Irrational numbers and Limits: The Golden ratio Question

[Unity+]

I think most know as two Fibonnaci numbers are divided together, dividing the smaller one by the larger one, you get closer to the golden ratio as the two Fibonnaci numbers get larger and larger.

 

Here is the equation that I want to use for this example:

 

(Correct me if this is an incorrect way of doing this).

 

Since at the limit of infinity there would, in fact, be a fraction, but of immeasurable magnitude, that would exist, therefore making this not an irrational number? I am not saying it isn't an irrational number, but I want to be sure of my logic on irrational numbers. Or would this immeasurable magnitude of the fraction make this an irrational number?

 

Also, the same question implies the question wouldn't this be considered multiplying two numbers, one being "rational" and one being irrational be considered making it rational? Or does this not apply to limits because of their properties?

 

Ask if you need clarification on the question.

 

 

[John]

The equation is

[math]\lim_{k \to \infty} \frac{F_{k+1}}{F_{k}} = \varphi[/math]

which, since the limit of a quotient is equal to the quotient of the limits of the numerator and denominator, ends up giving us

[math]\lim_{k \to \infty} F_{k+1} = \varphi \lim_{k \to \infty} F_{k}[/math].

This, of course, ends up with infinity on both sides of the equation.

What you must keep in mind when dealing with this limit is the index k never actually reaches infinity. It simply increases without bound. We can (theoretically) take arbitrarily large pairs of consecutive Fibonacci numbers and find their ratio, and the answer will be within some neighborhood of phi, but phi itself will still be irrational.

As for multiplying an irrational number by a rational number, let n be an irrational number. Multiplying it by 0, of course, will give us 0. So assume multiplying it by some non-zero rational number a/b will result in a rational number c/d. Then what we have is [math]\frac{a}{b}n = \frac{c}{d}[/math]. But this means [math]n = \frac{b}{a}\frac{c}{d}[/math]. Since [math]\frac{bc}{ad}[/math] is rational, n is therefore a rational number, which contradicts our assumption that n is irrational.

Edited November 17, 2013 by John

[Unity+]

The equation is

 

[math]\lim_{k \to \infty} \frac{F_{k+1}}{F_{k}} = \varphi[/math]

 

which, since the limit of a quotient is equal to the quotient of the limits of the numerator and denominator, ends up giving us

 

[math]\lim_{k \to \infty} F_{k+1} = \varphi \lim_{k \to \infty} F_{k}[/math].

 

This, of course, ends up with infinity on both sides of the equation.

 

What you must keep in mind when dealing with this limit is the index k never actually reaches infinity. It simply increases without bound. We can (theoretically) take arbitrarily large pairs of consecutive Fibonacci numbers and find their ratio, and the answer will be within some neighborhood of phi, but phi itself will still be irrational.

 

As for multiplying an irrational number by a rational number, let n be an irrational number. Multiplying it by 0, of course, will give us 0. So assume multiplying it by some non-zero rational number a/b will result in a rational number c/d. Then what we have is [math]\frac{a}{b}n = \frac{c}{d}[/math]. But this means [math]n = \frac{b}{a}\frac{c}{d}[/math]. Since [math]\frac{bc}{ad}[/math] is rational, n is therefore a rational number, which contradicts our assumption that n is irrational.

That would make sense. So, it depends on potentiality rather than what is initially in the equation. I think I understand.

 

Thanks for the help.

[mathematic]

Aside from this particular example, do you understand that every irrational number can be expressed as a limit of a sequence of rational numbers?

[studiot]

 

Aside from this particular example, do you understand that every irrational number can be expressed as a limit of a sequence of rational numbers?

 

 

Does this include trancendentals ?

[Unity+]

Aside from this particular example, do you understand that every irrational number can be expressed as a limit of a sequence of rational numbers?

Yes, each irrational number can be as a limit of a sequence of rational number. However, not all limits are expressing irrational numbers.

Edited November 17, 2013 by Unity+

[studiot]

 

Yes, each irrational number can be expressed an irrational number

 

I don't think you quite meant that?

 

[Unity+]

 

I don't think you quite meant that?

 

Hehe, my mistake. I don't know what I meant by that.

Edited November 17, 2013 by Unity+

[mathematic]

 

Does this include trancendentals ?

yes. Why would you think otherwise? The easiest way to see it is to consider the infinite decimal expansion for an irrational number X. Take the sequence Xn defined by truncating after n digits (which is rational). As n -> ∞, Xn -> X.