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I am currently in my last year of high-school and I try to learn Algebra on my own, one of my textbook exercise ask me to

Find elements:

$$ \left\{\;1,\;\frac12,\;\frac14,\;5,\;\sqrt{−2}\right\}\;\in\;\mathbb Z_5$$ I have no idea what $\mathbb Z_5$ neither what does it mean to "find elements" in this set. I tried to search on google but I found nothing about this particular notation. Can you please explain me what does this notation and "finding elements" mean without giving me the solution to this exercise as I want to do it on my own.

Thanks you very much

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  • $\begingroup$ $\mathbb Z_5$ is the field of integers modulo $5$, and consists of the elements $\{0,1,2,3,4\}$. So you're being asked to find, for example, for $\frac 12$, an integer k mod $5$ such that $2k \equiv 1 \pmod 5$ $\endgroup$ – Mathmo123 Sep 9 '14 at 0:33
  • $\begingroup$ The symbol $\sqrt{-2}$ here denotes any number $x$ such that $x^2=-2\color{grey}{=3}$. The symbol $\frac 1 x$ where $x\in \mathbb Z_5$ denotes the multiplicative inverse of $x$ in $\mathbb Z_5$, that is, it denotes the only number $y$ such that $xy=1$. Are you also asking what $\mathbb Z_5$ is? $\endgroup$ – Git Gud Sep 9 '14 at 0:36
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    $\begingroup$ What textbook are you using? From the sounds of it, it might be beneficial if you picked up an easier-to-read text. $\endgroup$ – Clayton Sep 9 '14 at 0:40
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In this context, $\mathbb{Z}_5$ is the finite field of $5$ elements. So $$\mathbb{Z}_5=\{0,1,2,3,4\},$$ and we have multiplication and addition performed on the elements modulo 5.

It is clear that $\frac{1}{2}$ and $\sqrt{-2}$ do not exist in our set. But, there are elements in our set, that "retain the essence" of the fractions and squareroots.

In the usual rational number field, we have $\dfrac{1}{2}$ defined as the number $2^{-1}$, that is, the multiplicative inverse of $2$, that is, the number $x$ such that $2\times x=1$. So, we need to find a corresponding element, $y$ say, in $\mathbb{Z}_5$ such that $2y\equiv 1\mod 5$.

In the complex number field, $\sqrt{-2}$ is defined as the number $x$ such that $x^2=-2$, so using that as an analogy, we need to find an element in $\mathbb{Z}_5$, say $y$, such that $y^2\equiv -2\equiv3 \mod 5$.

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I suspect that you either skipped over parts of the textbook, or picked up a very bad textbook. Any textbook should explain what it means by these things before asking you to solve these kind of problems.

But anyway.

$\mathbb{Z}_n$ is the set you get if you start with $\mathbb{Z}$ but then only treat two numbers as different if their difference isn't a multiple of $n$. So for example, $2$ and $3$ refer to distinct elements of $\mathbb{Z}_n$, because $2 - 3 = -1$, and that cannot be divided by 5. $2$ and $7$, on the other hand, are the same, because $7-2 = 5 = 1 \cdot 5$.

You can easily convince yourself that every number in $\mathbb{Z}$ is, over $\mathbb{Z}_n$ equal to one of the numbers $\left\{0,1,2,\ldots,n-1\right\}$, and that none of these numbers are the same as any other of these numbers. Thus, for most purposes, you can assume that $$ \mathbb{Z}_n = \left\{0,1,2,\ldots,n-1\right\}. $$ But note that this choice of number is not unique. There are other sets containing $n$ numbers which also have the property that - with the "difference divisible by $n$ means equal" from above - every number of $\mathbb{Z}$ is equal to exactly one of the $n$ numbers. You can, for example, also pick $$ \mathbb{Z}_n = \left\{0,n+1,2n+2,\ldots,(n-1)^2+n-1\right\}. $$

The interesting thing about $\mathbb{Z}_n$ is that you can do arithmetic within it, and it turns out that you don't need to be carefull which of the (very many!) "equal" numbers you do your computations with! The result stays the same (according to the weird "difference divisible by $n$ means equal" rule, of course), no matter which so-called "representative" you pick. Take for example the computation $3\cdot 4 + 7$. In normal arithmetic, this yields 19, which in $\mathbb{Z}_5$ is the same as $4$, because $19 - 4 = 15 = 3\cdot 5$. The same computation can be written as $3\cdot (-1) + 2$ over $\mathbb{Z}_5$ (convince yourself that over $\mathbb{Z}_5$, $-1 = 4$ and $2 = 7$), which yields $-1$, and that is over $\mathbb{Z}_5$ the same as $4$.

What the question wants you to do is to find elements in $\mathbb{Z}_5$ which e.g. correspond to $\frac{1}{2}$. The defining property of $\frac{1}{2}$ is that it's a number which, if multiplied with $2$, yields $1$ (this is often called the "multiplicative inverse of $2$"). Your job is to find a number in $\mathbb{Z}_5$ which satisfies this. Hint: Compute $2\cdot 3$ over $\mathbb{Z}_5$, and see what you get!)

For $\sqrt{-2}$, what they want you to do is to find a number in $\mathbb{Z}_5$ which, if squared, yields $-2$. But, since we're working in $\mathbb{Z}_5$, it doesn't literally have to yield $-2$, of course - that would be impossible for any real number -- it just needs to yield a number which is, with the weird equality rule of $\mathbb{Z}_5$, equal to $-2$. I won't give you a solution for $\sqrt{-2}$, but for $\sqrt[3]{2}$, a possible solution would be $3$, because $3^3 = 27$, which is the same as $2$ in $\mathbb{Z}_5$.

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