# Thread:

1. ## A sets proof

Hi again, another question I am having trouble starting:

Let $I$ be a nonempty subset of $\mathbb{Z}$ such that:

$(\forall x \in I)(\forall y \in I)[(x-y) \in I]$ and $(\forall z \in \mathbb{Z})(\forall x \in I)[z \cdot x \in I]$.

Show that for some $n \in I, I = \{z \in \mathbb{Z} \colon z = xn \ for \ some \ x \in \mathbb{Z}\}$.

Please no full solutions, but if someone could show me how to proceed

2. Originally Posted by nzmathman
Hi again, another question I am having trouble starting:

Let $I$ be a nonempty subset of $\mathbb{Z}$ such that:

$(\forall x \in I)(\forall y \in I)[(x-y) \in I]$ and $(\forall z \in \mathbb{Z})(\forall x \in I)[z \cdot x \in I]$.

Show that for some $n \in I, I = \{z \in \mathbb{Z} \colon z = xn \ for \ some \ x \in \mathbb{Z}\}$.

Please no full solutions, but if someone could show me how to proceed

1) Take a minimal positive element in I (why is there such an element?)

2) Applying Euclides algorithm show that any element in I is an integer multiple of the element you found in (1)

3) Go grab a beer and be happy.

Tonio

3. Originally Posted by tonio
1) Take a minimal positive element in I (why is there such an element?)

2) Applying Euclides algorithm show that any element in I is an integer multiple of the element you found in (1)

3) Go grab a beer and be happy.

Tonio
I know why a minimal element of I exists, but why can we conclude a minimal positive element exists?

Also, how would I apply Euclidean algorithm to something this abstract?

4. I know why a minimal element of I exists, but why can we conclude a minimal positive element exists?
There is no minimal element of $I$. Take any $x\in I$. Then the set $\{zx\mid z\in\mathbb{Z}\}\subseteq I$ does not have a minimal element.

To prove that there is a minimal positive element, it is sufficient to know that there is any positive element. This is natural numbers we are talking about

Also, how would I apply Euclidean algorithm to something this abstract?
Take any (w.l.o.g. positive) $x\in I$ and the minimal positive element $m$ of $I$. Then the GCD of $x$ and $m$ is a linear combination of $x$ and $m$ by Bézout's identity, an application of the Euclid's algorithm. From there it is easy to show that $m$ divides $x$.
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