Equivalent Norms on Rational Numbers

Theorem
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be norms on the rational numbers $\Q$.

Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent :
 * $\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$

Necessary Condition
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be equivalent.

By Norm is Power of Other Norm then:
 * $\exists \alpha \in \R_{\gt 0}: \forall q \in \Q: \norm q_1 = \norm q_2^\alpha$

In particular:
 * $\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$

Sufficient Condition
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
 * $\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$

By Norm of Negative then:
 * $\forall n \in \N: \norm {-n}_1 = \norm n_1 = \norm n_2^\alpha =\norm {-n}_2^\alpha$

Hence:
 * $\forall k \in \Z: \norm k_1 = \norm k_2^\alpha$

By Norm of Quotient then:
 * $\forall \dfrac a b \in \Q: \norm {\dfrac a b}_1 = \dfrac {\norm a_1} {\norm b_1} = \dfrac {\norm a_2^\alpha} {\norm b_2^\alpha} = \norm {\dfrac a b}_2^\alpha$

By Norm is Power of Other Norm then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent.