Dual wavelet

In number theory, Ostrowski's theorem, due to Alexander Ostrowski (1916), states that every non-trivial absolute value on the rational numbers Q is equivalent to either the usual real absolute value or a p-adic absolute value.

Definitions

Two absolute values $|\cdot |$ and $|\cdot |_{\ast }$ on a field K are defined to be equivalent if there exists a real number c > 0 such that

$|x|_{\ast }=|x|^{c}{\text{ for all }}x\in \mathbf {K} .$

The trivial absolute value on any field K is defined to be

$|x|_{0}:={\begin{cases}0,&{\text{if }}x=0\\1,&{\text{if }}x\neq 0.\end{cases}}$

The real absolute value on the rationals Q is the normal absolute value on the reals, defined to be

$|x|_{\infty }:={\begin{cases}x,&{\text{if }}x\geq 0\\-x,&{\text{if }}x<0.\end{cases}}$

This is sometimes written with a subscript 1 instead of infinity.

For a prime number p, the p-adic absolute value on Q is defined as follows: any non-zero rational x, can be written uniquely as $x=p^{n}{\dfrac {a}{b}}$ with a, b and p pairwise coprime and $n\in \mathbf {Z}$ some integer; so we define

$|x|_{p}:={\begin{cases}0,&{\text{if }}x=0\\p^{-n},&{\text{if }}x\neq 0.\end{cases}}$

Proof

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This is one of the only things that require you to spend a little money to make money. Just go make an account, get a paypal account, and start selling. To go one step beyond just affiliating products and services is to create your own and sell it through your blog. Not great if you really enjoy trying out all the themes. Talking in real time having a real person causes it to be personal helping me personally to sort out how to proceed. The first step I took was search for a discount code, as I did with HostGator. Using a HostGator coupon is a beneficial method to get started. As long as the necessities are able to preserve the horizontal functionality of your site, you would pretty much be fine. Consider a non-trivial absolute value on the rationals $(\mathbf {Q} ,|\cdot |_{\ast })$. We consider two cases, (i) $\exists {n\in \mathbf {N} },|n|_{\ast }>1$ and (ii) $\forall {n\in \mathbf {N} },|n|_{\ast }\leq 1$. It suffices for us to consider the valuation of integers greater than one. For if we find some $c\in \mathbf {R} ^{+}$ for which $|n|_{\ast }=|n|_{\ast \ast }^{c}$ for all naturals greater than one; then this relation trivially holds for 0 and 1, and for positive rationals $|m/n|_{\ast }=|m|_{\ast }/|n|_{\ast }=|m|_{\ast \ast }^{c}/|n|_{\ast \ast }^{c}=(|m|_{\ast \ast }/|n|_{\ast \ast })^{c}=|m/n|_{\ast \ast }^{c}$; and for negative rationals $|{-}x|_{\ast }=|x|_{\ast }=|x|_{\infty }^{c}=|{-}x|_{\infty }^{c}$.

Case I: ∃n ∈ N   |n| > 1

Consider the following calculation. Let $a,b\in \mathbf {N} ,>1$. Let $n\in \mathbf {N} ,>0$. Expressing bⁿ in base a yields $b^{n}=\Sigma _{i{<}m}c_{i}a^{i}$, where each $c_{i}\in \{0,1,\ldots ,a-1\}$ and $m\leq n\log b/\log a+1$. Then we see, by the properties of an absolute value:

${\begin{aligned}|b|_{\ast }^{n}=|b^{n}|_{\ast }&\leq am\max\{|a|_{\ast }^{m},1\}\\&\leq a(n\log _{a}b+1)\max\{|a|_{\ast }^{n\log _{a}b},1\}\\\Rightarrow |b|_{\ast }&\leq \underbrace {{\big (}a(n\log _{a}b+1){\big )}^{\frac {1}{n}}} _{\to 1{\text{ as }}n\to \infty }\max\{|a|_{\ast }^{\log _{a}b},1\}\\\Rightarrow |b|_{\ast }&\leq \max\{|a|_{\ast }^{\log _{a}b},1\}.\end{aligned}}$

Now choose $b\in \mathbf {N} ,>1$ such that $|b|_{\ast }>1$.Using this in the above ensures that $|a|_{\ast }>1$ regardless of the choice of a (else $|a|_{\ast }^{\log _{a}b}\leq 1$ implying $|b|_{\ast }\leq 1$). Thus for any choice of a, b > 1 above, we get $|b|_{\ast }\leq |a|_{\ast }^{\log b/\log a}$, i.e. $\log |b|_{\ast }/\log b\leq \log |a|_{\ast }/\log a$. By symmetry, this inequality is an equality.

Since a, b were arbitrary, there is a constant, $\lambda \in \mathbf {R} ^{+}$ for which $\log |n|_{\ast }=\lambda \log n$, i.e. $|n|_{\ast }=n^{\lambda }=|n|_{\infty }^{\lambda }$ for all naturals n > 1. As per the above remarks, we easily see that for all rationals, $|x|_{\ast }=|x|_{\infty }^{\lambda }$, thus demonstrating equivalence to the real absolute value.

Case II: ∀n ∈ N   |n| ≤ 1

As this valuation is non-trivial, there must be a natural number for which $|n|_{\ast }<1$. Factorising this natural, $n=\Pi _{i<r}p_{i}^{e_{i}}$ yields $|p|_{\ast }$ must be less than 1, for at least one of the prime factors p = p_j. We claim than in fact, that this is so for only one.

Suppose per contra that $p,q$ are distinct primes with absolute value less than 1. First, let $e\in \mathbf {N} ^{+}$ be such that $|p|_{\ast }^{e},|q|_{\ast }^{e}<1/2$. By the Euclidean algorithm, let $m,n\in \mathbf {Z}$ be integers for which $mp^{e}+nq^{e}=1$. This yields $1=|1|_{\ast }\leq |m|_{\ast }|p|_{\ast }^{e}+|n|_{\ast }|q|_{\ast }^{e}<{\frac {|m|_{\ast }+|n|_{\ast }}{2}}\leq 1$, a contradiction.

So must have $|p|_{\ast }=\alpha <1$ for some prime, and $|q|_{\ast }=1$ all other primes. Letting $c=-\log \alpha /\log p$, we see that for general positive naturals $n=\Pi _{i<r}p_{i}^{e_{i}}$; $|n|_{\ast }=\Pi _{i<r}|p_{i}|_{\ast }^{e_{i}}=|p_{j}|_{\ast }^{e_{j}}=(p^{-e_{j}})^{c}=|n|_{p}^{c}$. As per the above remarks we see that $|x|_{\ast }=|x|_{p}^{c}$ all rationals, implying the absolute value is equivalent to the p-adic one. Template:NumBlk

One can also show a stronger conclusion, namely that $|\cdot |_{\ast }:\mathbf {Q} \to \mathbf {R}$ is a nontrivial absolute value if and only if either $|\cdot |_{\ast }=|\cdot |_{\infty }^{c}$ for some $c\in (0,1]$ or $|\cdot |_{\ast }=|\cdot |_{p}^{c}$ for some $c\in (0,\infty ),p\in \mathbf {P}$.

Another Ostrowski's theorem

Another theorem states that any field, complete with respect to an archimedean absolute value, is (algebraically and topologically) isomorphic to either the real numbers or the complex numbers. This is sometimes also referred to as Ostrowski's theorem.

See also

• Valuation (algebra)
• Absolute value in general

References

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