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:

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


This gives:

\(\displaystyle m + j\) \(<\) \(\displaystyle n + j\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \paren {m + j}^+\) \(<\) \(\displaystyle \paren {n + j}^+\) from $(1)$ above
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle m + j^+\) \(<\) \(\displaystyle n + j^+\) Definition of Addition in Minimal Infinite Successor 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