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Multiplication and Exponentiation (Power) on Natural Numbers – Set Theory

Hi, infinite-set readers. Set theory is back.

The previous post we have discussed about the definition of addition on natural numbers. You can visit this link to read the previous post. This time, we will be discuss about the definition of the multiplication of two natural numbers. But before, the theorem will be given as a tool for defining the multiplication of two natural numbers. Happy reading.

(x) Theorem :

For any n\in\omega there is a single function P_{n}:\omega\rightarrow\omega such that :

    \[ P_{n}\left(0\right)=0\mbox{ and }P_{n}\left(x^{+}\right)=P_{n}\left(x\right)+n\]

Proof :

By taking E=\omega, e=0, g=P_{n}, and f is function from \omega to \omega with f\left(y\right)=y+n for all y\in\omega then by Finite Recursion Theorem, this theorem is proven.

The above Theorem (x) then becomes a tool for defining the multiplication of natural numbers.

Definition :

Given any n,m\in\omega and P_{n} are functions defined in the Theorem (x). Defined n.m with n.m=P_{n}\left(m\right).

In the defined of sum and multiplication above (and previous post), it applies associative, commutative, and distributive. But in this post will not be proved these. You can try it yourself to prove the above definition applies these characteristic.

Next will be discussed about the definition of the exponentiation (power) on natural numbers. But before that, will be given the theorem as a tool to define the powers.

(y) Theorem :

For any n\in\omega there is a single function H_{n}:\omega\rightarrow\omega such that :

    \[ H_{n}\left(0\right)=1\mbox{ and }H_{n}\left(x^{+}\right)=H_{n}\left(x\right).n\]

Proof : (more…)