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I thought of a way to prove conjectures in continuous settings, assuming they are true at the two ends points, To demonstrate it, I will show how it works with the power rule from diffrential calculus.$$$$ Just a basic recap, the power rule says that if $ r\in \mathbb{R}$, then $\frac{d(x ^ r)}{dx} = r x^{r-1}$. The algorithm goes like this, suppose that we somehow proved that the power rule work for two powers $a$ and $b$ (i.e $\frac{d(x ^ a)}{dx} = a x^{a-1}$ and $\frac{d(x ^ b)}{dx} = b x^{b-1}$), now we prove that this implies that the power rule works when the power equals $\frac{a + b}{2}$ (i.e that $\frac{d(x ^ \frac{a + b}{2})}{dx} = \frac{a + b}{2} x^{\frac{a + b}{2}-1}$).$$$$ This algorithm isn't meant to just prove the conjecture on the point on the middle of the interval, it's suppose to prove it on the entire inteval. If the power rule works for two powers, $c$ and $d$, and we proved the statement above, then the power rule works for every power $r$ such that $c < r < d$. Here's why:$$$$ through our proof, we know that the power rule works when the power equals $ \frac{c + d}{2}$ . So know we know that the power rule works when the power equals $c$ and $\frac{c + d}{2}$, this implies that it also work when the power equals $\frac{\frac{c + d}{2} + c}{2}$, and because it also works when the power equals $d$, it should also work when the power equals $\frac{\frac{c + d}{2} + d}{2}$.$$$$ According to my understanding, if I repeat the same argument, then it should prove that the power rule works for every power $r$ such that $c < r < d$. So I have three questions about this algorithm:

  1. Is it a valid way to prove that if the power rule works for any two powers $c$ and $d$, then it also works for any power $r$ such that $c < r < d?$ And if not, why$?$
  2. Could it be used to do proofs on continuous intervals for conjectures beyond the power rule$?$
  3. Assuming the answer to question $2$ is yes, could this algorithm be generalized, by replacing the denominator with any real number $r$ greater than $1$, and multiplying $a$ by $r - 1?$ I.e, in the example of the power rule, if we proved that if the power rule works for any two powers $a$ and $b$, then it also works for $ \frac{(r - 1)a + b}{r} $ (where $r \in \mathbb{R}$ and $ r > 1$), and we proved that the power rule works for two powers $c$ and $d$, can we thus conclude that the power works for any power $r$ such that $c < r < d?$ $$$$EDIT: I added an update to this question on my answer (I did it there, and not here, because I felt this question is getting too long).
Ofek Tevet
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1 Answers1

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I managed to find an example that uses this algorithm, to prove something that is false, so this algorithm alone isn't enough to prove conjectures. $$$$The counter example goes like this: Let's suppose that we want to prove that all the numbers between $1$ and $2$ are rational. To do that, we first prove that $1$ and $2$ are rationals (which they are based on the definition of a rational number), now we prove that if $a$ and $b$ are both rational, then $\frac{a + b}{2}$ is also rational, which is true, based on the closure under addition and division of the set of rational numbers.$$$$ According to my proof method, this should be enough to conclude that every number between $1$ and $2$ is rational, which is incorrect, $\sqrt{2}$ for example is between $1$ and $2$, and it isn't rational. $$$$Despite this, I haven't given up on this method, and I think that with one additonal proof, it can be saved. Let's suppose that we want to prove some conjecture where the parameter, $r \in [a, b]$ $$$$ According to my understanding, the reason this algorithm fails, is that unless $c = \frac{k}{2^n} a+ (1 - \frac{k}{2^n})b$ (where $k \in [0 .. 2^n]$ and $n \in \mathbb{N_0}$), then this algorithm doesn't prove that the conjecture is true when $r = c$, it proves that the conjecture is true, when the $r \to c$.$$$$ In the example of the power rule, if $c \neq \frac{k}{2^n} a+ (1 - \frac{k}{2^n})b$ (where again $k \in [0 .. 2^n]$ and $n \in \mathbb{N_0}$), it doesn't prove that $\frac{d}{dx}(x^c) = cx^{c-1}$, it proves that $\lim_{t\to c} \frac{d}{dx}(x^t) = \lim_{t \to c} tx^{t-1}$. $$$$If this is true, then the required modification to complete the proof is relatively straight forward, just prove that if the conjecture is true when $r \to c$, then the conjecture is true when $r = c$. In the example of the power rule, the required modification to complete the proof, is to prove that if $\lim_{t\to c} \frac{d}{dx}(x^t) = \lim_{t \to c} tx^{t-1}$, then $\frac{d}{dx}(x^c) = cx^{c-1}$. $$$$This modifcation is also the place where our proof that every number beween $1$ and $2$ is rational falls short, because if all we know is that $x \to c$ (where $c \in \mathbb{R}$), and that $x$ is rational, then we cannot conclude whetever $c$ is rational or not.$$$$I think with this modification, this algorithm will be valid to prove conjectures on continuous intervals, but I don't know for sure, is it$?$ And if it is, can it be generalized in the way I explained on the question$?$

Ofek Tevet
  • 348