Prove that the sum is equal to -1.

[evinda]

Hello! (Smile)

I want to prove that at the field $\mathbb{Q}_p$, where $p$ is a prime, it stands:

$$-1=\sum_{0}^{\infty} (p-1)p^i$$

That's what I have tried so far:$$-1=\sum_{i=0}^{\infty}(p-1)p^i =(p-1)+(p-1)p+(p-1)p^2+\dots \\ \Rightarrow 1-1=1+p-1+p^2-p+p^2-p^2+\dots \\ \Rightarrow 0=(1-1)+(p-p)+(p^2-p^2)+\dots$$

How can I continue? (Thinking)

 

[Euge]

Hello! (Smile)

I want to prove that at the field $\mathbb{Q}_p$, where $p$ is a prime, it stands:

$$-1=\sum_{0}^{\infty} (p-1)p^i$$

That's what I have tried so far:$$-1=\sum_{i=0}^{\infty}(p-1)p^i =(p-1)+(p-1)p+(p-1)p^2+\dots \\ \Rightarrow 1-1=1+p-1+p^2-p+p^2-p^2+\dots \\ \Rightarrow 0=(1-1)+(p-p)+(p^2-p^2)+\dots$$

How can I continue? (Thinking)

Hi evinda,

I think you're misunderstanding the meaning of the equation $-1 = \sum_0^\infty (p-1)p^i$. This is an equation in $\Bbb Q_p$, so to show that the formula holds, you need to prove

$$ \left|\sum_0^N (p-1)p^i - (-1)\right|_p \to 0$$ as $$N \to \infty.$$

Here, $|\cdot|_p$ denotes the $p$-adic norm on $\Bbb Q_p$. For all nonnegative integers $N$,

$$ \sum_0^N (p - 1)p^i = \sum_0^N (p^{i+1} - p^i) = p^{N+1} - 1.$$

Hence,

$$ \left|\sum_0^N (p-1)p^i - (-1)\right|_p = |p^{N+1}|_p = p^{-(N+1)} \to 0$$ as $$ N \to \infty.$$

 

[evinda]

Hi evinda,

I think you're misunderstanding the meaning of the equation $-1 = \sum_0^\infty (p-1)p^i$. This is an equation in $\Bbb Q_p$, so to show that the formula holds, you need to prove

$$ \left|\sum_0^N (p-1)p^i - (-1)\right|_p \to 0$$ as $$N \to \infty.$$

Here, $|\cdot|_p$ denotes the $p$-adic norm on $\Bbb Q_p$. For all nonnegative integers $N$,

$$ \sum_0^N (p - 1)p^i = \sum_0^N (p^{i+1} - p^i) = p^{N+1} - 1.$$

Hence,

$$ \left|\sum_0^N (p-1)p^i - (-1)\right|_p = |p^{N+1}|_p = p^{-(N+1)} \to 0$$ as $$ N \to \infty.$$

So, do we use this definition?

A $p-$ norm of $\mathbb{Q}_p$ is a function $||_p: \mathbb{Q}_p \to \mathbb{R}$

$$x \neq 0, x=p^{w_p(x)}u \mapsto p^{-w_p(x)}$$

$$\text{ For } x=0 \Leftrightarrow w_p(x)=\infty \\ p^{-\infty}=:0$$

Is so, is it $w_p(x)=N+1$ and $u=1$ ? (Worried)

 

[Euge]

So, do we use this definition?

A $p-$ norm of $\mathbb{Q}_p$ is a function $||_p: \mathbb{Q}_p \to \mathbb{R}$

$$x \neq 0, x=p^{w_p(x)}u \mapsto p^{-w_p(x)}$$

$$\text{ For } x=0 \Leftrightarrow w_p(x)=\infty \\ p^{-\infty}=:0$$

Is so, is it $w_p(x)=N+1$ and $u=1$ ? (Worried)

The $p$-adic norm on $\Bbb Q_p$ is defined by a map $|\cdot|_p : \Bbb Q \to \{p^n\, |\, n\in \Bbb Z\} \cup \{0\}$ such that $|0|_p = 0$ and $|x|_p = p^{-\text{ord}_p(x)}$ for $x \neq 0$. When $x = p^{N+1}$, $\text{ord}_p(x) = N+1$ and thus $|x|_p = p^{-(N+1)}$.

 

[evinda]

The $p$-adic norm on $\Bbb Q_p$ is defined by a map $|\cdot|_p : \Bbb Q \to \{p^n\, |\, n\in \Bbb Z\} \cup \{0\}$ such that $|0|_p = 0$ and $|x|_p = p^{-\text{ord}_p(x)}$ for $x \neq 0$. When $x = p^{N+1}$, $\text{ord}_p(x) = N+1$ and thus $|x|_p = p^{-(N+1)}$.

I understand! (Nod) Could you also give me an other example, with an other $x$, at which I can use the difinition of the $p-$adic norm? (Thinking)

 

[Euge]

I understand! (Nod) Could you also give me an other example, with an other $x$, at which I can use the difinition of the $p-$adic norm? (Thinking)

Sure. Let $p$ be a prime and let $n$ be an integer not divisible by $p$. Set $x = \frac{n}{p^2}$. Then

$$\text{ord}_p(x) = \text{ord}_p(n) - \text{ord}_p(p^2) = 0 - 2 = -2.$$

Hence, $|x|_p = p^{-\text{ord}_p(x)} = p^2$.

 

[evinda]

Sure. Let $p$ be a prime and let $n$ be an integer not divisible by $p$. Set $x = \frac{n}{p^2}$. Then

$$\text{ord}_p(x) = \text{ord}_p(n) - \text{ord}_p(p^2) = 0 - 2 = -2.$$

Hence, $|x|_p = p^{-\text{ord}_p(x)} = p^2$.

Why do we take the difference $\text{ord}_p(n) - \text{ord}_p(p^2)$ ? (Thinking)

 

[Euge]

Why do we take the difference $\text{ord}_p(n) - \text{ord}_p(p^2)$ ? (Thinking)

Recall that $\text{ord}_p(x)$ is by definition the highest power of $p$ which divides $x$ when $x\in \Bbb Z$ and $\text{ord}_p(a) - \text{ord}_p(b)$ when $x$ is of the form $\frac{a}{b}$, $a, b\in \Bbb Z$, $b \neq 0$. So in the case $x = \frac{n}{p^2}$, we have by definition $\text{ord}_p(x) = \text{ord}_p(n) - \text{ord}_p(p^2)$.