Proving the Sum of Additive Groups Z: (3/7)Z + (11/2)Z = (1/14)Z

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The discussion centers on proving the equality of additive groups, specifically that (3/7)Z + (11/2)Z = (1/14)Z. The proof begins by expressing the sum of the two groups as {(6k + 77m)/14 : k,m ∈ Z}. The participants confirm that while it is straightforward to show that (3/7)Z + (11/2)Z is a subset of (1/14)Z, demonstrating the converse requires proving that any element n in (1/14)Z can be represented in (3/7)Z + (11/2)Z. A successful example is provided where k=13 and m=-1 shows that 1/14 is indeed in the group (3/7)Z + (11/2)Z, confirming the equality.

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Z is the set of integers. Prove that (3/7)Z + (11/2)Z = (1/14)Z

Attempt:

By definition,
(3/7)Z+(11/2)Z={3k/7 + 11m/2 : k,m € Z} = {(6k + 77m)/14 : k,m € Z}.

Showing that 3/7Z+11/2Z is a subset of 1/14 Z is easy but I can't prove the converse. Can't show that whatever n€1/14Z I take satisfies that n€3/7Z+11/2Z.
 
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bedi said:
Z is the set of integers. Prove that (3/7)Z + (11/2)Z = (1/14)Z

Attempt:

By definition,
(3/7)Z+(11/2)Z={3k/7 + 11m/2 : k,m € Z} = {(6k + 77m)/14 : k,m € Z}.

Showing that 3/7Z+11/2Z is a subset of 1/14 Z is easy but I can't prove the converse. Can't show that whatever n€1/14Z I take satisfies that n€3/7Z+11/2Z.

If you could show 1/14 is in the group 3/7Z+11/2Z, then all multiples of 1/14 would also be in the group, right?
 
Yes! If I take k=13 and m=-1 then 1/14 is in the group 3/7Z+11/2Z. Thank you very much.
 

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