# 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 if and only if:

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

## Proof

### 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:

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

$\Box$

### 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.

$\blacksquare$