1+n = n+1 Well-Ordering Principle

Can u help with understanding how proof that $1+n = n+1$ using the well-ordering property?

The way that i take to solve it, is the following: Set $A= \{ n \in \N | 1+n \ne n+1 \}$ , $A \ne \oslash$ . and $m$ is, by the well ordering principle (WOP), the least element of $A$ such that $m < n$ .

Now, being $m-1 < m < n$ that satisfy

$1+ (m-1) = (m-1) + 1 \implies m = m$

Showing that $m$ is not the least element of $A$ in this way:

$\forall n < m \implies 1+n = n+1$

In this sense, a contradiction is reached with the assumption that

$A \ne \oslash$

Sorry for see this "horror logic math", im a beginner.

#NumberTheory

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Comments

I like the idea of this proof. It looks to prove a commutative property of the addition of natural numbers using the WOP.

I think the strategy for this kind of proof is to say m is the least element of $\bold{A}$ then show that this would imply that $p$ , a smaller number than $m$ e.g. $m-1$ also had to be in $\bold{A}$ . You would show that $1+p$ does not equal $p+1$ . This is a contradiction so $\bold{A}$ is an empty set proving the theorem.

lemma: $1-1 = (-1)+1$ (to be proven somehow)

$1+m \neq m+1$ (because $m$ is an element of $\bold{A}$ )

$\therefore 1+m-1 \neq m+1-1$ (subtract $1$ from both sides)

$\therefore 1+p \neq m-1+1$ (from $p=m-1$ and lemma)

$\therefore 1+p \neq p+1$ (a contradiction as $p < m$ )

It might help to explicitly state the base case for $n=1$ , i.e. that $1+1=1+1$ . This rules out the possibility that $m=1$ and $p = m-1$ is not a natural number.

I think the $1+(m-1)=(m-1)+1⟹m=m$ line in your proof is the most interesting. I think this implication assumes the associative property of addition. Associativity may be enough to prove this commutative property without invoking the WOP: e.g. $1+((n-1)+1)=1+n$ but, assuming associativity, this also equals $(1+(n-1))+1$ which equals $n+1$ .

Justin Travers - 6 months, 1 week ago

Hi Unai, thanks for your comment.

I think it is correct as it is written, not a typo. I would be interested in your thoughts.

Of course in the real world 1+p does equal p+1 but in the strange world where there are natural numbers in A (the set of natural numbers where 1+n does not equal n+1), the task is to show that the idea of a least element in A does not make sense because for every least element m, there is a lesser element p that also meets the criteria for membership of A. Because we know that natural numbers obey the WOP, this absence of a true least element means that A is in fact empty and this proves this commutative property of addition.

Not all "addition" is commutative. If we expand the idea of addition to include actions like concatenating stings of characters then "1"+"n"="1n"and "n"+"1"="n1". The task is to prove that addition of natural numbers is commutative like we think it should be.

Justin Travers - 5 months, 3 weeks ago

Thank you so much for the explanation, I got a bit confused but rereading with a more structured vision really helped. The proof from OP confused me quite a bit and it doesn't help that my education level atm is still quite low. Thanks :)

Unai Olalla - 5 months, 3 weeks ago

The solution is great, but I think there's a small typo: in the last line of the reasoning shouldn't it be 1+p not equal to p ?

Unai Olalla - 5 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$