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Do Gaussian integers form a group under multiplication?
Could you please show a full general proof checking for identity, closed, inverse and so on
PLEASE HELP AND SHOW WORKING !!!! :)
- DavidLv 710 years agoFavorite Answer
to make thing clear, i will assume in the following:
z = a+bi, where a,b are integers
w = c+di, where c,d are integers
u = h+ki, where h,k are integers
strictly speaking, the answer straight-away is no, only the non-zero numbers could possibly qualify, since 0 = 0+0i can not possibly have a an inverse. we will see later, that other elements do not, as well.
first, we check closure:
zw = (a+bi)(c+di) = a(c+di) + (bi)(c+di) = ac + adi + bci + bd(i^2)
= (ac-bd) + (ad+bc)i, wich is in Z[i], since ac-bd and ad+bc are both integers.
next, we check associativity:
(zw)u = [(a+bi)(c+di)](h+ki)
= [(ac-bd) + (ad+bc)i](h+ki)
= (ac-bd)h - (ad+bc)k + [(ac-bd)k + (ad+bc)h]i
= ach - bdh - adk + bck + (ack - bdk + adh + bch)i
= a(ch-dk) - b(ck+dh) + (a(ck+dh) + b(ch-dk))i
= (a+bi)[(ch-dk) + (ck+dh)i] = (a+bi)[(c+di)(h+ki)]
= z(wu). associativity is proved.
let 1 = 1+0i:
then z(1) = (a+bi)(1+0i) = a(1) - b(0) + (a(0) + b(1))i
= a+bi, while (1)z = (1+0i)(a+bi) = (1)a - (0)b + ((1)b + (0)a)i
= a+bi, so 1 = 1+0i is an identity.
unfortunately Z[i] does not contain all inverses.
let v = v+0i, where v is an integer greater than 1.
suppose v had an inverse 1/v = (m+ni).
then (v+0i)(m+ni) = 1+0i.
but (v+0i)(m+ni) = vm - 0n + (vn + 0m)i = vm + vni
thus vm = 1, and vn = 0.
since v ≠ 0, this means n = 0, so 1/v is of the form m = m+0i.
but vm = 1, so m must be 1/v.
since v is an integer > 1, 1/v is a positive integer less than 1,
but no such thing exists.
- AndyLv 410 years ago
Obviously they're closed under multiplication, and there is an identity element 1.
But there are no inverses. Consider the number z=3+4i, for instance. We have |z|=5, so if an inverse element z* exists then it must satisfy |z|=1/5, which is impossible since Gaussian integers have integer moduli.