What Are the Inverse Elements in Z_7 and Z_13?

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The discussion focuses on identifying the multiplicative inverses in the modular arithmetic systems Z_7 and Z_13. It concludes that all non-zero elements in both Z_7 and Z_13 have multiplicative inverses due to the primality of the moduli. Specifically, Z_7 has 6 elements with inverses (1, 2, 3, 4, 5, 6) and Z_13 has 12 elements with inverses (1 through 12). The key principle applied is that for a number n in Z_m, it has a multiplicative inverse if and only if gcd(m, n) = 1, which holds true for all elements in these prime modulus systems.

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Benny
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Hi I'm not sure what the following two questions (actually there are a few more but if I know how to do the following the others should be ok) is asking for.

In the following systems Z_n, write down the sets of elements that have inverses. ( (n) is equal to the number of elements in these sets.)

a) Z_7
d) Z_13

The answers to 'a' and 'd' are 4 and 12 respectively.

The question asks for "the sets of elements that have inverses" but 4 and 12 are just numbers, so how can they be the answers to the questions? I learned that if a number n(bar) is in z_m then n(bar) has a multiplicative inverse iff gcd(m,n) = 1 but I'm not sure how to apply this here. Can someone help me with this?
 
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I assume the problem is asking for "multiplicative" inverses, since all elements have additive inverses.

Problem (a) is asking for all elements of Z7, n, such that there is some m in Z7 so that mn= 1 (mod 7). One "brute strength" method is took look at every possible number: there are only 6 possibilities. One obvious answer is always "1".
What about 2? Is 2a= 1 (mod 7) for any a? That is, is there a number a such that 2a is of the form 7n+ 1 for some n? Again, an obvious candidate is "4": 2(4)= 8= 7+ 1. 2 has a multiplicative inverse: 4.
What about 3? Is 3a= 1 (mod 7)? 3(2)= 6, 3(3)= 9= 7+2, (3)(4)= 12= 7+ 5, 3(5)= 15= 2(7)+ 1. 3 has a multiplicative inverse: 5.
Of course, 4 has a multiplicative inverse: 2.
Of course, 5 has a multiplicative inverse: 3
That leaves 6. 6*6= 36= 5(7)+ 1. 6 is its own multiplicative inverse.
Every number in Z7 has a multiplicative inverse.

The key point is that 7 and 13 are both prime numbers. " if a number n(bar) is in z_m then n(bar) has a multiplicative inverse iff gcd(m,n) = 1". If n is prime then,
gcd(m,n)= 1 for every number less than n except 0.

The set of all elements of Z7 and Z13[/b] are just the elements of Z7 and Z13 except 0.
 
Thanks for the help HallsofIvy. The use of the theorem that you mentioned didn't come to me as I was working through the questions but it's good that you've mentioned it. I'll try to keep things like that in mind in future.
 
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