Is it possible to prove the reversibility exponent math?

Question

TL;DR: How do I prove x⁰=1 without proving by division of x?

I'm somewhat new to calculus, so I don't have much experience with complex proofs. That being said, I ran into someone recently who argues that x⁰=0 (where x is any number). I showed this person several videos by Eddie Woo and blackpenredpen, but they say that the explanations used "patterns instead of straight-foward thinking." I want to prove that x⁰=1, but I can't find any videos that prove this without using proof-by-reversibility patterns (dividing by the base to get 1).

I've tried using the Fundamental Theorem of Calculus to prove that even complex math is reversible (and can be used to prove by patterns), but this person just says that this thinking is flawed and is based on circular reasoning (they also suggested that calculators are programmed to display x⁰=1 to affirm and satiate mathematitians).

Any help is appreciated!

Answer 1

There can be problems with $x^0=1$, so someone who merely disagrees with that could have a point. But someone who claims conversely that $x^0=0$ has to have an explanation that deals with

$$ 2 = 2^1 = 2^{1+0} = 2^1 2^0 = 2^1 \times 0 = 0$$

I'd ask them which equality step they want to give up on, or if they think $2$ does equal $0$.

(BTW, the weakest step in that chain is where we used $a^{bc} = a^b a^c$, as that requires us to put some sort of restrictions on $a, b$, and $c$ to be valid.)

Answer 2

they also suggested that calculators are programmed to display x0=1 to affirm and satiate mathematitians

This is essentially correct. Similarly, English is written left-to-right to satisfy and satiate English speakers, while Hebrew and Arabic are written right-to-left to satisfy and satiate Hebrew and Arabic speakers. That $x^0 = 1$ is a convention, not something you can prove. The powers of $x$ can be defined as $$x^n = \begin{cases} 1 & \text{ if $n = 0$,}\\ x * x^{n-1} & \text{ if $n > 0.$} \end{cases}$$ This is an inductive definition. Similarly, if you are familiar with the Fibonacci numbers, they are defined $$F_n = \begin{cases} 1 & \text{ if $n = 0,1$,}\\ F_{n-1} + F_{n-2} & \text{ if $n > 1.$} \end{cases}$$

So how do you prove $F_0 = 1$? Well, you don't. It's a definition. In fact, sometimes the convention is that $F_0 = 0$. It depends on what text you are reading. But once we decide on the definition above, we don't get to argue that $F_2 = 7$. The statement $F_2 = 1+1 = 2$ is then provable, and we don't get to decide it.

So, tell your friend that they are in fact free to decide that $x^0 = 0$. It's just that this will confuse everyone else they talk to, because virtually all mathematicians agree on the convention $x^0 = 1$. They're free to write a whole book in which $x^0 = 0$, developing this as an alternate form of math - just like there is a book out there that defines $-1 * -1 = -1$. It's just that they will probably have a hard time finding readers. You could also publish a map that refers to New York City as Moscow, if you so wish.

But now why do most mathematicians write $x^0=1$? Well, that's a more interesting question. As I said, there is no proof, but there are many explanations as to why this is a good choice, as in the demonstrations you have alluded to. It turns out to be very convenient.