# Inductive proof for all naturals (including 0), x, starting from one of two given starting points

The given starting points for this inductive proof are the following:

  1) the formulas, G(2x-2) = G(x)^2 - G(x-2)^2 and, G(2x-1) = G(x+1)G(x) - G(x-1)G(x-2)


or 2) the single formula for G(x) given above, in terms of the number a = ((5^.5) - 1)/2

So starting from 1), because 0 is the smallest natural, the base cases would be:

G(-2) = G(0)^2 - G(-2)^2 and G(-1) = G(1)G(0) - G(-1)G(-2)

But is this correct so far? I know my teacher defined natural numbers, in his class, to begin with x=0, but in this question he gave us, wouldn't it make more sense if the base cases started with x=1?

Also, G defines Fibonacci Numbers, and we are given the following:

G(0) = 0; G(1) = 1; for any x>=2, G(x) = G(x-1) + G(x-2)

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You can choose your base case wherever you like. Here you are confusing the $x$ used in the recursion formula with the index of $G$. It only makes sense to start with $x=2$ for the first definition, in fact, because you don't have values $G(-1)$ and $G(-2)$. The second one would require $G(-1)$ for $x=1$ as well. Then the first one says $G(2)=G(2)^2-G(0)^2$. The second says $G(3)=G(3)G(2)-G(1)G(0)$. We will need starting values for $G(0),G(1)$ and maybe a selection of which possible value for $G(2)$ to use.
I don't understand what you mean by 2). You haven't said what closed form you think $G(n)$ obeys. Does it involve this $a$?