# Proof: $1\neq m\in\mathbb{N}, m-1\in\mathbb{N}$

1. Sep 27, 2015

### nuuskur

1. The problem statement, all variables and given/known data
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.
2. Relevant 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 $x\in S$

3. 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 $m\neq 1$

Last edited: Sep 27, 2015
2. Sep 27, 2015

### Fredrik

Staff Emeritus
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".

3. Sep 27, 2015

### nuuskur

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.

4. Sep 27, 2015

### Fredrik

Staff Emeritus
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.

5. Sep 27, 2015

### nuuskur

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.

6. Sep 27, 2015

### Fredrik

Staff Emeritus
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.

7. Sep 27, 2015

### 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.

8. Sep 27, 2015

### Fredrik

Staff Emeritus
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 $s$?

9. Sep 27, 2015

### nuuskur

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