# Really Simple Equations

Posted on 2017-04-01 by nbloomf

In this post we will take a break and solve some equations in the natural numbers. These equations will be really simple, but we have to start somewhere! In the more familiar notation we will solve the following equations: $\begin{array}{rrr} a+b=0 & \quad a+b=1 & \quad a+b=2 \\ ab=0 & \quad ab=1 & \quad ab=2 \end{array}$

Our strategy is based on two of the Peano axioms. Specifically, (1) $$\zero = \next(m)$$ has no solution $$m \in \nats$$, and (2) every element of $$\nats$$ is either $$\zero$$ or of the form $$\next(m)$$ for some $$m \in \nats$$. Property (2) can be used to perform case analysis on $$\nats$$, like so:

Let $$a,b \in \nats$$. If $$\nplus(a,b) = \zero$$, then $$a = b = \zero$$.

Suppose $$\nplus(a,b) = \zero$$. Note that either $$a = \zero$$ or $$a = \next(m)$$ for some $$m \in \nats$$. If $$a = \next(m)$$, then $\zero = \nplus(a,b) = \nplus(\next(m),b) = \next(\nplus(m,b)),$ a contradiction. So we have $$a = \zero$$. But now $$\zero = \nplus(\zero,b) = b$$ as claimed.

The following generalization will be handy.

We have the following.

1. Every natural number is either $$\zero$$, $$\next(\zero)$$, or of the form $$\next(\next(m))$$ for some $$m \in \nats$$.
2. Every natural number is either $$\zero$$, $$\next(\zero)$$, $$\next(\next(\zero))$$, or of the form $$\next(\next(\next(m)))$$ for some $$m \in \nats$$.
1. Let $$n \in \nats$$. Either $$n = \zero$$ or $$n = \next(k)$$ for some $$k \in \nats$$; but now either $$k = \zero$$ or $$k = \next(m)$$ for some $$m \in \nats$$. So either $$n = \zero$$, $$n = \next(\zero)$$, of $$n = \next(\next(m))$$ for some $$m \in \nats$$.
2. Similar to (1).

This will allow us to use more detailed case analysis on $$\nats$$.

Let $$a,b \in \nats$$. If $$\nplus(a,b) = \next(\zero)$$, then $$(a,b)$$ is either $$(\zero,\next(\zero))$$ or $$(\next(\zero),\zero)$$.

Note that either $$a = \zero$$, $$a = \next(\zero)$$, or $$a = \next(\next(m))$$ for some $$m \in \nats$$.

• If $$a = \next(\next(m))$$ for some $$m \in \nats$$, we have $\begin{eqnarray*} & & \next(\zero) \\ & \let{\nplus(a,b) = \next(\zero)} = & \nplus(a,b) \\ & \hyp{a = \next(\next(m))} = & \nplus(\next(\next(m)),b) \\ & \href{/posts/arithmetic-made-difficult/Plus.html#cor-plus-up-next} = & \next(\nplus(\next(m),b)) \\ & \href{/posts/arithmetic-made-difficult/Plus.html#cor-plus-up-next} = & \next(\next(\nplus(m,b))) \end{eqnarray*}$ so that $$\zero = \next(\nplus(m,b))$$, a contradiction.
• If $$a = \next(\zero)$$, then we have $$\nplus(\next(\zero),\zero) = \next(\zero) = \nplus(\next(\zero),b)$$, so that $$b = \zero$$ as claimed.
• If $$a = \zero$$, then $$\next(\zero) = \nplus(\zero,b) = b$$ as claimed.

Ain’t this fun!

Let $$a,b \in \nats$$. If $$\nplus(a,b) = \next(\next(\zero))$$, then $$(a,b)$$ is either $$(\zero,\next(\next(\zero)))$$ or $$(\next(\zero),\next(\zero))$$ or $$(\next(\next(\zero)),\zero)$$.

Note that either $$a = \zero$$, $$a = \next(\zero)$$, $$a = \next(\next(\zero))$$, or $$a = \next(\next(\next(m)))$$ for some $$m \in \nats$$.

• If $$a = \next(\next(\next(m)))$$ for some $$m \in \nats$$, we have $\begin{eqnarray*} & & \next(\next(\zero)) \\ & \let{\nplus(a,b) = \next(\next(\zero))} = & \nplus(a,b) \\ & \hyp{a = \next(\next(\next(m)))} = & \nplus(\next(\next(\next(m))),b) \\ & \href{/posts/arithmetic-made-difficult/Plus.html#cor-plus-up-next} = & \next(\nplus(\next(\next(m)),b)) \\ & \href{/posts/arithmetic-made-difficult/Plus.html#cor-plus-up-next} = & \next(\next(\nplus(\next(m),b))) \\ & \href{/posts/arithmetic-made-difficult/Plus.html#cor-plus-up-next} = & \next(\next(\next(\nplus(m,b)))) \end{eqnarray*}$ so that $$\zero = \next(\nplus(m,b))$$, a contradiction.
• If $$a = \next(\next(\zero))$$, then we have $\begin{eqnarray*} & & \nplus(\next(\next(\zero)),\zero) \\ & \href{/posts/arithmetic-made-difficult/Plus.html#thm-plus-zero-right} = & \next(\next(\zero)) \\ & = & \nplus(\next(\next(\zero)),b) \end{eqnarray*}$ so that $$b = \zero$$ as claimed.
• If $$a = \next(\zero)$$, then we have $\next(\next(\zero)) = \nplus(\next(\zero),b) = \next(\nplus(\zero,b)) = \next(b)$ so that $$b = \next(\zero)$$ as claimed.
• If $$a = \zero$$, then $$\next(\next(\zero)) = \nplus(\zero,b) = b$$ as claimed.

And another.

Let $$a,b \in \nats$$. If $$\nplus(a,b) = \next(\next(\next(\zero)))$$, then $$(a,b)$$ is either $$(\zero,\next(\next(\next(\zero))))$$ or $$(\next(\zero),\next(\next(\zero)))$$ or $$(\next(\next(\zero)),\next(\zero))$$, or $$(\next(\next(\next(\zero))),\zero)$$.

If $$a = \zero$$, then $$b = \next(\next(\next(\zero)))$$ as claimed.

Suppose instead that $$a = \next(m)$$ for some $$m$$. Now $\next(\nplus(m,b)) = \next(\next(\next(\zero))),$ so that $$\nplus(m,b) = \next(\next(\zero))$$. There are then three possibilities.

If $$m = \zero$$, then $$a = \next(\zero)$$ and $$b = \next(\next(\zero))$$.

If $$m = \next(\zero)$$, then $$a = \next(\next(\zero))$$ and $$b = \next(\zero)$$.

If $$m = \next(\next(\zero))$$, then $$a = \next(\next(\next(\zero)))$$ and $$b = \zero$$.

Okay this is boring. How about some equations with $$\ntimes$$?

Let $$a,b \in \nats$$. If $$\ntimes(a,b) = \zero$$, then either $$a = \zero$$ or $$b = \zero$$.

Suppose that neither $$a$$ nor $$b$$ is $$\zero$$; then we have $$a = \next(h)$$ and $$b = \next(k)$$ for some $$h,k \in \nats$$. Now $\begin{eqnarray*} & & \zero \\ & = & \ntimes(a,b) \\ & = & \ntimes(\next(h),\next(k)) \\ & \href{/posts/arithmetic-made-difficult/Times.html#cor-times-up-next} = & \nplus(\ntimes(h,\next(k)),\next(k)) \\ & \href{/posts/arithmetic-made-difficult/Plus.html#thm-plus-next-right} = & \next(\nplus(\ntimes(h,\next(k)),k)) \end{eqnarray*}$ a contradiction. So we must have either $$a = \zero$$ or $$b = \zero$$. Note also that in either case we indeed have $$\ntimes(a,b) = \zero$$.

woo

Let $$a,b \in \nats$$. If $$\ntimes(a,b) = \next(\zero)$$, then $$(a,b)$$ is $$(\next(\zero),\next(\zero))$$.

Suppose we have $$\ntimes(a,b) = \next(\zero)$$. Note that either $$a = \zero$$ or $$a = \next(\zero)$$ or $$a = \next(\next(m))$$ for some $$m \in \nats$$.

• If $$a = \zero$$ we have $$\next(\zero) = \ntimes(a,b) = \zero$$, a contradiction.
• If $$a = \next(\zero)$$, we have $\next(\zero) = \ntimes(\next(\zero),b) = b$ as claimed.
• If $$a = \next(\next(m))$$ for some $$m \in \nats$$, we have $\begin{eqnarray*} & & \next(\zero) \\ & = & \ntimes(a,b) \\ & = & \ntimes(\next(\next(m)),b) \\ & = & \nplus(b,\ntimes(\next(m),b)). \end{eqnarray*}$ There are now two possibilties for $$b$$. If $$b = \zero$$, we have $$\zero = \next(\zero)$$, a contradiction. If $$b = \next(k)$$ with $$k \in \nats$$, we have $\next(\zero) = \nplus(\next(k),\ntimes(\next(m),\next(k)))$ so that $\zero = \nplus(k,\ntimes(\next(m),\next(k))).$ But now we have $$k = \zero$$ and $\zero = \ntimes(\next(m),\next(\zero)) = \next(m),$ a contradiction.

One more:

Let $$a,b \in \nats$$. If $$\ntimes(a,b) = \next(\next(\zero))$$, then $$(a,b)$$ is either $$(\next(\next(\zero)),\next(\zero))$$ or $$(\next(\zero),\next(\next(\zero)))$$.

Suppose we have $$\ntimes(a,b) = \next(\next(\zero))$$. Note that either $$a = \zero$$ or $$a = \next(\zero)$$ or $$a = \next(\next(\zero))$$ or $$a = \next(\next(\next(m)))$$ for some $$m \in \nats$$.

• If $$a = \zero$$ we have $$\next(\zero) = \ntimes(a,b) = \zero$$, a contradiction.
• If $$a = \next(\zero)$$, we have $\next(\next(\zero)) = \ntimes(\next(\zero),b) = b$ as claimed.
• If $$a = \next(\next(\zero))$$, we have $\ntimes(\next(\next(\zero)),\next(\zero)) = \next(\next(\zero)) = \ntimes(\next(\next(\zero)),b)$ so that $$b = \next(\zero)$$ as claimed.
• Suppose we have $$a = \next(\next(\next(m)))$$ for some $$m \in \nats$$. Now $\begin{eqnarray*} & & \next(\next(\zero)) \\ & = & \ntimes(\next(\next(\next(m))),b) \\ & \href{/posts/arithmetic-made-difficult/Times.html#cor-times-up-next} = & \nplus(\ntimes(\next(\next(m)),b),b) \end{eqnarray*}$ If $$b = \zero$$, then we have $$\next(\next(\zero)) = \zero$$, a contradiction. Thus $$b = \next(k)$$ for some $$k \in \nats$$. This leaves two possibilities: $(\ntimes(\next(\next(m)),\next(k)),\next(k))$ is either $$(\next(\zero),\next(\zero))$$ or $$(\zero,\next(\next(\zero)))$$. If $$b = \next(\zero)$$, we have $$\next(\next(\zero)) = \next(\next(\next(m)))$$ for some $$m$$, a contradiction. If $$b = \next(\next(\zero))$$ and $$\ntimes(\next(\next(m)),b) = \zero$$, we have either $$b = \zero$$ (a contradiction) or $$\next(\next(m)) = \zero$$ (also a contradiction).

This is kind of ridiculous! But we’re doing number theory from first principles, so ridiculousness is kind of the point.