
#1
Oct407, 01:48 AM

P: 11

The golden ratio is irrational. Do you know any clever proofs for this fact? I put this here, because it's not homeworkonly more of a discussion.




#2
Oct407, 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?




#3
Oct407, 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.




#4
Oct407, 09:21 AM

P: 461

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.




#5
Oct407, 09:56 PM

P: 891

[tex]\frac{F_{2n}}{F_{2n1}} < Phi < \frac{F_{2n+1}}{F_{2n}}[/tex] If you assume [tex]phi = a/b[/tex] then the above inequality conflicts with that. 



#6
Oct507, 07:04 AM

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Nice proof, ramsey2879.




#7
Oct607, 05:05 AM

P: 688

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.




#8
Oct607, 05:30 AM

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Not a gazillion, Dodo. Infinitely many in fact.




#9
Oct607, 09:01 AM

P: 891

But [tex] phi^{n} = F_{n1} + 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. 



#10
Dec1307, 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 



#11
Dec1307, 07:32 AM

P: 258

could someone prove that irrational OP integer = irrational, OP = operations +, , / and *




#12
Dec1307, 11:53 AM

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x + z is irrational (else a/b  z = (abz)/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 



#13
Dec1307, 01:27 PM

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Thanks
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#15
Dec1307, 11:19 PM

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How about the continued fraction form for phi?




#16
Dec1707, 08:15 AM

P: 258

Thank you, really very simple.
Another simple proof: proves that if the nthroot 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^(n1)*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 


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