The integers are a Euclidean Domain

The Mathematician — Sun, 28 May 2023 13:00:37 +0000

How to prove that the integers are a Euclidean Domain

To show that the integers are a Euclidean domain (or possess a Division Algorithm), finding one norm only to verify the conditions to be a Euclidean Domain is enough.

Proof that the integers are Euclidean Domain

Proof: Define the following norm:

\begin{equation*} N(x) = \lvert x \rvert, \end{equation*}

where

x \in \mathbb{Z}

. We need to show that for any two elements

a,b \in \mathbb{Z}

with

b\neq 0

there exists elements

q

and

r

\mathbb{Z}

such that

\begin{equation*} a = qb + r \quad \text{with} \quad r = 0 \quad \text{or} \quad N(r) < N(q). \end{equation*}

We let

a

and

b

be nonzero elements and we will check two cases:

b

0

and

b

0

For $b$ > $0$ , let $a \in [qb, (q + 1)b)$ . Then $a = qb + r$ where $r \in [0, \lvert b \rvert)$ since $r = a - qb$ < $(q + 1)b - qb = b$ . Therefore, $N(r)$ < $N(b) = \lvert b \rvert$ or $r = 0$ .

For $b$ < $0$ , we take $a \in [-qb, -(q + 1)b)$ since $-b$ > $0$ . Then $a = q\cdot (-b) + r$ where $r = a + qb$ < $-(q + 1)b - qb = -b$ . Obviously, since -b > 0, we have that $\lvert r \rvert = N(r)$ < $N(-b) = \lvert -b \rvert$ or $r = 0$ .

We haven't included the case when the remainder is negative in all the above cases. The approach is similar to the above. Take the elements $b$ > $0$ and use the same properties we used there and $r > 0$ (note that we don't take this remainder to be negative!). Now let $a = q'b + r'$ , where $q' = q + 1$ and $r' = r - b$ . Then $r'$ < $0$ since $r \in [0, \lvert b \rvert)$ . Now we can perform the rest as $r - b \in (-b, b)$ :

\begin{align*} r - b < 0 &\iff r < b \\ &\iff r - b < b \\ &\iff r' < b \\ &\iff \lvert r' \rvert < \lvert b \rvert \\ &\iff N(r') < N(b), \end{align*}

which ends the proof by showing that the integers are indeed a Euclidean Domain. We could also prove everything with induction, but that is something that the reader can verify (but you can always leave a comment if you want know it).

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Prove that Z is not a group under multiplication

The Mathematician — Mon, 10 Apr 2023 13:00:41 +0000

Prove that $\mathbb{Z}$ is not a group under multiplication .

I do expect that when you are learning group theory and you search for an answer to the question above, you want to know how to tackle such a question. While this can be answered in one sentence, I will take you through the whole process of what you need to do.

Proof. Let

\mathbb{Z}

be the set of integers. To be a group under multiplication, i.e.,

(\mathbb{Z},\times)

, we must verify three things: associativity, identity, and inverse elements. Let

a,b,c \in \mathbb{Z}

.

It is straightforward by the definition of integers that

(a \times b)\times c = a \times (b \times c)

. So for all elements in

\mathbb{Z}

we know that the operation multiplication is associative.

Next, we need to find the identity element of

\mathbb{Z}

under multiplication. This is easy since

1

is identity element (note that

a \times 1 = 1 \times a = a

).

Finally, the last step is to find the inverse elements of

\mathbb{Z}

under multiplication. Now we take the element

a \in \mathbb{Z}

. Does this mean that there is an inverse element of

a

under multiplication? In other words, is there an element

a^{-1} \in \mathbb{Z}

such that:

\begin{align*} a \times a^{-1} = a^{-1} \times a = 1? \end{align*}

Assume it is true with

a \neq 1

. Then it means that

a^{-1} = \frac{1}{a} \in \mathbb{Z}

. But then

\frac{1}{a}

is a rational number, which isn’t contained in the set of integers. So

\mathbb{Z}

can’t be a group under multiplication since for all integers (expect for the identity element

1

) there is no inverse element.

A short counterexample you could give:

2

has no inverse element since

\frac{1}{2} \not \in \mathbb{Z}

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Z Archives - Epsilonify

The integers are a Euclidean Domain

How to prove that the integers are a Euclidean Domain

Proof that the integers are Euclidean Domain

Prove that Z is not a group under multiplication