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Homework Statement
Given that [itex]m\in\mathbb{N}[/itex] and [itex]m\neq 1[/itex], show that [itex]m-1\in\mathbb{N}[/itex]
For this problem assume [itex]\mathbb{N} = \{1,2,...\}[/itex], also known as the product of all inductive sets.
Homework Equations
Inductive subset: A subset [itex]S\in\mathbb{R}[/itex] is inductive if:
1) [itex]1\in S[/itex] (##1## is to be considered as the unit element for which ##1x = x1 = x## for every ##x##.
2) If [itex]x\in S[/itex], then [itex]x+1\in S[/itex] for every [itex]x\in S[/itex]
The Attempt at a Solution
Going to use the method ([itex]A\Rightarrow B\Leftrightarrow \neg B\Rightarrow\neg A[/itex]) (kontraposition?)
Assume [itex]m\in\mathbb{N}\setminus\{1\}[/itex]. If [itex]m-1\notin\mathbb{N}[/itex] then [itex]\mathbb{N}\setminus\{m\}=:M[/itex] is inductive.
If the above implication holds then we can say that if [itex]M[/itex] is not inductive then [itex]m-1\in\mathbb{N}[/itex]
Then let [itex]m\neq 1[/itex]
Def 1) We can say that [itex]1\in M[/itex]
Def 2) Let [itex]n\in M[/itex], then [itex]n\in\mathbb{N}[/itex] and because [itex]\mathbb{N}[/itex] is inductive, [itex]n+1\in\mathbb{N}[/itex]. We can also say that [itex]n+1\neq m[/itex], from which [itex]n+1\in M[/itex].
Hence [itex]M[/itex] is inductive.
Problem is with the part where I say [itex]n+1\neq m[/itex], intuitively I know it's correct, but how do I justify this exactly?
Will it suffice to say that addition is an algebraic operation? Therefore [itex]n+1= m[/itex] is impossible.
But from the definition of inductive subset, [itex]M[/itex] cannot be inductive for [itex]\mathbb{N} = \mathbb{N}\setminus\{m\}[/itex] is impossible (because every inductive subset has to contain [itex]\mathbb{N}[/itex]) and we have that [itex]m-1\in\mathbb{N}[/itex], if [itex]m\neq 1[/itex]
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