Mathematical Analysis I, 2004a

§§5–6. Natural Numbers. Induction 29
∗ Proof. Let S be the set of all those n ∈ N for which P (n) istrue,
S = {n ∈ N | P (n)}.
We have to show that actually each n ∈ N is in S, i.e., N ⊆ S.
First, we show that S is inductive.
Indeed, by assumption (i), P (1) is true; so 1 ∈ S.
Next, let x ∈ S. This means that P (x) is true. By assumption (ii), however,
this implies P (x + 1), i.e., x +1∈ S. Thus
1 ∈ S and (∀ x ∈ S) x +1∈ S;
S is inductive.
Then, by Theorem 1 (second clause), N ⊆ S, and all is proved.
□
This theorem is used to prove various properties of N “by induction.”
Examples.
(a) If m, n ∈ N, then also m + n ∈ N and mn ∈ N.
To prove the first property, fix any m ∈ N. LetP (n) mean
m + n ∈ N
(n ∈ N).
Then
(i) P (1) is true, for as m ∈ N, the definition of N yields m +1∈ N,
i.e., P (1).
(ii) P (k) ⇒ P (k +1) for k ∈ N. Indeed,
P (k) ⇒ m + k ∈ N ⇒ (m + k)+1∈ N
⇒ m +(k +1)∈ N ⇒ P (k +1).
Thus, by Theorem 1 ′ , P (n) holdsforall n; i.e.,
(∀ n ∈ N) m + n ∈ N
for any m ∈ N.
To prove the same for mn, weletP (n) mean
mn ∈ N (n ∈ N)
and proceed similarly.
(b) If n ∈ N, then n − 1=0or n − 1 ∈ N.
For an inductive proof, let P (n) mean
n − 1=0orn − 1 ∈ N
(n ∈ N).
Then proceed as in (a).