I EXAMPLE 7 The ring of Gaussian integers Z[i] = {a + bi | a, bE Z} is a Euclidean domain with d(a + bi) = a² + b°. Unlike the previous two examples, in this example the function d does not obviously satisfy the necessary conditions. That d(x) < d(xy) for x, y E Z[i] follows di- rectly from the fact that d(xy) = d(x)d(y) (Exercise 7). To verify that condition 2 holds, observe that if x, y E Z[i] and y + 0, then xy- E Ql], the field of quotients of Z[i] (Exercise 57 in Chapter 15). Say xy-l = s + ti, where s, tE Q. Now let m be the integer nearest s, and let n be the integer nearest t. (These integers may not be uniquely determined, but that does not matter.) Thus, Im – s] < 1/2 and |n – t| < 1/2. Then xy- = s + ti = (m – m + s) + (n – n + t)i = (m + ni) + [(s – m) + (t – n)i]. So, x = (m + ni)y + [(s – m) + (t – n)i]y. We claim that the division condition of the definition of a Euclidean domain is satisfied with q = m + ni and r= [(s – m) + (t – n)i]y. Clearly, q belongs to Z[i], and since r = x – qy, so does r. Finally, d(r) = d([(s – m) + (t – n)i])d(y) = [(s – m)? + (t – n)²]d(y) (1 + d(y)< d(y).
I EXAMPLE 7 The ring of Gaussian integers Z[i] = {a + bi | a, bE Z} is a Euclidean domain with d(a + bi) = a² + b°. Unlike the previous two examples, in this example the function d does not obviously satisfy the necessary conditions. That d(x) < d(xy) for x, y E Z[i] follows di- rectly from the fact that d(xy) = d(x)d(y) (Exercise 7). To verify that condition 2 holds, observe that if x, y E Z[i] and y + 0, then xy- E Ql], the field of quotients of Z[i] (Exercise 57 in Chapter 15). Say xy-l = s + ti, where s, tE Q. Now let m be the integer nearest s, and let n be the integer nearest t. (These integers may not be uniquely determined, but that does not matter.) Thus, Im – s] < 1/2 and |n – t| < 1/2. Then xy- = s + ti = (m – m + s) + (n – n + t)i = (m + ni) + [(s – m) + (t – n)i]. So, x = (m + ni)y + [(s – m) + (t – n)i]y. We claim that the division condition of the definition of a Euclidean domain is satisfied with q = m + ni and r= [(s – m) + (t – n)i]y. Clearly, q belongs to Z[i], and since r = x – qy, so does r. Finally, d(r) = d([(s – m) + (t – n)i])d(y) = [(s – m)? + (t – n)²]d(y) (1 + d(y)< d(y).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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In the notation of Example 7, show that d(xy) = d(x)d(y).
![I EXAMPLE 7 The ring of Gaussian integers
Z[i] = {a + bi | a, bE Z}
is a Euclidean domain with d(a + bi) = a² + b°. Unlike the previous
two examples, in this example the function d does not obviously satisfy
the necessary conditions. That d(x) < d(xy) for x, y E Z[i] follows di-
rectly from the fact that d(xy) = d(x)d(y) (Exercise 7). To verify that
condition 2 holds, observe that if x, y E Z[i] and y + 0, then xy- E
Ql], the field of quotients of Z[i] (Exercise 57 in Chapter 15). Say
xy-l = s + ti, where s, tE Q. Now let m be the integer nearest s, and let
n be the integer nearest t. (These integers may not be uniquely
determined, but that does not matter.) Thus, Im – s] < 1/2 and |n – t| <
1/2. Then
xy- = s + ti = (m – m + s) + (n – n + t)i
= (m + ni) + [(s – m) + (t – n)i].
So,
x = (m + ni)y + [(s – m) + (t – n)i]y.
We claim that the division condition of the definition of a Euclidean
domain is satisfied with q = m + ni and
r= [(s – m) + (t – n)i]y.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1c401a4d-5531-4107-bc93-6031cf60f754%2Ffa1dd7f3-5774-4b6c-bb67-068e4931bdc1%2Fdjx9xe.jpeg&w=3840&q=75)
Transcribed Image Text:I EXAMPLE 7 The ring of Gaussian integers
Z[i] = {a + bi | a, bE Z}
is a Euclidean domain with d(a + bi) = a² + b°. Unlike the previous
two examples, in this example the function d does not obviously satisfy
the necessary conditions. That d(x) < d(xy) for x, y E Z[i] follows di-
rectly from the fact that d(xy) = d(x)d(y) (Exercise 7). To verify that
condition 2 holds, observe that if x, y E Z[i] and y + 0, then xy- E
Ql], the field of quotients of Z[i] (Exercise 57 in Chapter 15). Say
xy-l = s + ti, where s, tE Q. Now let m be the integer nearest s, and let
n be the integer nearest t. (These integers may not be uniquely
determined, but that does not matter.) Thus, Im – s] < 1/2 and |n – t| <
1/2. Then
xy- = s + ti = (m – m + s) + (n – n + t)i
= (m + ni) + [(s – m) + (t – n)i].
So,
x = (m + ni)y + [(s – m) + (t – n)i]y.
We claim that the division condition of the definition of a Euclidean
domain is satisfied with q = m + ni and
r= [(s – m) + (t – n)i]y.
![Clearly, q belongs to Z[i], and since r = x – qy, so does r. Finally,
d(r) = d([(s – m) + (t – n)i])d(y)
= [(s – m)? + (t – n)²]d(y)
(1
+
d(y)< d(y).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1c401a4d-5531-4107-bc93-6031cf60f754%2Ffa1dd7f3-5774-4b6c-bb67-068e4931bdc1%2Fn7zza6.jpeg&w=3840&q=75)
Transcribed Image Text:Clearly, q belongs to Z[i], and since r = x – qy, so does r. Finally,
d(r) = d([(s – m) + (t – n)i])d(y)
= [(s – m)? + (t – n)²]d(y)
(1
+
d(y)< d(y).
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