A proof of the irrationality
of
Leslie Lamport
December 1, 1993
Abstract:
Theorem There does not exist in Q such that .
|
Proof Sketch: We assume for and obtain a contradiction.
Writing , where and have no common divisors (step
<1>1),
we deduce from and the lemma that both and
must be divisible by 2 (<1>2 and <1>3).
Assume:
-
-
Prove:False
<1>1: Choose , in Z such that
-
-
|
<2>1: Choose , in Z such that and .
|
Proof: By assumption <0>:1.
|
Let:
<2>2:
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Proof: <2>1 and definition of and .
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<2>3:
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Proof: [Definition of and ]
= [Simple algebra]
= [By <2>1]
|
<2>4:
Proof: By the definition of the gcd, it suffices
to:
Assume:
- divides
- divides
Prove:
|
<3>1:
divides .
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Proof: <2>:1 and the definition of .
|
<3>2:
divides .
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Proof: <2>:2 and definition of .
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<3>3Q.E.D.
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Proof: <3>1, <3>2, and the definition of gcd.
|
|
<2>5Q.E.D.
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<1>2: 2 divides .
|
<2>1:
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Proof: <1>1.1 implies .
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<2>2Q.E.D.
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Proof: By <2>1 and the lemma.
|
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<1>3: 2 divides .
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<2>1: Choose in Z such that .
<2>2:
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Proof: 2 = [<1>1.2 and <0>:2]
= [<2>1]
= [Algebra]
from which the result follows easily by algebra.
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<2>3Q.E.D.
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Proof: By <2>2 and the lemma.
|
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<1>4Q.E.D.
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- 1
- Leslie Lamport, 1993, How to write a proof. In
Global Analysis of Modern Mathematics, pp. 311-321. Publish or
Perish, Houston, Texas, February 1993. A symposium in honor of
Richard Palais' sixtieth birthday (also published as SRC Research
Report 94).
http://research.microsoft.com/users/lamport/proofs/src94.ps.Z
A proof of the irrationality
of
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