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I was watching my math lecture for a course and we were doing inverse matrices in finite fields in $\mathbb{Z}_5$. The determinant was $1 / (6-4)$ so $1/2$ which apparently means it is 3. I have no idea where 3 comes from, is $1$ over $2$ the same as division or does that mean something else?

I am entirely new to fields and have very limited knowledge on them, but my searches never revealed anything about division or what this would be for the determinant.

Robert Z
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Patrick
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2 Answers2

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Note that $2\cdot 3=6=5+1\equiv 1\pmod{5}$. This means that in the ring $\mathbb{Z}_5$, the element $2$ is invertible and the inverse of $2$ (what you call $1/2$), is $3$.

Robert Z
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    Ah, right, so it's like a double inverse sort of thing. Since 1/2 would have an inverse of 2 and 2 would have an inverse of 3 and since if you take the inverse of something twice you would end up with the same something. Does this mean 1/3 in Z5 would be 2? If so you have answered my question – Patrick Sep 01 '17 at 11:39
  • @Patrick Yes, you are correct $2^{-1}\equiv 3$ and $3^{-1}\equiv 2$ modulo $55. – Robert Z Sep 01 '17 at 11:50
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Division is defined the same way in modular arithmetic as it is on the regular number line. In other words, $\frac12$ is the unique number such that $2\cdot \frac12 = 1$. However, since modular arithmetic works differently from regular arithmetic, the interpretation of the above definition is very different: We have $2\cdot 3 = 1$, which means that $3$ actually fulfills the defining property of $\frac12$. We can also easily check that there are no other such numbers, so it is unique. That is why $\frac12 = 3$.

Note for later: If you ever have to solve quadratic equations in modular arithmetic (at least as long as it's modulo an odd prime), the formula $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ still works, except once again, the definition of $\sqrt{{}\cdot{}}$ is the same (sortof; there isn't anything like positive / negative to distinguish the two square roots of a number), but the interpretation is different, with, for instance, $\sqrt{-1} = \pm 2$ modulo $5$.

Arthur
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