Connection between "proof by induction" and "arbitrary element" arguments

Question

I'm guessing that this is more of a metalogic question; I am not very familiar with this area so I apologize if the terminology is lackluster.

As I advance through my math curriculum, I notice that proof by mathematical induction is becoming quite a common tool. Alternatively, I also see that the method of "arbitrary element" (not sure what the actual name is) is quite common. Only very rarely do I see examples where one can use both strategies to prove a proposition. For example, consider proving the statement that $\frac{n(n+1)}{2} = 1 + 2 + ... +n$... or, more formally:

$$f: \mathbb N \to \mathbb N$$

$$f(n) = \frac{n(n+1)}{2}$$

$$g: \mathbb N \to \mathbb N$$

$$g(n) = \sum_{i=1}^{n}i$$

$$\forall n \big(n \in \mathbb N \land n\geq 1 \implies f(n)=g(n)\big)$$

There is the traditional proof by induction (which is widely available)...but can be found on the website here: Proving the sum of the first $n$ natural numbers by induction

Then there is the method of "arbitrary element" which could be described as follows:

Choose an arbitrary element $n^*$.

Consider $g(n^*) = 1 + 2 + ... n^*$

Multiply this by $2$ and, for illustrative purposes, strategically arrange the sum as follows:

$2g(n^*)= \big (\color{blue}{1}+\color{green}{2}+...\color{red}{n^*} \big) + \big( \color{blue}{n^*}+\color{green}{n^*-1}+...+\color{red}{1} \big) = n^*(n^*+1)$

$g(n^*) = \frac{n^*(n^*+1)}{2}$

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As demonstrated above, the proposition $\forall n \big(n \in \mathbb N \land n\geq 1 \implies f(n)=g(n)\big)$ can be proven both ways. However, I find that this seems to be a rare quality...or at least the ease with which one can find both methods seems to vary from proposition to proposition.

So my question is the following: If one can find success with a proof by induction argument for a given proposition, must there exist an "arbitrary element" argument for that same proposition? Similarly, if one can find success with an "arbitrary element" argument for a proposition, must there exist a proof by induction argument?

The two strategies seem fundamentally "different"...i.e. you could not assign each step a symbol in the arguments and form some sort of mapping strategy between each symbol to say that the arguments are actually the "same".

Thanks!

Answer 1

The "arbitrary element" proof here implicitly makes use of several theorems about the commutativity and associativity of finite series on $N$ that would require inductive proofs in a more formal mathematical setting.

Answer 2

First, the 'arbitrary element' proof is better known as a universal proof.

And yes, if there is an inductive proof, then there is always a universal proof, and vice versa. It is just that there may not be a 'nice' or 'elegant' proof of the 'other' kind.

But yes, here is a trivial way to turn any univesl proof into an inductive proof (warning: it's going to be a 'groaner'!):

Simply define $P(n)$ to be the very claim $P$ that you are trying to prove, i.e. that you have already proven by the universal proof. Then the inductive proof is:

Base: First, we'll show that $P(0)$ is true. Well, $P(0)$ is just $P$, and we have a nice universal proof for $P$. Check!

Step: Assume that $P(n)$ is true. Well, $P(n)$ is simply $P$. But $P(n+1)$ is $P$ as well, and so we have shown $P(n+1)$. Check!

Ok, so now we have shown $\forall n P$. But since $P$ has no free variable $n$, that is simply the same as $P$, and so we are done.

Ok, so now let's (equally groanworthy!) transform any inductive proof of $P$ into a universal proof that $P$. Well, take any arbitrary number $n$. By the inductive proof, we can show $P$. Since $n$ was arbitrary, that means $\forall n P$. But since $P$ has no free variables $n$, that means $P$. Done.