Product Formula for Norms on Non-zero Rationals

Theorem
Let $\Q_{\ne 0}$ be the set of non-zero rational numbers.

Let $a \in \Q_{\ne 0}$.

Then:
 * $\size a \times \displaystyle\prod_{p \mbox{ } prime}^{} \norm a_p = 1$

where:
 * $\size {\,\cdot\,}$ is the absolute value on $\Q$
 * $\norm {\,\cdot\,}_p$ is the $p$-adic norm on $\Q$ for prime number $p$

Proof
First it is shown that the theorem holds for any positive integer.

Let $a \in \Z_{\gt 0}$.

By Fundamental Theorem of Arithmetic then we can factor $a$ as a product of one or more primes:
 * $a = p_1^{b_1} p_2^{b_2} \dots p_k^{b_k}$

Then:
 * $\begin{cases}

\norm a_q \, = \, 1 & \mbox {if } q \ne p_i \,\,\forall i = 1, \dots ,k\\ \\ \norm a_{p_i} = \, p_i^{-b_i} & \mbox {for } i = 1, \dots ,k \\ \\ \,\,\size a \,\,\,\, = \, p_1^{b_1} p_2^{b_2} \dots p_k^{b_k} \\ \end{cases}$

Hence:
 * $\size a \times \displaystyle\prod_{p \mbox{ } prime}^{} \norm a_p = \paren{p_1^{b_1} p_2^{b_2} \dots p_k^{b_k}} \times \paren{p_1^{-b_1} p_2^{-b_2} \dots p_k^{-b_k}} = 1$

Let $a \in \Z_{\lt 0}$.

So $-a \in \Z_{\gt 0}$.

By Norm of Negative then:
 * $\displaystyle \size a \times \prod_{p \mbox{ } prime}^{} \norm a_p = \size {-a} \times \prod_{p \mbox{ } prime}^{} \norm {-a}_p = 1$

So for all $a \in \Z_{\ne 0}$ then:
 * $\size a \times \displaystyle\prod_{p \mbox{ } prime}^{} \norm a_p = 1$

Let $\dfrac b c \in \Q_{\ne 0}$, where $b, c \in \Z_{\ne 0}$.

By Norm of Quotient then:
 * $\size {\dfrac b c} \times \displaystyle\prod_{p \mbox{ } prime}^{} \norm {\dfrac b c}_p = \dfrac {\size b} {\size c} \times \displaystyle\prod_{p \mbox{ } prime}^{} \dfrac {\norm b_p}{\norm c_p} = \dfrac {\size b \times \displaystyle\prod_{p \mbox{ } prime}^{} \norm b_p} {\size c \times \displaystyle\prod_{p \mbox{ } prime}^{} \norm c_p} = \dfrac 1 1 = 1$