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.
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.
Yeah, but phi is the root of a *monic* polynomial. Matt meant to say it is an algebraic integer which is not an integer.
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.
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.
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
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
Those are "two fractions". Those are two sequences of fraction. Phi is between every pair of corresponding numbers in those sequences.
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