Let $m,n \in\mathbb Z$. If $m \le n \le m$ then $m = n$.

Question

I an learning proofs with $\mathbb{N}$. Here are my axioms:

a)If $m,n \in\mathbb N$ then $m + n \in\mathbb N$

b)If $m,n \in\mathbb N$ then $mn \in\mathbb N$

c) $0 \notin\ \mathbb N$

d) For every $m \in\mathbb Z$, we have $m \in\mathbb N$ or $m = 0$ or $-m \in\mathbb N$

Definition: $m > n \Leftrightarrow m - n \in\mathbb N$.

By the way, I would greatly appreciate if someone could please explain to me why this is. My strategy is to use a contradiction.

Proposition 1: (that I have proven)

For $m \in\mathbb Z$, one and only one of the following is true: $m \in\mathbb N$, $-m \in\mathbb N$, $m = 0$.

I have proven $1 \in\mathbb N$ and: Let $m,n,p \in\mathbb Z$. If $m < n$ and $n < p$ then $m < p$ by deriving $p - m = (p - n) + (n - m) \in\mathbb N$.

I have also proven: For each $n \in\mathbb N$ there exists $m \in\mathbb N$ such that $m > n$

Proposition 2:

Let $m,n \in\mathbb Z$. If $m \le n \le m$ then $m = n$.

I get confused because of the equality within the $\leq$. How could I approach this? By separating $<$ and $=$ ?

Any hints would be greatly appreciated.

Answer 1

First proposition

• If $m \in \mathbb N$ and $-m \in \mathbb N$ then $ 0 \in \mathbb N$, absurd.

• If $m \in \mathbb N$ and $ m = 0$, absurd.

• Similar to $-m$.

Therefore it folows the affirmation.

Second proposition

If $m \leq n \leq m$ then $m - n \in \mathbb Z$ and by the first proposition only one of those $3$ holds, then if $m - n \in \mathbb N$ then $m>n$ by definition, which is a contradiction. Similar if $n-m \in \mathbb N$.

Thus the only possibility is $m - n = 0$.

Answer 2

Suppose $m \ne n$. That means: $m < n < m$. Then by definition, $m−n∈N$ and $n-m∈N$. By axiom (a): $0=m-n+n-m∈N$. Came to contradiction by axiom (c), so $m$ is indeed equal to $n$.

Answer 3

For your reference, here is a more 'logical' version of Aaron Maroja's argument, spelling out all steps in full formal detail, and referring only to the axioms, not to e.g. your proposition 1.

$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\mbox{#2}\unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\followsfrom}{\Leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} $As discussed in the comments for this question, the answer to your question

I get confused because of the equality within the $\leq$. How could I approach this? By separating $<$ and $=$ ?

is: by applying the definition of $\;\le\;$. And since you've not been provided with such a definition, we will use $$ m \le n \;\equiv\; \lnot(m>n) $$

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We can now start at the most complex side of the theorem, and calculate as follows:

$$\calc m \le n \le m \calcop\equiv{definition of $\;\le\;$, twice; definition of $\;<\;$, twice} \lnot(m - n \in \mathbb N) \;\land\; \lnot (n - m \in \mathbb N) \calcop\then{axiom $\ref d$, twice -- the only way forward using only the axioms} (m - n = 0 \;\lor\; -(m - n) \in \mathbb N) \;\land\; (n - m = 0 \;\lor\; -(n - m) \in \mathbb N) \calcop\equiv{arithmetic: simplify; logic: extract common disjunct} m = n \;\lor\; (n - m \in \mathbb N \;\land\; m - n \in \mathbb N) \calcop{\tag{*} \then}{axiom $\ref a$ -- this is the key step in this proof} m = n \;\lor\; (n - m) + (m-n) \in \mathbb N \calcop\equiv{arithmetic: simplify} m = n \;\lor\; 0 \in \mathbb N \calcop\equiv{axiom $\ref c$; logic: simplify} m = n \endcalc$$

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Note that except for the key step $\ref *$, all the steps in the above calculation are almost forced by the desire to simplify, and keeping our goal in mind.