# Phi is irrational

1. Oct 4, 2007

### 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.

2. Oct 4, 2007

### matt grime

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

3. Oct 4, 2007

### 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 $a^2=5b^2$ mod 25.

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

4. Oct 4, 2007

### 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.

5. Oct 4, 2007

### 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.

Last edited: Oct 4, 2007
6. Oct 5, 2007

### CRGreathouse

Nice proof, ramsey2879.

7. Oct 6, 2007

### 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.

8. Oct 6, 2007

### matt grime

Not a gazillion, Dodo. Infinitely many in fact.

9. Oct 6, 2007

### ramsey2879

your right
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.

10. Dec 13, 2007

### 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) $$\neq$$ 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

11. Dec 13, 2007

### al-mahed

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

12. Dec 13, 2007

### 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

13. Dec 13, 2007

### HallsofIvy

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

14. Dec 13, 2007

### Gib Z

That seems to be the most common definition for phi.

15. Dec 13, 2007

### CRGreathouse

How about the continued fraction form for phi?

16. Dec 17, 2007

### 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

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