TheoremBase

The reading of ℕ₀ in ℝ is the canonical embedding, under which equality and the order agree, giving injectivity and the order; Pascal's rule is carried over by the additivity of the reading.

Proof

Throughout, R\mathbb{R} is a commutative ring, being a field, so The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products applies with R=RR=\mathbb{R}; and for n∈N0n\in\mathbb{N}_{0}, the image nRn_{\mathbb{R}} is the element of R\mathbb{R} that nn denotes when read in R\mathbb{R}, by Commutative Rings, Fields and Ordered Fields: Standard Notation (iv), which is κ(n)\kappa(n) for the canonical embedding κ\kappa of The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (iv). Thus (i) and (ii) are The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vi), for equality and the order of N0\mathbb{N}_{0}, written for this reading.

Injectivity and order. By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vi), equality and the order agree whether they are formed in N0\mathbb{N}_{0} or in R\mathbb{R}:

m=n  ⟺  mR=nR,m≤n  ⟺  mR≤nR.m=n\iff m_{\mathbb{R}}=n_{\mathbb{R}},\qquad m\le n\iff m_{\mathbb{R}}\le n_{\mathbb{R}}.

This proves (i) and the first equivalence of (ii). In N0\mathbb{N}_{0}, m<nm<n means m≤nm\le n and m≠nm\neq n, by The Order on Omega Is a Well-Order with Membership as Its Strict Order, and Nothing Lies between n and Its Successor (ii), and likewise in R\mathbb{R}; so the two displayed equivalences give the second equivalence of (ii).

Pascal's rule. By The Binomial Coefficient (i) and Binomial Coefficients: Existence and Uniqueness by Pascal's Recursion, Vanishing above the Diagonal, Diagonal and First Values, the Factorial Formula and Symmetry (i), in N0\mathbb{N}_{0} we have (n0)=1\binom{n}{0}=1, (n+1k+1)=(nk)+(nk+1)\binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1}, and (0k+1)=0\binom{0}{k+1}=0, the last because 1≤k+11\le k+1: indeed k+1k+1 is the successor of kk by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication (v), hence a natural number by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication (i), and 1≤k+11\le k+1 by Arithmetic and Order of the Natural Numbers (x). Taking images, 0R0_{\mathbb{R}} and 1R1_{\mathbb{R}} are the zero and the unit of R\mathbb{R} by The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products (i), and the image of the sum is the sum of the images by The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products (ii); these are the three equations in R\mathbb{R}.

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