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Let $\mathbb{Z_p}$ denotes the set of $p$-adic integer. Then any element in it is of the from $(....,x_n,....,x_2,x_1)$ where $x_n \in \mathbb{Z/}p^n\mathbb{Z}$ ($p$= prime) and having a homomorphic map between any two adjacent elements.

Now we define $\mathbb{Q}_p$ to be the fraction field of $\mathbb{Z_p}$ as $ = \left\lbrace\frac{x}{y} \space \bigg{|} \space x,y \in \mathbb{Z_p} , y\neq 0 \right\rbrace$

So are the elements in $\mathbb{Q}_p$ of the from $\Big(.... , \frac{x_n}{y_n} ,...., \frac{x_2}{y_2} , \frac{x_1}{y_1} \Big )$? If not then how can we express them?

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    @JeanMarie My bad! I fixed it..Thanks.2017-02-18

1 Answers 1

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You can express them as Cauchy sequences (in the $ p $-adic norm) of rationals, since $ \mathbf Q $ is dense in $ \mathbf Q_p $. Alternatively, you may note that since $ \mathbf Q_p $ is the fraction field of $ \mathbf Z_p $, any $ p $-adic number is expressible as a ratio of $ p $-adic integers; and since any $ p $-adic integer can be expressed as $ p^k u $ where $ u $ is a $ p $-adic unit and $ k \geq 0 $, it follows that you may write a $ p $-adic number in the form $ p^k u $, where $ k $ is not restricted to be greater than $ 0 $, and $ u $ remains a $ p $-adic unit. In effect, you only need powers of $ p $ in the denominator. If you phrase this in terms of base $ p $ expansions, a general $ p $-adic number has the form (with the $ c_i $ lying in a fixed set of representatives for the residue class field $ \mathbf Z/p \mathbf Z $)

$$ \ldots c_2 c_1 c_0.c_{-1} c_{-2} \ldots c_{-n} $$

For a $ p $-adic integer, we have $ n = 0 $.

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    This is definitely my preferred way of looking at elements of $\Bbb Q_p$.2017-02-18