Showing S = ℤ Using 13 & 1000 in Subring

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Discussion Overview

The discussion revolves around demonstrating that a subring S of the integers ℤ, containing the elements 13 and 1000, must equal ℤ. The focus is on exploring methods to establish the presence of the identity element in S, which is crucial for extending S to the entirety of ℤ.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant suggests constructing a bijection using the numbers 13 and 1000 to show that S equals ℤ.
  • Another participant notes the necessity of showing that 1 ∈ S, proposing that since 12 can be expressed as a combination of 1000 and 13, it implies S has an identity element.
  • A third participant mentions that since 13 and 1000 are relatively prime, it follows that 1 is in the subring generated by these numbers.

Areas of Agreement / Disagreement

There is no consensus on the approach to take, and multiple viewpoints regarding the proof exist among participants.

Contextual Notes

The discussion does not resolve the mathematical steps necessary to fully establish the claim, and the implications of the relationships between the numbers are not fully explored.

Bachelier
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Let S be a subring of the ring of integers ℤ. Show that if 13 ∈ S and 1000 ∈ S, then S equals ℤ.

I'm thinking we should construct a bijection using both numbers. Any ideas? thanks
 
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I see that ∃ no takers. I know the problem looks weird but it is a problem I read in an old paper.

The only thing I can think of is that we need to show that 1 ∈ S. To do this I can use 12 ∈ S because 12 = 1000 - 13*76, therefore 13 - 12 ∈ S and hence S has an identity and therefore we can extend to Z.
 
Bachelier said:
Let S be a subring of the ring of integers ℤ. Show that if 13 ∈ S and 1000 ∈ S, then S equals ℤ.

I'm thinking we should construct a bijection using both numbers. Any ideas? thanks

13 and 1000 are relatively prime so 1 is in the subring that they generate.
 
Thank you Lavinia. You're becoming my favorite Science adviser. :)
 

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