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Would you please let me know how do I improve this proof?

The numbers $$F_{0},F_{1},F_{2},...$$ are defined as follows (this is a definition by mathematical induction, by the way): $$F_{0} = 0, F_{1} = 1, F_{n+2} = F_{n+1} + F_{n} \ for \ n=0,1,2,...$$

Prove that for any $$n≥0$$ we have

$$F_{n} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{n−1}$$

proof

Let P(n) be the proposition that

$$F_{n} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{n−1}$$

And we know

$$F_{0} = 0, F_{1} = 1, F_{n+2} = F_{n+1} + F_{n} \ for \ n=0,1,2,...$$

Then set up the base cases P(0),P(1),P(2) as

$$F_{0} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{0−1},F_{1} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{1−1},F_{2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{2−1}$$

we know from the fibonacci sequence and the given information, $$F_{0} = 0,F_{1} = 1,F_{2} = 1$$

Therefore, P(0),p(1),p(2) is true.

Assume that there is an $$k, (k \in \mathbf{Z})$$, then p(k) is

$$F_{k} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k−1}$$

And because $$F_{n+2}>F_{n+1}>F_{n} \ and \ F_{n+2} = F_{n+1} + F_{n} \ for \ n=0,1,2,...$$

And then by PMI, $$p(k) \implies p(k+1) \implies p(k+2)$$ so that

$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k+2−1}=\left(\frac{1 + \sqrt {5}}{2}\right)^{k+1}$$ and because

$$F_{k+2} ＝ F_{k+1} ＋ F_{k}$$

Therefore $$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}+\left(\frac{1 + \sqrt {5}}{2}\right)^{k-1}$$
Therefore $$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}+\left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(\frac{1 + \sqrt {5}}{2}\right)^{-1}$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(1+\left(\frac{1 + \sqrt {5}}{2}\right)^{-1}\right)$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(1+\frac{2}{1 + \sqrt {5}}\right)$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(\frac{1 + \sqrt {5}}{1 + \sqrt {5}}+\frac{2}{1 + \sqrt {5}}\right)$$
$$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k}\left(\frac{3+ \sqrt {5}}{1 + \sqrt {5}}\right)$$
because $$\left(\frac{3+ \sqrt {5}}{1 + \sqrt {5}}\right)=\left(\frac{1 + \sqrt {5}}{2}\right)$$
Therefore $$F_{k+2} ≤ \left(\frac{1 + \sqrt {5}}{2}\right)^{k+1}$$
In conclusion, $$p(k) \implies p(k+1) \implies p(k+2)$$ is true. By PMI, $$p(n) \implies p(n+1) \implies p(n+2) \ and p(n)$$ is true.

  • 3
    I don't really understand the downvotes on this one. – 2017-01-27
  • 2
    @Ovi me too. Probably the question is too long for the downvoters; I do not see another reason – 2017-01-27
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    Did you try to use the formula $$F_n=\frac{\alpha^n-\beta^n}{\sqrt{5}}$$ where $\alpha=\frac{\sqrt{5}+1}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}$ ? – 2017-01-27
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    You don't need to display every math expression. Changing some of the double-dollars to single-dollars will shorten the vertical length and make it easier to read. Also, you might want to use the `align*` environment to chain together all your inequalities that start with $F_{k+2}$. – 2017-01-27

1 Answers 1

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Base cases are fine.

At the inductive hypothesis you must assume that $P(k)$ and $P(k-1)$ are true. You have only said to assume $P(k)$

You could use "Strong induction" and assume that for all $i\le k, P(i)$ is true.

And then you seem to spin a while, to get to the point.

Show that $P(k+1)$ is true based on the assumption $P(k)$ and $P(k-1)$ are true

let $\phi = \frac {1 + \sqrt 5}{2}$

Show that

$F_{k-1} < \phi^{k-2} , F_{k} < \phi^{k-1} \implies F_{k+1}<\phi^{k}$

$F_{k+1} = F_k + F_{k-1}$

$F_k + F_{k-1}<\phi^{k-1} + \phi^{k-2}$

$F_{k+1}<\phi^{k-2} (\phi+1)$

I say $\phi^2 = \phi+1$

$\left(\frac {1+\sqrt 5}{2}\right)^2 = \frac {6+2\sqrt 5}{4} = 1+\frac {1+\sqrt 5}{2}$

$F_{k+1}<\phi^{k}$

QED