Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x = -\frac{1}{3}$????

share|improve this question
$3\times4=-1\pmod{13}$. –  Did Jun 17 '12 at 15:45
@did I get that, but not seeing the connection to x=-1/3? Could you explain? –  yiyi Jun 17 '12 at 15:47
If $x\equiv-1/3\pmod{13}$, then $3x\equiv-1\pmod{13}$. You know that $3\cdot4\equiv-1\pmod{13}$, so ... –  Brian M. Scott Jun 17 '12 at 15:48
For a beginner, one may also have to add to Brain's hint that $13$ is a prime, so inverses $\mod 13$ are unique. –  Ragib Zaman Jun 17 '12 at 15:50
@RagibZaman I know that 13 is prime, it is also the smallest prime number which when you reverse the digits, ie 31, is also a prime number. –  yiyi Jun 17 '12 at 16:00

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 $\mathbb{Z}_{13}.$

share|improve this answer
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
@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
It is incorrect to say that "$1/3$ is not a fraction in $\mathbb Z/13$". In fact all rationals with denominator coprime to $13$ do exist mod $13$, and, just as for integers, it is quite convenient to be able to work with such fractions. Their arithmetic is the same as in $\rm\mathbb Q$. The reason that this works will be understood better when one studies fraction rings from a universal viewpoint (so-called localizations). $\qquad$ $\quad$ $\ $ –  Bill Dubuque Jun 17 '12 at 16:04
@BillDubuque thanks, for that insight. I'll have to find a book on localizations. Do you know one that is not too far advance. –  yiyi Jun 17 '12 at 16:08
@MaoYiyi Although localization could be introduced in an (advanced) elementary number theory course, ususally it is not studied till after a course in abstract algebra. The point is that fraction arithmetic is universal, e.g. the equation $\rm\:1/2 - 1/3 = 1/6\:$ will remain true in any ring where $2$ and $3$ have inverses, e.g. in $\rm\:\mathbb Z/m\:$ for all $\rm\:m\:$ coprime to $6$, e.g. it becomes $\rm\:7\! -\! 9 \equiv 11\ (mod\ 13).\:$ $\quad$ $\quad$ $\ $ $\ $ $\ $ $\ $ $\, $ –  Bill Dubuque Jun 17 '12 at 16:13

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

share|improve this answer

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!

share|improve this answer

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.

share|improve this answer
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
@Cameron Actually one can do fractional arithmetic in $\rm\:\mathbb Z/13,\:$ as long as one works in the subring of rationals with denominator coprime to $13$. See my comments to Ragib's answer. –  Bill Dubuque Jun 17 '12 at 16:35
@Bill: Interesting! I haven't yet dealt with localizations, but that makes sense. –  Cameron Buie Jun 17 '12 at 16:40

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 $x=4 \mod 13$.

share|improve this answer
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 $0 \le n \lt 13$?

share|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.