# Norms Equivalent to Absolute Value on Rational Numbers

## Theorem

Let $\alpha \in \R_{\gt 0}$.

Let $\norm{\,\cdot\,}:\Q \to \R$ be the mapping defined by:

$\forall x \in \Q: \norm{x} = \size {x}^\alpha$

where $\size {x}$ is the absolute value of $x$ in $\Q$.

Then:

$\norm{\,\cdot\,}$ is a norm on $\Q$ if and only if $\,\,\alpha \le 1$

## Proof

### Necessary Condition

The contrapositive is proved.

Let $\alpha \gt 1$.

The norm axiom (N3) (Triangle Inequality) is not satisfied:

 $\displaystyle \norm {1 + 1}$ $=$ $\displaystyle \size {1 + 1}^\alpha$ $\displaystyle$ $=$ $\displaystyle 2^\alpha$ $\displaystyle$ $\gt$ $\displaystyle 2$ Real Power Function on Base Greater than One is Strictly Increasing $\displaystyle$ $=$ $\displaystyle \size{1}^\alpha + \size{1}^\alpha$ $\displaystyle$ $=$ $\displaystyle \norm{1} + \norm{1}$

By Rule of Transposition the result follows.

$\Box$

### Sufficient Condition

Suppose $\alpha \le 1$.

It is shown that $\norm{\,\cdot\,}$ satisfies the norm axioms (N1)-(N3).

#### (N1) Positive Definiteness

Let $x \in \Q$.

 $\displaystyle \norm {x} = 0$ $\iff$ $\displaystyle \size {x}^\alpha = 0$ Definition of $\norm{\,\cdot\,}$ $\displaystyle$ $\iff$ $\displaystyle \size {x} = 0$ Definition of $a$ to the power of $r$ for $a \in \R_{\ge 0}$ and $r \in \R_{\gt 0}$ $\displaystyle$ $\iff$ $\displaystyle x = 0$ Absolute Value is Norm and norm axiom (N1) (Positive Definiteness)

$\Box$

#### (N2) Multiplicativity

Let $x, y \in \Q$.

Then:

 $\displaystyle \norm {x y}$ $=$ $\displaystyle \size {x y}^\alpha$ Definition of $\norm{\,\cdot\,}$ $\displaystyle$ $=$ $\displaystyle \paren {\size {x} \size {y} }^\alpha$ Absolute Value is Norm and norm axiom (N2) (Multiplicativity) $\displaystyle$ $=$ $\displaystyle \size {x}^\alpha \size {y}^\alpha$ Power of product $\displaystyle$ $=$ $\displaystyle \norm {x} \norm {y}$ Definition of $\norm{\,\cdot\,}$

$\Box$

#### (N3) Triangle Inequality

Let $x, y \in \Q$.

Without loss of generality let $\norm y \lt \norm x$.

If $\norm x = 0$ then $\norm y = 0$.

By (N1) above, $x = y = 0$.

Hence:

 $\displaystyle \norm{x + y}$ $=$ $\displaystyle \norm 0$ $\displaystyle$ $=$ $\displaystyle 0$ $\displaystyle$ $=$ $\displaystyle 0 + 0$ $\displaystyle$ $=$ $\displaystyle \norm x + \norm y$

If $\norm x \gt 0$ then:

$\norm x \gt 0 \iff \size {x}^\alpha \gt 0 \iff \size x \gt 0$

Hence:

 $\displaystyle \norm{x + y}$ $=$ $\displaystyle \size{x + y}^\alpha$ $\displaystyle$ $\le$ $\displaystyle \paren{\size x + \size y}^\alpha$ Norm axiom (N3) (Triangle Inequality) $\displaystyle$ $=$ $\displaystyle \size {x}^\alpha \paren{1 + \dfrac {\size y} {\size x} }^\alpha$ $\displaystyle$ $\le$ $\displaystyle \size {x}^\alpha \paren{1 + \dfrac {\size y} {\size x} }$ Real Power Function on base Greater than One is Strictly Increasing $\displaystyle$ $\le$ $\displaystyle \size {x}^\alpha \paren{1 + \dfrac {\size {y}^\alpha} {\size {x}^\alpha} }$ Real Power Function on base between Zero and One is Strictly Decreasing $\displaystyle$ $\le$ $\displaystyle \size {x}^\alpha + \size {y}^\alpha$ $\displaystyle$ $=$ $\displaystyle \norm {x} + \norm {y}$

$\blacksquare$