## Phi is irrational

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.
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 Recognitions: Homework Help Science Advisor Clever? Does the fact that it is an algebraic number that is not an integer count as clever?

Recognitions:
Homework Help
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 $a^2=5b^2$ mod 25.

 Quote by matt grime 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.

## 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.

Blog Entries: 2
 Quote by 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.

$$\frac{F_{2n}}{F_{2n-1}} < Phi < \frac{F_{2n+1}}{F_{2n}}$$

If you assume $$phi = a/b$$ then the above inequality conflicts with that.
 Recognitions: Homework Help Science Advisor Nice proof, ramsey2879.
 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.
 Recognitions: Homework Help Science Advisor Not a gazillion, Dodo. Infinitely many in fact.

Blog Entries: 2
 Quote by 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.
But $$phi^{n} = F_{n-1} + F_{n}phi$$

Then

$$1+phi = phi^2$$

so we have phi is a root of x^2- x -1 but the discriminate is $$\sqrt{5}$$ so phi is irrational.
 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) $$\neq$$ 1 ==> q^2 = 5x^2 ==> q = sqrt(5)x ==> contradiction if gcd(5,q) $$\neq$$ 1 ==> 125y^2 = p^2 ==> 25y = p ==> gcd(5,p) $$\neq$$ 1 ==> contradiction ** the two contradictions shows up because gcd(p,q)=1 so gcd(5,p) = gcd(5,q) = 1 ==> gcd(p,q) $$\neq$$ 1 ==> contradiction sqrt(5) is irrational
 could someone prove that irrational OP integer = irrational, OP = operations +, -, / and *

Recognitions:
Homework Help
 Quote by al-mahed 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

Recognitions:
Gold Member
Staff Emeritus
 Quote by ramsey2879 How about $$\frac{F_{2n}}{F_{2n-1}} < Phi < \frac{F_{2n+1}}{F_{2n}}$$ If you assume $$phi = a/b$$ then the above inequality conflicts with that.
 Quote by 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.
Those are "two fractions". Those are two sequences of fraction. Phi is between every pair of corresponding numbers in those sequences.

Recognitions:
Homework Help
 Quote by ramsey2879 so we have phi is a root of x^2- x -1 but the discriminate is $$\sqrt{5}$$ so phi is irrational.
That seems to be the most common definition for phi.
 Recognitions: Homework Help Science Advisor How about the continued fraction form for phi?

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 by 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