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