Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $p$ be a prime number. Define the p-adic absolute value function $|\cdot|_{p}$ on $\mathbb{Q}$:

$|x|_{p}=\left\{ \begin{array}{ll} 0 & \text{if }x=0\\ p^{-k} & \text{if }x=p^{k}\frac{m}{n}\text{ and }\gcd(p,mn)=1\end{array}\right.$

where $m,n\in\mathbb{Z\setminus}\{0\}$ and $p\nmid m$ and $p\nmid n$. Show that for $x,y\in\mathbb{Q}$,

$|x+y|_p \leq max\left\{ |x|_{p},|y|_{p}\right\}$

How do I express $|x+y|_p$?

share|cite|improve this question

If $x$ or $y$ are equal to $0$, then the inequality is easy.

Assume $xy\neq 0$. Write $x = p^a\frac{r}{s}$ and $y=p^b\frac{t}{u}$, where $\gcd(p,rstu)=1$, and $a$ and $b$ are integers; that is, $|x|_p = p^{-a}$, $|y|_p = p^{-b}$.

Assume, without loss of generality, that $a\leq b$. Then $b-a\geq 0$, and $$\begin{align*} x+y &= p^a\frac{r}{s} + p^b\frac{t}{u}\\ &= p^a\left(\frac{r}{s} + p^{b-a}\frac{t}{u}\right)\\ &= p^a\left(\frac{ru + p^{b-a}ts}{su}\right). \end{align*} $$ Can you take it from here? You'll have two cases, depending on whether $a\lt b$ or $a=b$.

share|cite|improve this answer

Suppose that $x = p^{k_x}\frac{m_x}{n_x}$ and $y = p^{k_y}\frac{m_y}{n_y}$; then $x+y = p^{k_x}\frac{m_x}{n_x} + p^{k_y}\frac{m_y}{n_y}$. Suppose that $k_x \le k_y$; then $x+y = p^{k_x}\left(\frac{m_x}{n_x} + p^{k_y-k_x}\frac{m_y}{n_y} \right) = p^{k_x}\frac{m_xn_y+p^{k_y-k_x}m_yn_x}{n_xn_y}$. Does the denominator of that fraction have $p$ as a factor? If not, what can you conclude about the number of factors of $p$ in $x+y$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.