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The title is exercise 2.2 in The Fundamental Theorem of Algebra.

The hint for the problem is: Find the value of $\frac{1}{3}$ in $\mathbb{Z}_{13}$

(please realize that my knowledge of the subject is what I read in Chapter 2)

I have gotten that $x = -\frac{1}{3}$.

I know that $\mathbb{Z}_{13}$ is the integers modulo 13. Thus $x \equiv n \pmod{13}$? for some integer n, $0\le n < 13$.

Thus for $x – n = km$ for some integer $k$, with $m \neq 0$ and an integer.

How does m divide $(x-n)$, when $(x-n)$ is not an integer, if $x = -\frac{1}{3}$????

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6 Answers 6

up vote 4 down vote accepted

The notation can indeed be confusing the first few times one encounters in. When one says $x=-1/3$ in the context of $\mathbb{Z}_{13}$, it is simply short-hand for saying that $x$ is some number from $0$ to $12$ such that $-3x \equiv 1 \pmod{13}.$ So remember that $x$ is not the usual real number $-1/3$, but we use this notation because it behaves like $-1/3$ would, in that $-3x=1$ in $\mathbb{Z}_{13}.$

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thanks. I was getting very lost because of the (1/3), a fraction. I have looked on wikipedia and rutgers, but neither used method you explained. thanks alot. –  yiyi Jun 17 '12 at 15:50
1  
@MaoYiyi You are welcome. For some practice, you should find out what $1/4$ is in $\mathbb{Z}_7$ and what is $-1/2$ in $\mathbb{Z}_5$? Also see what is $1/2$ in $\mathbb{Z}_{16}.$ Is there more than one answer there? –  Ragib Zaman Jun 17 '12 at 15:53

Hint $\rm\ mod\,\ 3n\!+\!1\!:\ -1\equiv 3n\:\Rightarrow\:\dfrac{-1}3\,\equiv\, \dfrac{3n}3 \,\equiv\, n$

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What you could have done also $$ 3x + 7 = 6\pmod{13} \implies 3x = -1 \pmod{13} \implies 3x = 12 \pmod{13}$$ At this point it should be pretty easy!

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Note that $3x\equiv -1$ iff $3x\equiv 12$ modulo $13$. From there, it should be simple.

$x=-1/3$ only in the sense that it is the member of $\Bbb Z_{13}$ such that $3x\equiv -1$ modulo $13$. This is not the same as saying $x=-1/3$ in the sense of real numbers.

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how is that 1/3? in mod 13? Modular arithmetic doesn't have fractions. –  yiyi Jun 17 '12 at 15:48
    
True, there are no fractions, here. However, there are multiplicative inverses. Perhaps it would be better if the book had said "find the value of $3^{-1}$ in $\Bbb Z_{13}$", instead? Moreover, since $13$ is a prime number, then $\Bbb Z_{13}$ is a field, and so every non-$0$ element of $\Bbb Z_{13}$ has a unique multiplicative inverse. We can't do this in $\Bbb Z_n$ for general $n$. –  Cameron Buie Jun 17 '12 at 15:52

We look for an integer $x$ so that 3x=-1 mod 13. This is the same as $3x+1=0 \mod 13$ or $13 | 3x+1$. Now in this case we can just run through the integers $0\le x\le 12$ to find the answer $x=4 \mod 13$.

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You might point out that just "running through" the integers $2\leq p-1$ isn't a great plan for general primes $p$, as $p$ could be rather large.... –  Cameron Buie Jun 17 '12 at 15:57

Hint: what is $-1 \pmod {13}$ in the normal set of representatives $0 \le n \lt 13$?

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