Theorem 4.8.1 Irrationality of V2 V2 is irrational.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Suggest you refer to Theorem 4.8.1.
Please help me prove this using Theorem 4.8.1
![Theorem 4.8.1 Irrationality of V2
V2 is irrational.
Proof (by contradiction): [We take the negation and suppose it to be true.] Suppose
not. That is, suppose V2 is rational. Then there are integers m and n with no com-
mon factors such that
m
4.8.1
n
[by dividing m and n by any common factors if necessary]. [We must derive a contradic-
tion.] Squaring both sides of equation (4.8.1) gives
bonsh
m2
2 =
Or, equivalently,
m2 = 2n².
4.8.2
Note that equation (4.8.2) implies that m is even (by definition of even). It follows
that m is even (by Proposition 4.7.4). We file this fact away for future reference and
also deduce (by definition of even) that
2k for some integer k.
4.8.3
m =
Substituting equation (4.8.3) into equation (4.8.2), we see that
² = (2k) = 4k = 2n².
m
Dividing both sides of the right-most equation by 2 gives
n? = 2k?.
Consequently, n is even, and so n is even (by Proposition 4.7.4). But we also know
that m is even. [This is the fact we filed away.] Hence both m andn have a common
factor of 2. But this contradicts the supposition that m and n have no common fac-
tors. [Hence the supposition is false and so the theorem is true.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd34ec7b4-6ecf-40a8-b01d-5afafc11be89%2Fb1463257-e324-4b5a-a6ba-8103ce49f224%2Fzzjbgi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 4.8.1 Irrationality of V2
V2 is irrational.
Proof (by contradiction): [We take the negation and suppose it to be true.] Suppose
not. That is, suppose V2 is rational. Then there are integers m and n with no com-
mon factors such that
m
4.8.1
n
[by dividing m and n by any common factors if necessary]. [We must derive a contradic-
tion.] Squaring both sides of equation (4.8.1) gives
bonsh
m2
2 =
Or, equivalently,
m2 = 2n².
4.8.2
Note that equation (4.8.2) implies that m is even (by definition of even). It follows
that m is even (by Proposition 4.7.4). We file this fact away for future reference and
also deduce (by definition of even) that
2k for some integer k.
4.8.3
m =
Substituting equation (4.8.3) into equation (4.8.2), we see that
² = (2k) = 4k = 2n².
m
Dividing both sides of the right-most equation by 2 gives
n? = 2k?.
Consequently, n is even, and so n is even (by Proposition 4.7.4). But we also know
that m is even. [This is the fact we filed away.] Hence both m andn have a common
factor of 2. But this contradicts the supposition that m and n have no common fac-
tors. [Hence the supposition is false and so the theorem is true.]
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