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Statement

For all integers $b\ge a\ge1$, a model for $(a,b)$ exists.

Proof

We first record the integer division step. If $0<p<b$, there are integers $k\ge0$ and $0\le r<p$ such that

$$b+r=(k+2)p. \tag{1}$$

Write $b=qp+s$ with $0\le s<p$. If $s=0$, then $b>p$ gives $q\ge2$; take $k=q-2$ and $r=0$. If $s>0$, then $q\ge1$; take $k=q-1$ and $r=p-s$, which satisfies $0<r<p$. These choices prove (1) in both cases.

Now use strong induction on $b$. If $a=b$, the diagonal base model applies. If $a<b$, put $p=b-a$. Then $0<p<b$. Choose $k,r$ by (1). Since $1\le p-r\le p<b$, the induction hypothesis supplies a model for $(p-r,p)$. Apply Proposition 4.2 with parameter $k$. The new pair is

$$\begin{aligned} (p-r)+kp&=(k+1)p-r=b-p=a,\\ (p-r)+(k+1)p&=(k+2)p-r=b. \end{aligned}$$

Thus it is a model for $(a,b)$. The induction strictly reduces the second parameter, and $k=0$ is allowed through the separately proved suspension case. No coprimality assumption is used or needed.

Source and scope

Exposition, Lemma 5.1, p. 6; UniversalHubModels.negative_step and all_models, pinned Lean lines 10307–10347. The parameter $r$ here is a remainder correction and is unrelated to the path length used in the pruning lemmas.

Used by. Theorem 1.1.

Bears on. #571.