Is x=2 the Correct Solution to x+5=1 in ℤ6?

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Math9999

Homework Statement


Solve x+5=1 in ℤ6.

Homework Equations


None.

The Attempt at a Solution


6+1=7
7-5=2
Is this the right way to solve this problem?
 
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I'm not sure what your thoughts were here, so I don't know, whether this is the right way. There are more than one way to get the correct result. In ##\mathbb{Z}_6## there is neither an element ##6## nor ##7##; at least not in its usual representation. But you can use the normal way to solve such equations:
$$
x+5=1\,\vert \,-5 \\
x = 1 - 5 \,=\, -4
$$
and then determine what ##-4 \in \mathbb{Z}_6## is.
 
It's modular arithmetic, the answer is x=2.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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