Natural Number Addition Commutes with Zero

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\N$ be the natural numbers.

Then:

$\forall n \in \N: 0 + n = n = n + 0$


Proof

Proof by induction:


From definition of addition:

\(\ds \forall m, n \in \N: \, \) \(\ds m + 0\) \(=\) \(\ds m\)
\(\ds m + n^+\) \(=\) \(\ds \paren {m + n}^+\)


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

$0 + n = n = n + 0$


Basis for the Induction

By definition, we have:

$0 + 0 = 0 = 0 + 0$

Thus $\map P 0$ is seen to be true.


This is our basis for the induction.


Induction Hypothesis

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


So this is our induction hypothesis $\map P k$:

$0 + k = k = k + 0$


Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:

$0 + k^+ = k^+ = k^+ + 0$


Induction Step

This is our induction step:


\(\ds 0 + k^+\) \(=\) \(\ds \paren {0 + k}^+\) Definition of Addition
\(\ds \) \(=\) \(\ds \paren {k + 0}^+\) Induction Hypothesis
\(\ds \) \(=\) \(\ds k^+\) Definition of Addition: $k + 0 = k$

By definition:

$k^+ + 0 = k^+$

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


Therefore:

$\forall n \in \N: 0 + n = n = n + 0$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Natural_Number_Addition_Commutes_with_Zero&oldid=570619"