# A sets proof

• Aug 24th 2010, 01:39 PM
nzmathman
A sets proof
Hi again, another question I am having trouble starting:

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

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

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

Please no full solutions, but if someone could show me how to proceed (Nod)
• Aug 24th 2010, 06:50 PM
tonio
Quote:

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

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

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

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

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

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
• Aug 25th 2010, 01:39 AM
nzmathman
Quote:

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?
• Aug 25th 2010, 03:47 AM
emakarov
Quote:

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 $\displaystyle I$. Take any $\displaystyle x\in I$. Then the set $\displaystyle \{zx\mid z\in\mathbb{Z}\}\subseteq I$ 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 (Bigsmile)

Quote:

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