Phi is irrational

at3rg0

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.

matt grime

Clever? Does the fact that it is an algebraic number that is not an integer count as clever?

CRGreathouse

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.

The root of 4x-3=0 is algebraic but rational.

DeadWolfe

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

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

Nice proof, ramsey2879.

dodo

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

Not a gazillion, Dodo. Infinitely many in fact.

ramsey2879

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

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

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

CRGreathouse

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

Those are "two fractions". Those are two sequences of fraction. Phi is between every pair of corresponding numbers in those sequences.

Gib Z

That seems to be the most common definition for phi.

CRGreathouse

How about the continued fraction form for phi?

al-mahed

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