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I can't solve the following equation system using $\mathbb{Z}_3$: \begin{align} &x+y=1,\\ &2x+y=0. \end{align} $\mathbb{Z}_3$ is a field with the numbers $(0,1,2)$ and modular operations.

What I tried: I added $2$ to the first equation and got:

$$ x+y+2= 0 = 2x + y,$$

$$x+2=2x.$$

And I don't know what to do from here, i know it seems obvious $x=2$ but we just learned all the $11$ axioms of a field, and I can only use them or the $5$ properties that derived from them. My problem is that I don't know how to express the $2x$ in a way that can help solve it.

Thanks for help.

Start wearing purple
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Michael
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  • How would you solve this equation over the reals? You do know that moving a term to the other side amounts to subtracting it from both sides. The field axioms imply that the outcome is unique, and the process is reversible. Hence it leads to an equivalent equation. In general, the elementary row operations used in inverting a matrix or solving a linear system work over any field (check the derivation of the method - it only used field axioms!!!). You just need to remember to use the multiplicatio/division of the appropriate field at all steps. – Jyrki Lahtonen Oct 19 '13 at 17:32
  • Thanks for help, i solved the equations. I did not know i could do other things except using the axioms. – Michael Oct 19 '13 at 18:44
  • See this question and its answers for a discussion about inverting a 3x3 matrix (or solving a linear system of three equations and three unknowns) over $\Bbb{Z}_{29}$. – Jyrki Lahtonen Oct 21 '13 at 20:40

3 Answers3

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You can solve it in exactly the same way as you would in the rational or real numbers; solve for $y$ the first equation, $y=1-x$, and substitute in the second one: $$ 2x+(1-x)=0 $$ that becomes $$ x=-1 $$ or $x=2$, since you are in $\mathbb{Z}/3\mathbb{Z}$. Now you get $y=1-(-1)=2$.

Or sum the two equations, getting $x+2x+y+y=1$, so $2y=1$. Since $2^{-1}=2$, because $2\cdot 2=1$, you have again $y=2$. Subtracting the first equation from the second gives $x=-1$, that is, $x=2$.

In general you can compute inverses in fields of the form $\mathbb{Z}/p\mathbb{Z}$ ($p$ a prime) using the Euclidean algorithm and Bézout's theorem.

egreg
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Since $\frac{\mathbb{Z}}{3\mathbb{Z}}$ is a field, we know that $ab=ba$ for every $a,b \in \frac{\mathbb{Z}}{3\mathbb{Z}}$ and that this product is zero if and only if one of $a$ or $b$ is zero. So, we know that $2x=2(2)$ cannot be zero in this field. Now we just need to deduce what it is! This is the same as observing that for every nonzero $a \in \frac{\mathbb{Z}}{3\mathbb{Z}}$ there is a unique element $b$ so that $ab=1=ba$ (in larger fields you would have a harder time and would actually need to pull out the largest multiple of $p$, where your field is of the form $\frac{\mathbb{Z}}{p\mathbb{Z}}$, but that is equivalent for this field). So, if we look at the nonzero elements, we have the set $\lbrace 1,2 \rbrace$. Well, we know that for any number $a$, $1a=a=a1$, so it remains for you to go through and check the results with $a=2$ in order to find what the inverse of $a$ is (note that the inverse of $a$ is in the set $\lbrace 1,2 \rbrace$). This will tell you immediately what $2x=2(2)$ is in $\frac{\mathbb{Z}}{3\mathbb{Z}}$.

Geoff
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  • Thanks for help, so you are saying that in small fields i should just check all the numbers i have? I mean after i simplified the equations – Michael Oct 19 '13 at 18:47
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Ok, you must work $(\mod 3).$ So the first equation is written $y=1-x=1+2x$ in ${\bf Z}_3.$ Now the second equation is written as $2x+(1+2x)=0$ thus $4x+1=0$ or $x+1=0$ or $x=-1=2.$ So $y=1+2\cdot 2=1+4=5=2.$ The desire solution is $(x,y)=(2,2).$

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