Proof methods w/o Contradiction (Intuitionistic Frameworks of Math)

Question

Regarding the standard formulation of the division algorithm:

If $a,b$ are integers with $b > 0$, then there exists unique $q,r$ such that $a = bq + r$ with $0 \leq r < b$.

The standard proof of existence is to invoke the well ordering principle on the set $$S = \{ \, j \in \mathbb{N} \, | \, j = a - bn \text{ for some } n \in \mathbb{Z} \, \}.$$ Here we are taking the naturals to include 0. One can easily show that the set $S$ is nonempty in both cases, $a \geq 0$ and $a < 0$, and since defined to be a subset of $\mathbb{N}$, the Well Ordering Principle (when modified to be for sets of nonnegative numbers instead of positive numbers) guarantees a least element, say $r = a - bq$. By definition $r \geq 0$, so to finish the existence proof we need to show $r < b$. To show that $r < b$ one usually proceeds by contradiction, and easily obtains the result.

Edit: In particular, one does the following: We have $r = a-bq$ which is minimal over the set $S$. We want to show $r < b$. Suppose on the contrary $r = a - bq \geq b$. Then $a \geq b + bq = b(1+q)$ which means $a - b(1+q) \geq 0$, but we can see that this contradicts the minimality of $r = a-bq$.

My question is: can the division algorithm (the step of existence where $r < b$) be proven without contradiction?

Thinking about this question got me wondering - in order to precisely answer this question you would either need to:

1. Exhibit a proof that is not by contradiction
2. Somehow prove that a direct proof exists, without exhibiting it
3. Prove that no direct proof exists.

Although my knowledge is naive and scarce, I am aware the concept of proving independence, thus showing that no proof of a certain proposition (say, The Continuum Hypothesis) exists in ZF/ZFC. But is there any knowledge of independence for a certain type of proof ? Has anyone done work on showing, perhaps, that certain propositions cannot be proven without contradiction, or other proof styles? In that case I suppose it would mean that in the axiomatic framework (ZF/ZFC or PA) together with constructive logic, then the statement is independent?

Answer 1

the standard method is to invoke the well ordering principle on a properly defined set.

This is a bad proof, as it is non-constructive (as you noticed) without good reason. But there are several good constructive proofs. In my short-ish teaching career, I've written up two inductive proofs. Often, the case $a \geq 0$ is treated first (as a lemma), and then the general case is derived from it. There are two ways to prove the $a \geq 0$ case: either by induction on $a$ (going from $a$ to $a+1$ either increments $r$ without changing $q$, or resets $r$ to $0$ while incrementing $q$ -- and you can probably tell which case happens when), or by strong induction on $a$ (separating between the $a < b$ case, in which case you can just set $q = 0$ and $r = b$, and the $a \geq b$ case, in which case you can subtract $b$ from $a$ and increment $q$). There are also two ways to deduce the general case from the $a \geq 0$ case: either you add a sufficiently large multiple of $b$ to $a$ so that $a$ becomes nonnegative, or you replace $a$ by $-a$ and mess around with $q$ and $r$ (somewhat awkwardly requiring two cases). Oh, and there is also a way to prove the general case by two-sided induction, thus avoiding the need to factor out the $a \geq 0$ case as a lemma (but at the cost of having two induction steps, one upward and one downward).

For full details, see

• the proof of Proposition 2.150 in my Notes on the combinatorial fundamentals of algebra, version of 10 January 2019, or

• the proof of Theorem 2.6.1 in my Introduction to Modern Algebra (UMN Spring 2019 Math 4281 notes).

(The first source uses two-sided induction; the second uses strong induction.)

That said, even the non-constructive proof using the well-ordering principle can be salvaged with some nontrivial work.

Indeed, assume that $R$ is a subset of $\mathbb{N}$ that is inhabited (i.e., there exists some element of $R$; this is the constructively useful version of nonemptiness) and has a decidable membership (i.e., there is a function $f : \mathbb{N} \to \left\{0,1\right\}$ such that any $n \in \mathbb{N}$ satisfies $n \in R$ if and only if $f\left(n\right) = 1$). Then, $R$ has a smallest element (constructively). To prove this, you just pick the witness $r \in R$ for the inhabitedness of $R$, and compute $f\left(0\right), f\left(1\right), \ldots, f\left(r\right)$; the first time you obtain $1$, you have found the smallest element of $R$.

Now, I assume you want to apply this to $R = \left\{ r \in \mathbb{N} \mid a = qb+r \text{ for some } q \in \mathbb{Z} \right\}$. It is easy to see that this set $R$ is constructively inhabited (by $r = a$ if $a \in \mathbb{N}$, and otherwise, e.g., by $r = \left(1-b\right)a$). Does $R$ have decidable membership? Yes, but I believe that this takes a while to prove if you don't already have division with remainder in your backpack. It boils down to checking whether an integer $u$ is divisible by $b$. Well, if it is, then the quotient has absolute value $\leq \left|u\right|$, so there are only finitely many options to check. Not very nice, but constructive.

Answer 2

That use of contradiction arises essentially due to use of induction in the form of well-ordering, i.e. descent to "minimal criminal". We can eliminate this use of contradiction by reformulating the induction in ascent form - as a strong induction on $\,a,\,$ i.e.

if $\, a < b\,$ then $\, a\, = \,b\cdot 0 + a,\,$ else by induction $\, a-b\, =\, bq+r\,$ $\Rightarrow\,a\, =\, b(q\!+\!1) + r$

Interpreted constructively we obtain a recursive remainder algorithm on naturals $\,a,b>0$

$r(a,b)\ :=\ $ if $\,a < b\,$ then $\,a\,$ else $\,r(a\!-\!b,b)\ \ $ ; $\, r = $ remainder of $\,a\div b$

which repeatedly subtracts $\,b\,$ till we land in the sought residue range $\,0\le r< b\,$ for $\,r = a\bmod b$

Remark $ $ For another common example of descent of $\,\lnot P\,$ vs. ascent of $\,P\,$ see here on the proof of existence of prime factorizations. Alas, it's quite common that proofs that are innately constructive are presented in nonconstructive form (a ubiquitous example is Euclid's constructive proof that there are infinitely many primes, which sadly is quite commonly presented as a proof by contradiction).