TheoremBase

The Sum of nn Ones is Strictly Increasing in nn

Statement

Let N\mathbb{N} be the set of natural numbers, with successor map SS and order relations << and ≤\le, and for n∈Nn\in\mathbb{N} let [n][n] be the initial segment it determines. Let R\mathbb{R} be the ordered field of real numbers, with additive identity 00 and multiplicative identity 11.

For n∈Nn\in\mathbb{N} let 1(n)∈Rn\mathbf{1}^{(n)}\in\mathbb{R}^{n} be the nn-tuple with every component equal to 11, and let σ:N→R\sigma:\mathbb{N}\to\mathbb{R} be given by the finite sum

σ(n)=∑k=1n1k(n).\sigma(n)=\sum_{k=1}^{n}\mathbf{1}^{(n)}_{k}.

Then the following hold for all m,n∈Nm,n\in\mathbb{N}, inequalities between natural numbers being those of N\mathbb{N} and inequalities between values of σ\sigma those of R\mathbb{R}.

1. (Recursion) σ(1)=1\sigma(1)=1 and σ(S(n))=σ(n)+1\sigma(S(n))=\sigma(n)+1.

2. (Strict monotonicity) If m<nm<n, then σ(m)+1≤σ(n)\sigma(m)+1\le\sigma(n); in particular σ(m)≤σ(n)\sigma(m)\le\sigma(n) and σ(m)≠σ(n)\sigma(m)\ne\sigma(n).

3. (Injectivity) If σ(m)=σ(n)\sigma(m)=\sigma(n), then m=nm=n.

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