π is irrational
James A. Swenson
University of Wisconsin – Platteville
[email protected]
April 29, 2011
MAA-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 b
are whole numbers, was given in 1761 by Johann Lambert of Switzerland.
Some of Lambert’s other ideas
cosh and sinh in trig
area of hyperbolic triangles
map projections
the hygrometer
star systems
http://en.wikipedia.org/wiki/File:JHLambert.jpg
James A. Swenson (UWP)
π is irrational
4/29/11
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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
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Easier proofs
Mary Cartwright (Cambridge, 1945) and Ivan Niven (Oregon, 1947)
produced proofs that don’t require any methods beyond Calc II.
image: Konrad Jacobs
http://www.lms.ac.uk/newsletter/339/339 12.html
James A. Swenson (UWP)
π is irrational
4/29/11
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Source of this talk:
[1] Li Zhou and Lubomir Markov, Recurrent proofs of the irrationality of certain
trigonometric values, Amer. Math. Monthly 117 (2010), no. 4, 360–362.
Li Zhou, Polk St. College (FL)
http://search.intelius.com/Li-Zhou-MlEq61qz
James A. Swenson (UWP)
Lubomir Markov, Barry Univ. (FL)
http://www.barry.edu/marc/faculty/Default.htm
π is irrational
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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 are
positive integers.
1
Define a clever sequence {zn }.
2
Prove: zn is always a positive integer.
3
Prove: lim zn = 0.
n→∞
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 are
positive integers.
1
Define a clever sequence {zn }.
2
Prove: zn is always a positive integer.
3
Prove: lim zn = 0.
n→∞
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 are
positive integers.
1
Define a clever sequence {zn }.
2
Prove: zn is always a positive integer.
3
Prove: lim zn = 0.
n→∞
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 are
positive integers.
1
Define a clever sequence {zn }.
2
Prove: zn is always a positive integer.
3
Prove: lim zn = 0.
n→∞
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 are
positive integers.
1
Define a clever sequence {zn }.
2
Prove: zn is always a positive integer.
3
Prove: lim zn = 0.
n→∞
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
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Initial definitions: fn (x)
Definition
πx − x 2
For any integer n ≥ 0, let fn (x) =
n!
James A. Swenson (UWP)
π is irrational
n
.
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Initial definitions: fn (x)
Remark
On the interval [0, π], fn (x) ≥ 0.
James A. Swenson (UWP)
π is irrational
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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
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Initial definitions: In
Definition
π
Z
For any integer n ≥ 0, let In =
fn (x) sin x dx.
0
James A. Swenson (UWP)
π is irrational
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What is In ?
I0 = 2
π
Z
f0 (x) sin(x) dx
I0 =
Z0 π
=
0
(πx−x 2 )
0!
0
sin x dx
π
Z
=
sin x dx
0
= − cos π + cos 0
= 1 + 1.
James A. Swenson (UWP)
π is irrational
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What is In ?
I1 = 4
Z
I1
π
=
1!
0
IBP
=
IBP
=
sin x dx
Z
(
((x(
π +
(cos
(x(−
πx)
((
(
0
2
(x|π +
((
(π(−(2x)
sin
0
(
(
(
=
2I0
=
4.
James A. Swenson (UWP)
1
(πx−x 2 )
(π − 2x) cos x dx
Z
π is irrational
π
0
π
2 sin x dx
0
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What is In ?
I1 = 4
Z
I1
π
=
1!
0
IBP
=
IBP
=
sin x dx
Z
(
((x(
π +
(cos
(x(−
πx)
((
(
0
2
(x|π +
((
(π(−(2x)
sin
0
(
(
(
=
2I0
=
4.
James A. Swenson (UWP)
1
(πx−x 2 )
(π − 2x) cos x dx
Z
π is irrational
π
0
π
2 sin x dx
0
4/29/11
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What is In ?
I1 = 4
Z
I1
π
=
1!
0
IBP
=
IBP
=
sin x dx
Z
(
((x(
π +
(cos
(x(−
πx)
((
(
0
2
(x|π +
((
(π(−(2x)
sin
0
(
(
(
=
2I0
=
4.
James A. Swenson (UWP)
1
(πx−x 2 )
(π − 2x) cos x dx
Z
π is irrational
π
0
π
2 sin x dx
0
4/29/11
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What is In ?
I1 = 4
Z
I1
π
=
1!
0
IBP
=
IBP
=
sin x dx
Z
(
((x(
π +
(cos
(x(−
πx)
((
(
0
2
(x|π +
((
(π(−(2x)
sin
0
(
(
(
=
2I0
=
4.
James A. Swenson (UWP)
1
(πx−x 2 )
(π − 2x) cos x dx
Z
π is irrational
π
0
π
2 sin x dx
0
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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 − π 2 In−2 .
For example:
I2 = (4 · 2 − 2)I1 − π 2 I0 = 24 − 2π 2
I3 = (4 · 3 − 2)I2 − π 2 I1 = 240 − 24π 2
I4 = (4 · 4 − 2)I3 − π 2 I2 = 3360 − 360π 2 + 2π 4
James A. Swenson (UWP)
π is irrational
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The form of In
Theorem
For each n ≥ 0: there is a polynomial gn (x), of degree ≤ n, having integer
coefficients, such that In = gn (π). [Moreover, gn (x) is always even.]
What does this mean?
n
0
1
2
3
4
In
2
4
24 − 2π 2
240 − 24π 2
3360 − 360π 2 + 2π 4
gn (x)
2
4
24 − 2x 2
240 − 24x 2
3360 − 360x 2 + 2x 4
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 − π 2 In−2
James A. Swenson (UWP)
π is irrational
4/29/11
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Outline
1
History
2
Plan & Setup
3
Proof
James A. Swenson (UWP)
π is irrational
4/29/11
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π is irrational: step 1
What if π were rational?
Now suppose, for the sake of contradiction, that π is the rational number
a/b, where a and b are positive integers. Then, for example,
I4 = 3360 − 360π 2 + 2π 4 =
3360b 4 − 360a2 b 2 + 2a4
b4
Thus b 4 I4 is the integer 3360b 4 − 360a2 b 2 + 2a4 .
Definition
For each n, let zn = b n In .
James A. Swenson (UWP)
π is irrational
4/29/11
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π is irrational: step 1
What if π were rational?
Now suppose, for the sake of contradiction, that π is the rational number
a/b, where a and b are positive integers. Then, for example,
I4 = 3360 − 360π 2 + 2π 4 =
3360b 4 − 360a2 b 2 + 2a4
b4
Thus b 4 I4 is the integer 3360b 4 − 360a2 b 2 + 2a4 .
Definition
For each n, let zn = b n In .
James A. Swenson (UWP)
π is irrational
4/29/11
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π is irrational: step 2
Theorem
For each n, zn = b n In is a positive integer.
Proof.
(x(π − x))n
Recall that In =
fn (x) sin x dx, where fn (x) =
. When
n!
0
0 < x < π, both fn (x) and sin x are positive. Thus In > 0, and so zn > 0.
Z
π
Next, recall that by our earlier theorem, In = gn (π), where gn is a
polynomial with integer coefficients. We assumed that π = a/b; since
gn (x) has degree ≤ n, multiplying gn (π) by b n clears the denominators, so
zn is an integer.
James A. Swenson (UWP)
π is irrational
4/29/11
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π is irrational: step 2
Theorem
For each n, zn = b n In is a positive integer.
Proof.
(x(π − x))n
Recall that In =
fn (x) sin x dx, where fn (x) =
. When
n!
0
0 < x < π, both fn (x) and sin x are positive. Thus In > 0, and so zn > 0.
Z
π
Next, recall that by our earlier theorem, In = gn (π), where gn is a
polynomial with integer coefficients. We assumed that π = a/b; since
gn (x) has degree ≤ n, multiplying gn (π) by b n clears the denominators, so
zn 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 = b n In is a positive integer.
Proof.
(x(π − x))n
Recall that In =
fn (x) sin x dx, where fn (x) =
. When
n!
0
0 < x < π, both fn (x) and sin x are positive. Thus In > 0, and so zn > 0.
Z
π
Next, recall that by our earlier theorem, In = gn (π), where gn is a
polynomial with integer coefficients. We assumed that π = a/b; since
gn (x) has degree ≤ n, multiplying gn (π) by b n clears the denominators, so
zn is an integer.
James A. Swenson (UWP)
π is irrational
4/29/11
18 / 23
π is irrational: step 3
Theorem
lim zn = 0.
n→∞
Proof by the Squeeze Theorem.
0
<
zn = b
n
Z
π
Z 0π
(πx − x 2 )n
sin x dx
n!
b n (πx − x 2 )n dx
Z π π π 2 n
n
1
dx
≤ n!
b π· −
2
2
0
(since the integrand is greatest when x = π/2)
Z π 2 n
π
1
= n!
bn
dx
4
0
≤
James A. Swenson (UWP)
1
n!
0
π is irrational
4/29/11
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π is irrational: step 3
Theorem
lim zn = 0.
n→∞
Proof by the Squeeze Theorem.
0
<
zn = b
n
Z
π
Z 0π
(πx − x 2 )n
sin x dx
n!
b n (πx − x 2 )n dx
Z π π π 2 n
n
1
dx
≤ n!
b π· −
2
2
0
(since the integrand is greatest when x = π/2)
Z π 2 n
π
1
= n!
bn
dx
4
0
≤
James A. Swenson (UWP)
1
n!
0
π is irrational
4/29/11
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π is irrational: step 3
Theorem
lim zn = 0.
n→∞
Proof by the Squeeze Theorem.
0
<
zn = b
n
Z
π
Z 0π
(πx − x 2 )n
sin x dx
n!
b n (πx − x 2 )n dx
Z π π π 2 n
n
1
dx
≤ n!
b π· −
2
2
0
(since the integrand is greatest when x = π/2)
Z π 2 n
π
1
= n!
bn
dx
4
0
≤
James A. Swenson (UWP)
1
n!
0
π is irrational
4/29/11
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π is irrational: step 3
Theorem
lim zn = 0.
n→∞
Proof by the Squeeze Theorem.
0
<
zn = b
n
Z
π
Z 0π
(πx − x 2 )n
sin x dx
n!
b n (πx − x 2 )n dx
Z π π π 2 n
n
1
dx
≤ n!
b π· −
2
2
0
(since the integrand is greatest when x = π/2)
Z π 2 n
π
1
= n!
bn
dx
4
0
≤
James A. Swenson (UWP)
1
n!
0
π is irrational
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π is irrational: step 3
Proof (continued).
0
<
zn ≤
=
=
=
=
n
π2
dx
b
4
0
Z π 2 n
π b
1
dx
n!
4
0
2 n
1
π b
·
π
·
n!
4
2 n
1
a
a
n! · b · 4b
1
n!
Z
π
n
a (a2 /4b)n
·
,
b
n!
which converges to 0 as n → ∞. By the Squeeze Theorem,
lim zn = 0.
n→∞
James A. Swenson (UWP)
π is irrational
4/29/11
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π is irrational: step 3
Proof (continued).
0
<
zn ≤
=
=
=
=
n
π2
dx
b
4
0
Z π 2 n
π b
1
dx
n!
4
0
2 n
1
π b
·
π
·
n!
4
2 n
1
a
a
n! · b · 4b
1
n!
Z
π
n
a (a2 /4b)n
·
,
b
n!
which converges to 0 as n → ∞. By the Squeeze Theorem,
lim zn = 0.
n→∞
James A. Swenson (UWP)
π is irrational
4/29/11
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π is irrational: step 3
Proof (continued).
0
<
zn ≤
=
=
=
=
n
π2
dx
b
4
0
Z π 2 n
π b
1
dx
n!
4
0
2 n
1
π b
·
π
·
n!
4
2 n
1
a
a
n! · b · 4b
1
n!
Z
π
n
a (a2 /4b)n
·
,
b
n!
which converges to 0 as n → ∞. By the Squeeze Theorem,
lim zn = 0.
n→∞
James A. Swenson (UWP)
π is irrational
4/29/11
20 / 23
π is irrational: step 3
Proof (continued).
0
<
zn ≤
=
=
=
=
n
π2
dx
b
4
0
Z π 2 n
π b
1
dx
n!
4
0
2 n
1
π b
·
π
·
n!
4
2 n
1
a
a
n! · b · 4b
1
n!
Z
π
n
a (a2 /4b)n
·
,
b
n!
which converges to 0 as n → ∞. By the Squeeze Theorem,
lim zn = 0.
n→∞
James A. Swenson (UWP)
π is irrational
4/29/11
20 / 23
π is irrational: step 3
Proof (continued).
0
<
zn ≤
=
=
=
=
n
π2
dx
b
4
0
Z π 2 n
π b
1
dx
n!
4
0
2 n
1
π b
·
π
·
n!
4
2 n
1
a
a
n! · b · 4b
1
n!
Z
π
n
a (a2 /4b)n
·
,
b
n!
which converges to 0 as n → ∞. By the Squeeze Theorem,
lim zn = 0.
n→∞
James A. Swenson (UWP)
π is irrational
4/29/11
20 / 23
π is irrational: step 3
Proof (continued).
0
<
zn ≤
=
=
=
=
n
π2
dx
b
4
0
Z π 2 n
π b
1
dx
n!
4
0
2 n
1
π b
·
π
·
n!
4
2 n
1
a
a
n! · b · 4b
1
n!
Z
π
n
a (a2 /4b)n
·
,
b
n!
which converges to 0 as n → ∞. By the Squeeze Theorem,
lim zn = 0.
n→∞
James A. Swenson (UWP)
π is irrational
4/29/11
20 / 23
π is irrational: step 3
Proof (continued).
0
<
zn ≤
=
=
=
=
n
π2
dx
b
4
0
Z π 2 n
π b
1
dx
n!
4
0
2 n
1
π b
·
π
·
n!
4
2 n
1
a
a
n! · b · 4b
1
n!
Z
π
n
a (a2 /4b)n
·
,
b
n!
which converges to 0 as n → ∞. By the Squeeze Theorem,
lim zn = 0.
n→∞
James A. Swenson (UWP)
π is irrational
4/29/11
20 / 23
Next steps?
Other things you can prove by this method
1
2
3
4
5
π 2 is irrational. (Since gn (x) is even.)
Z 1
(x − x 2 )n x
e dx.)
e is irrational. (Use In =
n!
0
If r 6= 0 is rational,
then e r is irrational.
Z r
(r x − x 2 )n x
(Use In =
e dx.)
n!
0
If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
If r 2 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
2
3
4
5
π 2 is irrational. (Since gn (x) is even.)
Z 1
(x − x 2 )n x
e dx.)
e is irrational. (Use In =
n!
0
If r 6= 0 is rational,
then e r is irrational.
Z r
(r x − x 2 )n x
(Use In =
e dx.)
n!
0
If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
If r 2 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
2
3
4
5
π 2 is irrational. (Since gn (x) is even.)
Z 1
(x − x 2 )n x
e dx.)
e is irrational. (Use In =
n!
0
If r 6= 0 is rational,
then e r is irrational.
Z r
(r x − x 2 )n x
(Use In =
e dx.)
n!
0
If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
If r 2 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
2
3
4
5
π 2 is irrational. (Since gn (x) is even.)
Z 1
(x − x 2 )n x
e dx.)
e is irrational. (Use In =
n!
0
If r 6= 0 is rational,
then e r is irrational.
Z r
(r x − x 2 )n x
(Use In =
e dx.)
n!
0
If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
If r 2 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
2
3
4
5
π 2 is irrational. (Since gn (x) is even.)
Z 1
(x − x 2 )n x
e dx.)
e is irrational. (Use In =
n!
0
If r 6= 0 is rational,
then e r is irrational.
Z r
(r x − x 2 )n x
(Use In =
e dx.)
n!
0
If r 6= 0 is rational, then tan r is irrational. (Harder; see [3].)
If r 2 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
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Thanks!
[1] Johann Heinrich Lambert, Mémoire sur quelques propriétés remarquables des
quantités transcendentes circulaires et logarithmiques, Histoire de l’Académie Royale
des 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 certain
trigonometric values, Amer. Math. Monthly 117 (2010), no. 4, 360–362.
Online resources
Berlin-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
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