Proof: [itex]1\neq m\in\mathbb{N}, m-1\in\mathbb{N}[/itex]

[nuuskur]

Homework Statement

Given that m\in\mathbb{N} and m\neq 1, show that m-1\in\mathbb{N}
For this problem assume \mathbb{N} = \{1,2,...\}, also known as the product of all inductive sets.

Homework Equations

Inductive subset: A subset S\in\mathbb{R} is inductive if:
1) 1\in S ($1$ is to be considered as the unit element for which $1x = x1 = x$ for every $x$.
2) If x\in S, then x+1\in S for every x\in S

The Attempt at a Solution

Going to use the method (A\Rightarrow B\Leftrightarrow \neg B\Rightarrow\neg A) (kontraposition?)
Assume m\in\mathbb{N}\setminus\{1\}. If m-1\notin\mathbb{N} then \mathbb{N}\setminus\{m\}=:M is inductive.
If the above implication holds then we can say that if M is not inductive then m-1\in\mathbb{N}

Then let m\neq 1
Def 1) We can say that 1\in M
Def 2) Let n\in M, then n\in\mathbb{N} and because \mathbb{N} is inductive, n+1\in\mathbb{N}. We can also say that n+1\neq m, from which n+1\in M.
Hence M is inductive.

Problem is with the part where I say n+1\neq m, 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 n+1= m is impossible.

But from the definition of inductive subset, M cannot be inductive for \mathbb{N} = \mathbb{N}\setminus\{m\} is impossible (because every inductive subset has to contain \mathbb{N}) and we have that m-1\in\mathbb{N}, if m\neq 1

 

[Fredrik]

You have left m and n essentially arbitrary. So m=3 and n=2 is a possibility. This would make n+1=m.

I think you have to use that $\mathbb N$ has the property that there are no elements in that set other than the ones that are mentioned in the definition of "inductive set".

 

[nuuskur]

You have left m and n essentially arbitrary. So m=3 and n=2 is a possibility. This would make n+1=m.

I think you have to use that $\mathbb N$ has the property that there are no elements in that set other than the ones that are mentioned in the definition of "inductive set".

The set M does not contain 3, Therefore 2+1, which is an algebraic operation, for both 2 and 1 are in M, has to be contained within the set M.

 

[Fredrik]

Sorry, that was a silly mistake by me. I'm not sure I have time to think about how to figure out how to turn your approach into a complete proof, but I see a simpler one. It's based on the idea that the statement $m\in\mathbb N$ implies one of two mutually exclusive possibilities, one of which is $m=1$. What I have in mind for the other one is something more useful than the obvious $m\neq 1$. Since the $m=1$ possibility is ruled out by assumption, the other one must be a true statement.

 

[nuuskur]

Sorry, that was a silly mistake by me. I'm not sure I have time to think about how to figure out how to turn your approach into a complete proof, but I see a simpler one. It's based on the idea that the statement $m\in\mathbb N$ implies one of two mutually exclusive possibilities, one of which is $m=1$. What I have in mind for the other one is something more useful than the obvious $m\neq 1$. Since the $m=1$ possibility is ruled out by assumption, the other one must be a true statement.

Assuming I get this right you mean to get at: $m =1\vee m\in\mathbb{N}\setminus\{1\}$ We know that $m\in\mathbb{N}\setminus\{1\} =: A$, but then what is $m+(-1)$? This, at least, is the first thing I tried, but because it isn't an algebraic operation I can't say anything about $m-1$ so I couldn't get anywhere with it.

 

[Fredrik]

Think of a property that every element of $\mathbb N$ except 1 has. If $m$ is an element of $\mathbb N$ such that $m\neq 1$, then $m$ must have that property.

 

[nuuskur]

The only thing I can think of right now is if $m\neq 1$ then there always exist $p,q\in\mathbb{N}$ such that $p< m < q$, but then again it sounds like something that needs to be proven first.

 

[Fredrik]

It's hard to tell you something without giving away the complete solution, but if we define $s(x)=x+1$ for each $x\in\mathbb N$, then what is the range of the function $s$?

 

[nuuskur]

It's hard to tell you something without giving away the complete solution, but if we define $s(x)=x+1$ for each $x\in\mathbb N$, then what is the range of the function $s$?

$[2,\infty)$.. right, I get where this is going, thanks :D