Proofs Without Words
Dana C. Ernst
Northern Arizona University
Mathematics & Statistics Department
http://danaernst.com
MAT 123
October 23, 2017
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Warning!
Pictures can be misleading!
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Warning!
Pictures can be misleading!
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Warning!
Pictures can be misleading!
Theorem?
Hmmm, it looks like 32.5 = 31.5.
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Play Time
Let’s play a game.
• I’ll show you a picture,
• You see if you can figure out what mathematical fact it
describes or proofs.
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Theorem
For all n ∈ N, 1 + 3 + 5 + · · · + (2n − 1) = n2.
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This the same as the previous theorem, but with a different visual
proof.
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This the same as the previous theorem, but with a different visual
proof.
Theorem
For all n ∈ N, 1 + 3 + 5 + · · · + (2n − 1) = n2.
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Theorem
For all n ∈ N, 1 + 2 + · · · + n =
n(n + 1)
2
.
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Theorem
For all n ∈ N, 1 + 2 + · · · + n =
n(n + 1)
2
.
Note
The numbers Tn
:= 1 + 2 + · · · + n are called triangular numbers.
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Theorem
For all n ∈ N, 1 + 2 + · · · + n = C(n + 1, 2) :=
(n + 1)!
2!(n − 1)!
.
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Theorem
For all n ∈ N, 1 + 2 + · · · + n = C(n + 1, 2) :=
(n + 1)!
2!(n − 1)!
.
Corollary
For all n ∈ N, C(n + 1, 2) =
n(n + 1)
2
.
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The nth pentagonal number is defined to be Pn
:=
3n2 − n
2
.
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The nth pentagonal number is defined to be Pn
:=
3n2 − n
2
.
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The nth pentagonal number is defined to be Pn
:=
3n2 − n
2
.
Theorem
Pn
= 3Tn−1
+ n.
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Theorem (Nicomachus’ Theorem)
For all n ∈ N, 13 + 23 + · · · + n3 = (1 + 2 + · · · + n)2.
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Theorem (Nicomachus’ Theorem)
For all n ∈ N, 13 + 23 + · · · + n3 = (1 + 2 + · · · + n)2.
Corollary
For all n ∈ N, 13 + 23 + · · · + n3 =
(
n(n + 1)
2
)
2
.
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Theorem
2π > 6
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Theorem (Pythagorean Theorem)
If a, b, c ∈ N are the lengths of the sides of a right triangle, where
c the length of the hypotenuse, then a2 + b2 = c2.
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Theorem (Pythagorean Theorem)
If a, b, c ∈ N are the lengths of the sides of a right triangle, where
c the length of the hypotenuse, then a2 + b2 = c2.
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Theorem
We have the following summation formula:
∞
∑
k=1
(
1
2
)
k
= 1.
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Hint
Focus on a single color.
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Hint
Focus on a single color.
Theorem
We have the following summation formula:
∞
∑
k=1
(
1
4
)
k
=
1
3
.
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Theorem
We have the following summation formula:
∞
∑
k=1
(
1
3
)
k
=
1
2
.
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Theorem
A circle of radius r has area πr2.
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Theorem
The alternating sum of the first n odd natural numbers is n. In
other words, for all n ∈ N,
n
∑
k=1
(−1)n−k(2k − 1) = n.
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Theorem
We have the following fact concerning integrals:
∫
π/2
0
sin2(x) dx =
π
4
=
∫
π/2
0
cos2(x) dx.
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Sources
MathOverflow:
mathoverflow.net/questions/8846/proofs-without-words
Art of Problem Solving:
artofproblemsolving.com/Wiki/index.php/Proofs_
without_words
Wikipedia:
en.wikipedia.org/wiki/Squared_triangular_number
Strogatz, NY Times:
opinionator.blogs.nytimes.com/2010/04/04/
take-it-to-the-limit/
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