Phi is irrational


by at3rg0
Tags: irrational
at3rg0
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#1
Oct4-07, 01:48 AM
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The golden ratio is irrational. Do you know any clever proofs for this fact? I put this here, because it's not homework--only more of a discussion.
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matt grime
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#2
Oct4-07, 02:51 AM
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Clever? Does the fact that it is an algebraic number that is not an integer count as clever?
CRGreathouse
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#3
Oct4-07, 07:23 AM
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Once you show that sqrt(5) is irrational it's pretty easy. You can use the standard proof for that -- suppose a/b = sqrt(5) with a/b in lowest terms, then consider [itex]a^2=5b^2[/itex] mod 25.

Quote Quote by matt grime View Post
Clever? Does the fact that it is an algebraic number that is not an integer count as clever?
The root of 4x-3=0 is algebraic but rational.

DeadWolfe
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#4
Oct4-07, 09:21 AM
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Phi is irrational


Yeah, but phi is the root of a *monic* polynomial. Matt meant to say it is an algebraic integer which is not an integer.
ramsey2879
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#5
Oct4-07, 09:56 PM
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Quote Quote by at3rg0 View Post
The golden ratio is irrational. Do you know any clever proofs for this fact? I put this here, because it's not homework--only more of a discussion.
How about

[tex]\frac{F_{2n}}{F_{2n-1}} < Phi < \frac{F_{2n+1}}{F_{2n}}[/tex]

If you assume [tex]phi = a/b[/tex] then the above inequality conflicts with that.
CRGreathouse
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#6
Oct5-07, 07:04 AM
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Nice proof, ramsey2879.
dodo
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#7
Oct6-07, 05:05 AM
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I don't get it. There has to be a gazillion rationals between those two fractions, no matter how big n is or how close to the limit you are.
matt grime
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#8
Oct6-07, 05:30 AM
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Not a gazillion, Dodo. Infinitely many in fact.
ramsey2879
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Oct6-07, 09:01 AM
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Quote Quote by Dodo View Post
I don't get it. There has to be a gazillion rationals between those two fractions, no matter how big n is or how close to the limit you are.
your right
But [tex] phi^{n} = F_{n-1} + F_{n}phi[/tex]

Then

[tex]1+phi = phi^2[/tex]

so we have phi is a root of x^2- x -1 but the discriminate is [tex]\sqrt{5}[/tex] so phi is irrational.
al-mahed
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#10
Dec13-07, 02:04 AM
P: 258
supose sqrt(5) = p/q in lowest terms ==> p,q integer such that gcd(p,q)=1 ==>

==> 5 = p^2/q^2 ==> 5q^2 = p^2

if gcd(5,p) [tex]\neq[/tex] 1 ==> q^2 = 5x^2 ==> q = sqrt(5)x ==> contradiction

if gcd(5,q) [tex]\neq[/tex] 1 ==> 125y^2 = p^2 ==> 25y = p ==> gcd(5,p) [tex]\neq[/tex] 1 ==> contradiction

** the two contradictions shows up because gcd(p,q)=1

so gcd(5,p) = gcd(5,q) = 1 ==> gcd(p,q) [tex]\neq[/tex] 1 ==> contradiction

sqrt(5) is irrational
al-mahed
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#11
Dec13-07, 07:32 AM
P: 258
could someone prove that irrational OP integer = irrational, OP = operations +, -, / and *
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Dec13-07, 11:53 AM
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Quote Quote by al-mahed View Post
could someone prove that irrational OP integer = irrational, OP = operations +, -, / and *
Let z be an integer, n be a positive integer, and x be an irrational number.

x + z is irrational (else a/b - z = (a-bz)/b which is rational)

x - z is irrational by the above.

x * n is irrational (else a/b / n = a/(bn) which is rational)

x / n is irrational (else a/b * n = (an)/b which is rational)

x * 0 is rational

x / 0 is undefined
HallsofIvy
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#13
Dec13-07, 01:27 PM
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Quote Quote by ramsey2879 View Post
How about

[tex]\frac{F_{2n}}{F_{2n-1}} < Phi < \frac{F_{2n+1}}{F_{2n}}[/tex]

If you assume [tex]phi = a/b[/tex] then the above inequality conflicts with that.
Quote Quote by Dodo View Post
I don't get it. There has to be a gazillion rationals between those two fractions, no matter how big n is or how close to the limit you are.
Those are "two fractions". Those are two sequences of fraction. Phi is between every pair of corresponding numbers in those sequences.
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#14
Dec13-07, 10:06 PM
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Quote Quote by ramsey2879 View Post
so we have phi is a root of x^2- x -1 but the discriminate is [tex]\sqrt{5}[/tex] so phi is irrational.
That seems to be the most common definition for phi.
CRGreathouse
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#15
Dec13-07, 11:19 PM
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How about the continued fraction form for phi?
al-mahed
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#16
Dec17-07, 08:15 AM
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Thank you, really very simple.

Another simple proof: proves that if the nth-root of a positive whole number will not be a positive whole number**, also will not be a rational number.

This should generalize our results.

consider gcd(p,q) = 1 (p/q in lowest terms)

k^1/n = p/q ==> kq^n = p^n ==> k | p ==> kq^n = (k^n)*(x^n) ==>

==> q^n = k^(n-1)*x^n ==> k | q ==> k | p and q ==> contradiction

** note that if k^1/n is a whole number ==> p/q will not be in lowest terms

Quote Quote by CRGreathouse View Post
Let z be an integer, n be a positive integer, and x be an irrational number.

x + z is irrational (else a/b - z = (a-bz)/b which is rational)

x - z is irrational by the above.

x * n is irrational (else a/b / n = a/(bn) which is rational)

x / n is irrational (else a/b * n = (an)/b which is rational)

x * 0 is rational

x / 0 is undefined


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