TAOCP 7.1.3 Exercise 11
Let $\mathbb{N} \subset \mathbf{No}$ denote the positive integers embedded in Conway’s field (the surreal numbers), and let $\otimes$ denote the field multiplication in $\mathbf{No}$.
Section 7.1.3: Bitwise Tricks and Techniques
Exercise 11. ▶ [M26] (H. W. Lenstra, 1978.) Find a simple way to characterize all pairs of positive integers $(m, n)$ for which $m \otimes n = mn$ in Conway's field.
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Let $\mathbb{N} \subset \mathbf{No}$ denote the positive integers embedded in Conway’s field (the surreal numbers), and let $\otimes$ denote the field multiplication in $\mathbf{No}$.
Claim
For all positive integers $m,n$,
$$ m \otimes n = mn. $$
Hence the characterization is that every pair of positive integers satisfies the identity.
Proof
In Conway’s construction of the surreal numbers, each integer $n$ is defined recursively by
$$ 0 = {;|;}, \quad 1 = {0 \mid }, \quad 2 = {1 \mid }, \quad \dots $$
and in general $n+1 = {n \mid }$. This embedding of the integers into $\mathbf{No}$ is the unique order-preserving ring embedding of $\mathbb{Z}$ into the surreal field.
A key structural fact about Conway’s field is that the embedding $\mathbb{Z} \hookrightarrow \mathbf{No}$ is a ring homomorphism. In particular, it preserves both addition and multiplication:
$$ (m+n){\mathbf{No}} = m{\mathbf{No}} + n_{\mathbf{No}}, \quad (mn){\mathbf{No}} = m{\mathbf{No}} \otimes n_{\mathbf{No}}. $$
To justify multiplication preservation directly, one may argue by induction on $m$. For fixed $n$, the recursive definition of surreal multiplication satisfies the distributive and compatibility properties that force
$$ 1 \otimes n = n, $$
and if $m \otimes n = mn$, then
$$ (m+1)\otimes n = (m \otimes n) + (1 \otimes n) = mn + n = (m+1)n. $$
Thus, by induction on $m$, the identity holds for all positive integers $m,n$.
Conclusion
Since the equality holds for every pair of positive integers, the complete characterization is:
$$ {(m,n) \in \mathbb{N}^2 : m \otimes n = mn} = \mathbb{N}^2. $$