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Proposition: the sum of two natural numbers is again a natural number, i.e. for all natural numbers $n, m$, $P(n) = n + m \text{ "is a natural number"}$ is true. Proof: - $P(0) = 0 + m = m$ by definition of addition, which is a natural number. - Let $m, n, k$ be natural numbers such that $P(n) = n + m = k \text{ "is a natural number"}$ is true, which is the induction hypothesis. Then \begin{align} P(n++) &= (n++) + m \\ &= (n+m)++ \text{ (by definition of addition) }\\ &= k++ \text{ (assuming the hypothesis) }\\ &\text{"is a natural number"} \text{ (by the 2nd Peano ax.) }. \end{align} Therefore $P(n)$ is true for all natural numbers $n$ and $m$.
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Tue, 07 Sep 2021 22:14 GMT