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I am reading the wiki page for p-adic numbers and it states that they are a field extension of the rationals so each member has to have a modular multiplicative inverse. So how would I take the inverse of, say, 35 in the ring of 2-adic numbers?

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To understand the $2$-adics better, first try finding the multiplicative inverse of $-7$. Note that $7=1-8$, and use the geometric series. –  g.castro Dec 3 '12 at 17:30
    
Long division... –  GEdgar Dec 3 '12 at 18:35
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4 Answers 4

up vote 5 down vote accepted

Long division. $1$ divided by $35$. First, $35$ base $2$ is $100011.$ So the problem is:

C10

Look at the right-most digits. $1$ goes into $1$ how many times? $1$:

C9

Multiply:

C8

Subtract:

C7

Next digit is $1$, so $1$ goes in the quotient:

C6

Multiply:

C5

Subtract:

C4

Next digit is zero, $0$ goes in the quotient. Next digit is $1$. Multiply:
C3

Subtract:

C2

Three zeros, then a 1:

C1

Continue. It is eventually periodic, of course.

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Following @RobertIsrael, I say: $35=1+34$. Using the geometric series for $1/(1+x)$, $$ 1/35=1-34+34^2-34^3+\cdots\,, $$ a $2$-adically convergent series.

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Another way: note that $2^{12} \equiv 1 \text{ mod }35$. Now $(2^{12} - 1)/35 = 117 = 1 + 2^2 + 2^4+2^5+2^6$ so $$ \dfrac{1}{35} = -\frac{117}{1-2^{12}} = - \sum_{j=0}^\infty\sum_{k \in \{0,2,4,5,6\}} 2^{12 j+k} = 1 + \sum_{j=0}^\infty\sum_{k \in \{1,3,7,8,9,10,11\}} 2^{12j+k}$$

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With Mathematica, the following command will provide you an answer for 20 digits :

BaseForm[PowerMod[35, -1, 2^20], 2]

= 10001010111110001011 (base 2)

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