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

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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