Identity Function is Completely Multiplicative
Jump to navigation
Jump to search
Theorem
Let $I_{\Z_{>0}}: \Z_{>0} \to \Z_{>0}$ be the identity function:
- $\forall n \in \Z_{>0}: I_{\Z_{>0}} \left({n}\right) = n$
Then $I_{\Z_{>0}}$ is completely multiplicative.
The validity of the material on this page is questionable. In particular: Completely multiplicativity is defined for fields but $\Z_{>0}$ is not a field You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Proof
- $\forall m, n \in \Z_{>0}: I_{\Z_{>0}} \left({m n}\right) = m n = I_{\Z_{>0}} \left({m}\right) I_{\Z_{>0}} \left({n}\right)$
$\blacksquare$