2

Where $n \in \mathbb{N}$

and

$$ F_n = \begin{cases} 0 & \text{ if } n = 0 \\ 1 & \text{ if } n = 1 \\ F_{n-1} + F_{n-2} & \text{ if } n > 1 \end{cases} $$

This is basically describing the famous Fibonacci sequence.

If we try the base case with 1, it works (for $0$ I am not sure...)

When $n = 1$

$2 \cdot0 - 1^2 = (-1)^1 = -1$

For the inductive hypothesis, we assume this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ is true.

Now, for the inductive step, we try to prove for $n+1$, so for $F_{n+2} \cdot F_{n} - F^2_{n+1} = (-1)^{n+1}$.

Since $n$ is always a natural number, and it will be always or even or odd, the $-1$ raised to $n$ will be always either $-1$ (when $n$ is odd) or $1$ (when $n$ is even).

Thus, $F_{n+1} \cdot F_{n-1} - F^2_n$ = -($F_{n+2} \cdot F_{n} - F^2_{n+1}$).

Or simply: $$ (-1)^n = -(-1)^{n+1} $$ $$ 1 \cdot (-1)^n = -(-1)\cdot(-1)^n $$ $$ 1 = -(-1) $$ Which is true.

I don't know if this is sufficient... I can arrive at the second step and say that we know this $(-1)^n$ is equals to $F_{n+1} \cdot F_{n-1} - F^2_n$, but I am not sure...

4 Answers4

5

The identity may be derived from the interesting fact that

$$\left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^k = \left ( \begin{array} \\ F_{k+1} & F_k\\F_k & F_{k-1} \\ \end{array} \right )$$

This may be proved via induction.

The result follows from taking the determinant of both sides.

Ron Gordon
  • 138,521
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Your proof is completely incorrect. You need to prove why $$F_{n+2} F_{n} - F_{n+1}^2 = -(F_{n+1} F_{n-1} - F_{n}^2)$$ to prove that $F_{n+2} F_{n} - F_{n+1}^2 = (-1)^{n+1}$.

You can check the base case. For the inductive step, we have \begin{align} F_{n+2} F_{n} - F_{n+1}^2 & = (F_{n+1}+F_{n})F_{n} - F_{n+1}^2 & \left(\because F_{n+2} = F_{n+1} + F_{n} \right)\\ & = F_{n+1} F_{n} + F_{n}^2 - F_{n+1}^2\\ & = F_{n+1} F_{n} - F_{n+1}^2 + F_{n}^2\\ & = F_{n+1}(F_{n}-F_{n+1}) + F_{n}^2\\ & = -F_{n+1} F_{n-1} + F_{n}^2 & \left(\because F_{n+1} = F_{n} + F_{n-1}\right)\\ & = -\left(F_{n+1}F_{n-1} - F_{n}^2\right) \end{align} Now you are done.

Adhvaitha
  • 5,441
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I think you have the right idea; you wrote down the important equality $$-(F_{n+1}\cdot F_{n-1}-F_n^2)=F_{n+2}\cdot F_n-F_{n+1}^2$$ but you got there going the wrong way - that is to say you assumed the thing you were trying to prove to get there. The rest of your proof from there is a non-sequitir which proves that $(-1)^{n+1}=-(-1)^n$, but that's not what you wanted.

What you ought to do is take the expression $$F_{n+2}\cdot F_n-F_{n+1}^2$$ and expand $F_{n+2}=F_{n}+F_{n+1}$ and $F_{n+1}=F_{n-1}+F_{n}$ and rearrange to get the expression $$-(F_{n+1}\cdot F_{n-1}-F_n^2)$$ which you can, using the inductive hypothesis, change to $$-(-1)^n=(-1)^{n+1}$$ completing the proof. Thus, you prove the important equality through purely algebraic means and only use the inductive hypothesis at the end.

Milo Brandt
  • 60,888
0

Note: Based on your answers and on what I have learnt, I will try to give my complete easy answer to this question/problem.

Ignoring the base case, which I think all we agree that I proved in my question, I will pass directly to the inductive step:

I will assume that $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ is true for $n$, and I will try to prove that also $F_{n+2} \cdot F_{n} - F^2_{n+1} = (-1)^{n+1}$ is true for $n + 1$.

The most important thing to note (immediately from the beginning) is that if $(-1)^n$ is even, then $(-1)^{n+1}$ is odd, and vice-versa, that is if $(-1)^n$ is odd, then $(-1)^{n+1}$ is even. This because, remember, a negative number raised to the power of an even number, becomes a positive number, but if raised to the power of an odd number, remains a negative number.

That said, we can try to verify if the following equation is true:

$$F_{n+1}\cdot F_{n-1}-F_n^2 = -(F_{n+2}\cdot F_n-F_{n+1}^2)$$

Now, we must have some imagination, or try some times to come out with something useful. In this case, I replaced $F_{n + 2}$ on the right side expression with $(F_{n + 1} + F_n)$, from the definition given in my question of $F_n$ (I must admit this step is one of the most difficult steps):

$$F_{n+1}\cdot F_{n-1}-F_n^2 = -((F_{n + 1} + F_n)\cdot F_n-F_{n+1}^2)$$

Now, I will just distribute the right side expression:

$$F_{n+1}\cdot F_{n-1}-F_n^2 = -(F_{n + 1} \cdot F_n + F^2_n - F_{n+1}^2)$$

In the following step, I will just exchange the position of $F_{n+1}^2$ with $F^2_n$, because it comes out that is the right thing to do (magical step, eh? this is why maths is considered difficult, because it's a lot of imagination and providences of god)

$$F_{n+1}\cdot F_{n-1}-F_n^2 = -(F_{n + 1} \cdot F_n - F_{n+1}^2 + F^2_n)$$

Picking from the first 2 terms of the right side expression $F_{n + 1}$, we have:

$$F_{n+1}\cdot F_{n-1}-F_n^2 = -(F_{n + 1}\cdot (F_n - F_{n+1}) + F^2_n)$$

Now, try to guess what $(F_n - F_{n+1})$ is?

(yes, don't be mad with me, I did not invent this)

From the definition of what $F_n$ is in my question, we arrive to say that:

$$F_n - F_{n+1} = -F_{n- 1}$$

$$-F_{n+1} = -F_{n- 1} - F_n $$

$$F_{n+1} = F_n + F_{n - 1} $$

Thus, $(F_n - F_{n+1})$ can be replaced with $-F_{n- 1}$ in my right side expression, and we obtain:

$$F_{n+1}\cdot F_{n-1}-F_n^2 = -( -F_{n + 1}\cdot F_{n- 1} + F^2_n)$$

Simplifying the minuses, we have:

$$F_{n+1}\cdot F_{n-1}-F_n^2 = F_{n + 1}\cdot F_{n- 1} - F^2_n$$

which is true, so we proved that if $P(n)$ is true, then also $P(n + 1)$, for all $n > 1$.