Ordering on Natural Numbers is Compatible with Addition

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Theorem

Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.

Then:

$m < n \iff m + k < n + k$


Corollary

Let $a, b, c, d \in \N$ where $\N$ is the set of natural numbers.

Then:

$a > b, c > d \implies a + c > b + d$


Proof

Proof by induction:

For all $k \in \N$, let $\map P k$ be the proposition:

$m < n \iff m + k < n + k$


$\map P 0$ is true, as this just says $m + 0 = m < n = n + 0$.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P j$ is true, where $j \ge 0$, then it logically follows that $\map P {j^+}$ is true.


So this is our induction hypothesis:

$m < n \iff m + j < n + j$


Then we need to show:

$m < n \iff m + j^+ < n + j^+$


Induction Step

This is our induction step:


Let $m < n$.

Then:

\(\ds m\) \(<\) \(\ds n\)
\(\ds \leadstoandfrom \ \ \) \(\ds m\) \(\subsetneq\) \(\ds n\) Element of Finite Ordinal iff Subset‎
\(\ds \leadstoandfrom \ \ \) \(\ds m^+\) \(\subset\) \(\ds n\) Definition of Successor Set
\(\ds \leadstoandfrom \ \ \) \(\ds m^+\) \(\subsetneq\) \(\ds n^+\) Definition of Successor Set
\(\text {(1)}: \quad\) \(\ds \leadstoandfrom \ \ \) \(\ds m^+\) \(<\) \(\ds n^+\) Element of Finite Ordinal iff Subset‎


This gives:

\(\ds m + j\) \(<\) \(\ds n + j\)
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {m + j}^+\) \(<\) \(\ds \paren {n + j}^+\) from $(1)$ above
\(\ds \leadstoandfrom \ \ \) \(\ds m + j^+\) \(<\) \(\ds n + j^+\) Definition of Addition in Minimally Inductive Set

So $\map P j \implies \map P {j^+}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall m, n, k \in \N: m < n \iff m + k < n + k$

$\blacksquare$


Sources

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