# Absolute values on Q (Part II)

Continuing on from absolute values on Q part I/ , I am going to show that the definition of p-adic absolute value satisfies the three properties of absolute values.

Property 1

Let $x=0$ and p be a fixed prime.  By definition then, $|x|_p = 0$.  This satisfies the first property ($|x|_p=0$ if and only if $x=0$).

Suppose now $x \ne 0$.  Then $|x|_p = p^{-ord_p(x)}$.  Using Conjecture 1, $|x|_p=|p^n \cdot a|_p =p^{-n}$.  Since $p^{-n}$ cannot be 0, $p^{-n} > 0$.

Property 2

Let $x=0$ and $y \ne 0$ and p be a fixed prime.  Then $|x \cdot y|_p = |0|_p = 0 = |x|_p \cdot |y|_p$.

Suppose $|x|_p = |p^n \cdot a|$ and $|y|_p = |p^n \cdot b|$.  In other words, x and y have the same exponent on p. Then, by definition,

$|x \cdot y|_p = |p^n \cdot a \cdot p^n \cdot b|_p = |p^{2n} \cdot ab|_p = p^{-2n}$.

By the second property,

$|x \cdot y|_p = |x|_p \cdot |y|_p = |p^n c\dot a|_p \cdot |p^n \cdot b|_p = p^{-n}p{-n}=p{-2n}$

Suppose x and y have different exponents on the p.  $|x|_p=|p^n \cdot a|_p and |y|_p=|p^m cdot b|$.  Multiply x by y….

$|xy|_p=|p^n \cdot a \cdot p^m \cdot b|_p=|p^{n+m} \cdot ab|_p=p^{-n-m}$

By the second property,

$|xy|_p=|x|_p \cdot |y|_p = p^{-n} \cdot p^{-m}=p^{-n-m}$.

Property 3

Let $x=y=0$ and p be a fixed prime.  Then by definition $|x+y|_p = 0$.  This partly satisfies the property for $0 \le 0+0$  Now consider $x \ne 0$.  Then $|x+y|_p = |x|_p = |p^n \cdot a|_p = p^{-n}$.    This contributes to satisfying the property for $p^{-n} \le p^{-n} + 0$.

Now let x and y have the same exponent on p.

$|x+y|_p = |p^n \cdot a + p^n \cdot b|_p = |p^n(a+b)|_p=p^{-n}$

If the definition satisfies the property, then $p^{-n}$ should be less than $|x|_p+|y|_p$.  In fact this is true because $|x|_p+|y|_p = p^{-n}+p^{-n}$ which is larger than $p^{-n}$.

Last case is when x and y have different exponents on p.

$|x+y|_p=|p^n \cdot a + p^m \cdot b|_p=p^{-n}$ where $n<m$

Now consider property 3.

$|x|_p+|y|_p=p^{-n}+p^{-m}$

Now that p-adic absolute values do satisfy the properties of absolute values, there is one more thing to look at:  Archimedean absolute values.  A little warning for you, the following definition is counter-intuitive.

Definition 2:  An absolute value on Q is said to be non-Archimedean if the properties of absolute values are satisfied along with an additional property:

$|x+y| \le max(|x|,|y|)$ for all x,y \in Q.

Absolute values that satisfy the three properties but not the fourth are said to be Archimedean.

Theorem 1:  p-adic absolute values are non-Archimedean.

It has already been shown that p-adic absolute values satisfy the first three conditions.  Now all that needs to be shown is that p-adic absolute values satisfy the fourth property.  Assume p is a fixed prime

Case 1: $x=y=0$

$|x+y|_p = |0|_p = 0 \le max(|x|_p,|y|_p) = 0$

Case 2: $x \ne 0, y=0$

$|x+y|_p = |x|_p = p^{-n} \le max(|x|_p,|y|_p) = max(p^{-n}, 0) = p^{-n}$

Case 3:  x and y have the same exponent on p

$|x+y|_p = p^{-n} \le max(|x|_p,|y|_p) = max(p^{-n}, p^{-n}) = p^{-n}$

Case 4:  x and y have different exponents on p (n<m)

$|x+y|_p = p^{-n} \le max(|x|_p,|y|_p) = max(p^{-n}, p^{-m}) = p^{-m}$

QED