Partition of Integers with mod

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Homework Statement


Are the following subsets partitions of the set of integers?

The set of integers divisible by 4, the set of integers equivalent to 1 mod 4, 2 mod 4, and 3 mod 4.


Homework Equations





The Attempt at a Solution


Yes, it is a partition of the set of integers. Consider 4/4 = 1, 5/4 = 1 R 1, 6/4 = 1 R 2, 7/4 = 1 R 3.

However, how would you create a negative number like -5?
 
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nicnicman said:

Homework Statement


Are the following subsets partitions of the set of integers?

The set of integers divisible by 4, the set of integers equivalent to 1 mod 4, 2 mod 4, and 3 mod 4.

Homework Equations


The Attempt at a Solution


Yes, it is a partition of the set of integers. Consider 4/4 = 1, 5/4 = 1 R 1, 6/4 = 1 R 2, 7/4 = 1 R 3.

However, how would you create a negative number like -5?

I'm not too clear on what your argument is supposed to mean. What's R? But -5=3 mod 4 since (-5)=(-2)*4+3. Hmm, I think I see. R means 'remainder', yes?
 
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Sorry, I forgot to mention this is far from a formal proof. R just means remainder.
 
nicnicman said:
Sorry, I forgot to mention this is far from a formal proof. R just means remainder.

Yeah, it is far from formal. But I see what you are doing. Are you supposed to give something formal or just answer yes or no?
 
Just answer yes or no. And, you answered my question. I'm pretty sure the answer is yes. Thanks.
 
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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