Showing S = ℤ Using 13 & 1000 in Subring

[Bachelier]

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

 

[Bachelier]

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.

 

[lavinia]

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.

 

[Bachelier]

Thank you Lavinia. You're becoming my favorite Science adviser. :)

 

[blue_raver22]

I would approach this problem by utilizing the properties of subrings and the integers. First, we know that a subring of the integers must contain the identity element 0, and must be closed under addition and multiplication.

Since 13 and 1000 are both elements of S, this means that they are both closed under addition and multiplication within S. Therefore, any integer can be written as a combination of 13 and 1000 through addition and multiplication, making S a subset of ℤ.

Next, we can consider the fact that 13 and 1000 are coprime, meaning they do not share any common factors other than 1. This means that any integer can also be written as a combination of 13 and 1000 through the process of finding common factors and using the Euclidean algorithm.

Combining these two observations, we can conclude that S is a subset of ℤ and also contains all integers through the use of 13 and 1000. Therefore, S must equal ℤ.

In other words, by utilizing the properties of subrings and the specific elements of 13 and 1000, we can show that S is a subring that is equal to the entire ring of integers ℤ.