole numbers, was given in 1761 by Johann Lambert of Switzerland.

http://en.wikipedia.org/wiki/File:JHLambert.jpg

Some of Lambert’s other ideas

cosh and sinh in trig

area of hyperbolic triangles

map projections

the hygrometer

star systems

James A. Swenson (UWP) π is irrational 4/29/11 4 / 23

Lambert’s proof

I’m not going to show you Lambert’s proof: 58 dense pages.

James A. Swenson (UWP) π is irrational 4/29/11 5 / 23

Easier proofs

Mary Cartwright (Cambridge, 1945) and Ivan Niven (Oregon, 1947)produced proofs that don’t require any methods beyond Calc II.

http://www.lms.ac.uk/newsletter/339/339 12.html image: Konrad Jacobs

James A. Swenson (UWP) π is irrational 4/29/11 6 / 23

Source of this talk:

[1] Li Zhou and Lubomir Markov, Recurrent proofs of the irrationality of certaintrigonometric values, Amer. Math. Monthly 117 (2010), no. 4, 360–362.

Li Zhou, Polk St. College (FL)

http://search.intelius.com/Li-Zhou-MlEq61qz

Lubomir Markov, Barry Univ. (FL)

http://www.barry.edu/marc/faculty/Default.htm

James A. Swenson (UWP) π is irrational 4/29/11 7 / 23

Outline

1 History

2 Plan & Setup

3 Proof

James A. Swenson (UWP) π is irrational 4/29/11 8 / 23

Proving that something is impossible

Theorem

π is not a rational number.

Proof.

Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.

1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.

3 Prove: limn→∞

zn = 0.

Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!

James A. Swenson (UWP) π is irrational 4/29/11 9 / 23

Proving that something is impossible

Theorem

π is not a rational number.

Proof.

Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.

1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.

3 Prove: limn→∞

zn = 0.

Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!

James A. Swenson (UWP) π is irrational 4/29/11 9 / 23

Proving that something is impossible

Theorem

π is not a rational number.

Proof.

Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.

1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.

3 Prove: limn→∞

zn = 0.

Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!

James A. Swenson (UWP) π is irrational 4/29/11 9 / 23

Proving that something is impossible

Theorem

π is not a rational number.

Proof.

Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.

1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.

3 Prove: limn→∞

zn = 0.

Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!

James A. Swenson (UWP) π is irrational 4/29/11 9 / 23

Proving that something is impossible

Theorem

π is not a rational number.

Proof.

Suppose, for the sake of contradiction, that π = a/b, where a and b arepositive integers.

1 Define a clever sequence {zn}.2 Prove: zn is always a positive integer.

3 Prove: limn→∞

zn = 0.

Since a sequence of positive integers can’t converge to 0. . .. . . this is a contradiction. This proves that π is irrational!

James A. Swenson (UWP) π is irrational 4/29/11 9 / 23

Initial definitions: fn(x)

Definition

For any integer n ≥ 0, let fn(x) =

(πx − x2

)nn!

.

James A. Swenson (UWP) π is irrational 4/29/11 10 / 23

Initial definitions: fn(x)

Remark

On the interval [0, π], fn(x) ≥ 0.

James A. Swenson (UWP) π is irrational 4/29/11 10 / 23

Initial definitions: fn(x)

Remark

On the interval [0, π], fn(x) has its maximum value at x = π/2.

James A. Swenson (UWP) π is irrational 4/29/11 10 / 23

Initial definitions: InDefinition

For any integer n ≥ 0, let In =

∫ π

0fn(x) sin x dx.

James A. Swenson (UWP) π is irrational 4/29/11 11 / 23

What is In?

I0 = 2

I0 =

∫ π

0f0(x) sin(x) dx

=

∫ π

0

(πx−x2)0

0! sin x dx

=

∫ π

0sin x dx

= − cosπ + cos 0

= 1 + 1.

James A. Swenson (UWP) π is irrational 4/29/11 12 / 23

What is In?

I1 = 4

I1 =

∫ π

0

(πx−x2)1

1! sin x dx

IBP= (((((((((

(x2 − πx) cos x∣∣π0

+

∫ π

0(π − 2x) cos x dx

IBP= ((((((((

(π − 2x) sin x |π0 +

∫ π

02 sin x dx

= 2I0

= 4.

James A. Swenson (UWP) π is irrational 4/29/11 13 / 23

What is In?

I1 = 4

I1 =

∫ π

0

(πx−x2)1

1! sin x dx

IBP= (((((((((

(x2 − πx) cos x∣∣π0

+

∫ π

0(π − 2x) cos x dx

IBP= ((((((((

(π − 2x) sin x |π0 +

∫ π

02 sin x dx

= 2I0

= 4.

James A. Swenson (UWP) π is irrational 4/29/11 13 / 23

What is In?

I1 = 4

I1 =

∫ π

0

(πx−x2)1

1! sin x dx

IBP= (((((((((

(x2 − πx) cos x∣∣π0

+

∫ π

0(π − 2x) cos x dx

IBP= ((((((((

(π − 2x) sin x |π0 +

∫ π

02 sin x dx

= 2I0

= 4.

James A. Swenson (UWP) π is irrational 4/29/11 13 / 23

What is In?

I1 = 4

I1 =

∫ π

0

(πx−x2)1

1! sin x dx

IBP= (((((((((

(x2 − πx) cos x∣∣π0

+

∫ π

0(π − 2x) cos x dx

IBP= ((((((((

(π − 2x) sin x |π0 +

∫ π

02 sin x dx

= 2I0

= 4.

James A. Swenson (UWP) π is irrational 4/29/11 13 / 23

What is In?

In in terms of previous values

By the same method (using IBP twice), we find out that when n ≥ 2,

In = (4n − 2)In−1 − π2In−2.

For example:

I2 = (4 · 2− 2)I1 − π2I0 = 24− 2π2

I3 = (4 · 3− 2)I2 − π2I1 = 240− 24π2

I4 = (4 · 4− 2)I3 − π2I2 = 3360− 360π2 + 2π4

James A. Swenson (UWP) π is irrational 4/29/11 14 / 23

The form of In

Theorem

For each n ≥ 0: there is a polynomial gn(x), of degree ≤ n, having integercoefficients, such that In = gn(π). [Moreover, gn(x) is always even.]

What does this mean?

n In gn(x)

0 2 21 4 42 24− 2π2 24− 2x2

3 240− 24π2 240− 24x2

4 3360− 360π2 + 2π4 3360− 360x2 + 2x4

The theorem says that In will always look “like this” – the proof is routine,by strong induction, based on our recurrence:

In = (4n − 2)In−1 − π2In−2

James A. Swenson (UWP) π is irrational 4/29/11 15 / 23

Outline

1 History

2 Plan & Setup

3 Proof

James A. Swenson (UWP) π is irrational 4/29/11 16 / 23

π is irrational: step 1

What if π were rational?

Now suppose, for the sake of contradiction, that π is the rational numbera/b, where a and b are positive integers. Then, for example,

I4 = 3360− 360π2 + 2π4 =3360b4 − 360a2b2 + 2a4

b4

Thus b4I4 is the integer 3360b4 − 360a2b2 + 2a4.

Definition

For each n, let zn = bnIn.

James A. Swenson (UWP) π is irrational 4/29/11 17 / 23

π is irrational: step 1

What if π were rational?

Now suppose, for the sake of contradiction, that π is the rational numbera/b, where a and b are positive integers. Then, for example,

I4 = 3360− 360π2 + 2π4 =3360b4 − 360a2b2 + 2a4

b4

Thus b4I4 is the integer 3360b4 − 360a2b2 + 2a4.

Definition

For each n, let zn = bnIn.

James A. Swenson (UWP) π is irrational 4/29/11 17 / 23

π is irrational: step 2

Theorem

For each n, zn = bnIn is a positive integer.

Proof.

Recall that In =

∫ π

0fn(x) sin x dx , where fn(x) =

(x(π − x))n

n!. When

0 < x < π, both fn(x) and sin x are positive. Thus In > 0, and so zn > 0.

Next, recall that by our earlier theorem, In = gn(π), where gn is apolynomial with integer coefficients. We assumed that π = a/b; sincegn(x) has degree ≤ n, multiplying gn(π) by bn clears the denominators, sozn is an integer.

James A. Swenson (UWP) π is irrational 4/29/11 18 / 23

π is irrational: step 2

Theorem

For each n, zn = bnIn is a positive integer.

Proof.

Recall that In =

∫ π

0fn(x) sin x dx , where fn(x) =

(x(π − x))n

n!. When

0 < x < π, both fn(x) and sin x are positive. Thus In > 0, and so zn > 0.

Next, recall that by our earlier theorem, In = gn(π), where gn is apolynomial with integer coefficients. We assumed that π = a/b; sincegn(x) has degree ≤ n, multiplying gn(π) by bn clears the denominators, sozn is an integer.

James A. Swenson (UWP) π is irrational 4/29/11 18 / 23

π is irrational: step 2

Theorem

For each n, zn = bnIn is a positive integer.

Proof.

Recall that In =

∫ π

0fn(x) sin x dx , where fn(x) =

(x(π − x))n

n!. When

0 < x < π, both fn(x) and sin x are positive. Thus In > 0, and so zn > 0.

Next, recall that by our earlier theorem, In = gn(π), where gn is apolynomial with integer coefficients. We assumed that π = a/b; sincegn(x) has degree ≤ n, multiplying gn(π) by bn clears the denominators, sozn is an integer.

James A. Swenson (UWP) π is irrational 4/29/11 18 / 23

π is irrational: step 3

Theorem

limn→∞

zn = 0.

Proof by the Squeeze Theorem.

0 < zn = bn

∫ π

0

(πx − x2)n

n!sin x dx

≤ 1n!

∫ π

0bn(πx − x2)n dx

≤ 1n!

∫ π

0bn

(π · π

2−(π

2

)2)n

dx

(since the integrand is greatest when x = π/2)

= 1n!

∫ π

0bn

(π2

4

)n

dx

James A. Swenson (UWP) π is irrational 4/29/11 19 / 23

π is irrational: step 3

Theorem

limn→∞

zn = 0.

Proof by the Squeeze Theorem.

0 < zn = bn

∫ π

0

(πx − x2)n

n!sin x dx

≤ 1n!

∫ π

0bn(πx − x2)n dx

≤ 1n!

∫ π

0bn

(π · π

2−(π

2

)2)n

dx

(since the integrand is greatest when x = π/2)

= 1n!

∫ π

0bn

(π2

4

)n

dx

James A. Swenson (UWP) π is irrational 4/29/11 19 / 23

π is irrational: step 3

Theorem

limn→∞

zn = 0.

Proof by the Squeeze Theorem.

0 < zn = bn

∫ π

0

(πx − x2)n

n!sin x dx

≤ 1n!

∫ π

0bn(πx − x2)n dx

≤ 1n!

∫ π

0bn

(π · π

2−(π

2

)2)n

dx

(since the integrand is greatest when x = π/2)

= 1n!

∫ π

0bn

(π2

4

)n

dx

James A. Swenson (UWP) π is irrational 4/29/11 19 / 23

π is irrational: step 3

Theorem

limn→∞

zn = 0.

Proof by the Squeeze Theorem.

0 < zn = bn

∫ π

0

(πx − x2)n

n!sin x dx

≤ 1n!

∫ π

0bn(πx − x2)n dx

≤ 1n!

∫ π

0bn

(π · π

2−(π

2

)2)n

dx

(since the integrand is greatest when x = π/2)

= 1n!

∫ π

0bn

(π2

4

)n

dx

James A. Swenson (UWP) π is irrational 4/29/11 19 / 23

π is irrational: step 3

Proof (continued).

0 < zn ≤ 1n!

∫ π

0bn

(π2

4

)n

dx

= 1n!

∫ π

0

(π2b

4

)n

dx

= 1n! · π ·

(π2b4

)n= 1

n! ·ab ·(

a2

4b

)n=

a

b· (a2/4b)n

n!,

which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞

zn = 0.

James A. Swenson (UWP) π is irrational 4/29/11 20 / 23

π is irrational: step 3

Proof (continued).

0 < zn ≤ 1n!

∫ π

0bn

(π2

4

)n

dx

= 1n!

∫ π

0

(π2b

4

)n

dx

= 1n! · π ·

(π2b4

)n= 1

n! ·ab ·(

a2

4b

)n=

a

b· (a2/4b)n

n!,

which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞

zn = 0.

James A. Swenson (UWP) π is irrational 4/29/11 20 / 23

π is irrational: step 3

Proof (continued).

0 < zn ≤ 1n!

∫ π

0bn

(π2

4

)n

dx

= 1n!

∫ π

0

(π2b

4

)n

dx

= 1n! · π ·

(π2b4

)n= 1

n! ·ab ·(

a2

4b

)n=

a

b· (a2/4b)n

n!,

which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞

zn = 0.

James A. Swenson (UWP) π is irrational 4/29/11 20 / 23

π is irrational: step 3

Proof (continued).

0 < zn ≤ 1n!

∫ π

0bn

(π2

4

)n

dx

= 1n!

∫ π

0

(π2b

4

)n

dx

= 1n! · π ·

(π2b4

)n= 1

n! ·ab ·(

a2

4b

)n=

a

b· (a2/4b)n

n!,

which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞

zn = 0.

James A. Swenson (UWP) π is irrational 4/29/11 20 / 23

π is irrational: step 3

Proof (continued).

0 < zn ≤ 1n!

∫ π

0bn

(π2

4

)n

dx

= 1n!

∫ π

0

(π2b

4

)n

dx

= 1n! · π ·

(π2b4

)n= 1

n! ·ab ·(

a2

4b

)n=

a

b· (a2/4b)n

n!,

which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞

zn = 0.

James A. Swenson (UWP) π is irrational 4/29/11 20 / 23

π is irrational: step 3

Proof (continued).

0 < zn ≤ 1n!

∫ π

0bn

(π2

4

)n

dx

= 1n!

∫ π

0

(π2b

4

)n

dx

= 1n! · π ·

(π2b4

)n= 1

n! ·ab ·(

a2

4b

)n=

a

b· (a2/4b)n

n!,

which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞

zn = 0.

James A. Swenson (UWP) π is irrational 4/29/11 20 / 23

π is irrational: step 3

Proof (continued).

0 < zn ≤ 1n!

∫ π

0bn

(π2

4

)n

dx

= 1n!

∫ π

0

(π2b

4

)n

dx

= 1n! · π ·

(π2b4

)n= 1

n! ·ab ·(

a2

4b

)n=

a

b· (a2/4b)n

n!,

which converges to 0 as n→∞. By the Squeeze Theorem,limn→∞

zn = 0.

James A. Swenson (UWP) π is irrational 4/29/11 20 / 23

Next steps?

Other things you can prove by this method

1 π2 is irrational. (Since gn(x) is even.)

2 e is irrational. (Use In =

∫ 1

0

(x − x2)n

n!ex dx .)

3 If r 6= 0 is rational, then er is irrational.

(Use In =

∫ r

0

(r x − x2)n

n!ex dx .)

4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)

5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)

James A. Swenson (UWP) π is irrational 4/29/11 21 / 23

Next steps?

Other things you can prove by this method

1 π2 is irrational. (Since gn(x) is even.)

2 e is irrational. (Use In =

∫ 1

0

(x − x2)n

n!ex dx .)

3 If r 6= 0 is rational, then er is irrational.

(Use In =

∫ r

0

(r x − x2)n

n!ex dx .)

4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)

5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)

James A. Swenson (UWP) π is irrational 4/29/11 21 / 23

Next steps?

Other things you can prove by this method

1 π2 is irrational. (Since gn(x) is even.)

2 e is irrational. (Use In =

∫ 1

0

(x − x2)n

n!ex dx .)

3 If r 6= 0 is rational, then er is irrational.

(Use In =

∫ r

0

(r x − x2)n

n!ex dx .)

4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)

5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)

James A. Swenson (UWP) π is irrational 4/29/11 21 / 23

Next steps?

Other things you can prove by this method

1 π2 is irrational. (Since gn(x) is even.)

2 e is irrational. (Use In =

∫ 1

0

(x − x2)n

n!ex dx .)

3 If r 6= 0 is rational, then er is irrational.

(Use In =

∫ r

0

(r x − x2)n

n!ex dx .)

4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)

5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)

James A. Swenson (UWP) π is irrational 4/29/11 21 / 23

Next steps?

Other things you can prove by this method

1 π2 is irrational. (Since gn(x) is even.)

2 e is irrational. (Use In =

∫ 1

0

(x − x2)n

n!ex dx .)

3 If r 6= 0 is rational, then er is irrational.

(Use In =

∫ r

0

(r x − x2)n

n!ex dx .)

4 If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)

5 If r2 6= 0 is rational, then sin2 r , cos2 r , and tan2 r are all irrational.(Again: see [3].)

James A. Swenson (UWP) π is irrational 4/29/11 21 / 23

Next steps?

One (true) thing you can’t prove by this method

π is transcendental. . .[1] F. Lindemann, Ueber die Zahl π, Math. Ann. 20 (1882), no. 2, 213–225

(German).

James A. Swenson (UWP) π is irrational 4/29/11 22 / 23

Thanks!

[1] Johann Heinrich Lambert, Memoire sur quelques proprietes remarquables desquantites transcendentes circulaires et logarithmiques, Histoire de l’Academie Royaledes Sciences et des Belles-Lettres de Berlin (1761), 265-322.

[2] Ivan Niven, A simple proof that π is irrational, Bull. Amer. Math. Soc. 53 (1947),509.

[3] Li Zhou and Lubomir Markov, Recurrent proofs of the irrationality of certaintrigonometric values, Amer. Math. Monthly 117 (2010), no. 4, 360–362.

Online resourcesBerlin-Brandenburgische Akademie der Wissenschaften: http://bibliothek.bbaw.de/

Spiral staircase: http://commons.wikimedia.org/wiki/File:Wfm mackintosh lighthouse.jpg

Ivan Niven: http://en.wikipedia.org/wiki/File:Ivan Niven.jpg

Mary Cartwright: http://en.wikipedia.org/wiki/File:Dame Mary Lucy Cartwright.jpg

Johann Lambert: http://en.wikipedia.org/wiki/Johann Heinrich Lambert

Ferdinand von Lindemann: http://en.wikipedia.org/wiki/File:Carl Louis Ferdinand von Lindemann.jpg

James A. Swenson (UWP) π is irrational 4/29/11 23 / 23