0

Recently I learned about interpolation with Newton polynomials and divided differences. Specifically we learned how $f[x_0,...,x_k] = \frac{f[x_1,...,x_k] - f[x_0,...,x_{k-1}]}{x_k - x_0}$, where $f[x_0,...,x_i]$ is the coefficient of $x^i$ of the unique polynomial of degree at most $i$ that interpolates $f$ at distinct $x_0,...x_i$. We were able to prove this by choosing an arbitrary $k$ and showing this theorem holds, without induction.

I was wondering what is implied about a theorem that can be proved without induction vs a theorem that requires induction to be proved. Does this have any relation to complexity classes maybe?

Spencer Kraisler
  • 2,882
  • 1
    Usually, it should mean that the proof works in a larger context than that of natural numbers. However, one typically uses some other theorem that holds only over $\Bbb N$ (presumably because its proof did involve induction) – Hagen von Eitzen Nov 16 '20 at 18:47

1 Answers1

1

Practically every interesting family of facts depending on a natural number parameter $n \in \mathbb{N}$ has to be proven "with induction" in the sense that if you didn't assume induction in your foundations you wouldn't be able to prove it. As a simple example, every time you assume the commutativity or associativity of addition in the integers, the rationals, the reals, or the complex numbers, that fact ultimately boils down to commutativity or associativity of addition of natural numbers, which must be proven by induction; in fact you need induction even to define addition.

On the other hand, there's a different more colloquial notion of what it means to prove something "with induction" or "without induction," which is whether you explicitly perform an induction in the proof. When you don't do this you are basically shoving the induction into another fundamental fact you've proven with induction at some point (such as the commutativity or associativity of addition).

For example, consider the classic induction exercise

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

Induction is required even to define the LHS so in that sense it's unavoidable. On the other hand, we can give a classic proof that does not require explicitly performing an induction, by instead considering

$$2(1 + 2 + \dots + n) = (1 + n) + (2 + (n-1)) + \dots + (n + 1) = n(n+1).$$

This proof implicitly uses the commutativity and associativity of addition many times, which as mentioned above ultimately depends on induction (Edit: in two ways! Each individual use of commutativity or associativity requires induction, and using them repeatedly also requires induction!) But it doesn't perform any new inductions; all the induction necessary is subsumed into commutativity and associativity, which are usually invoked without explicitly even naming them, let alone noting that they require induction to prove.

Qiaochu Yuan
  • 419,620