Integral Domains: Strange Integers
Z is the set of all integers ..., 2, 1, 0, 1, 2, ... Form the set
As a subset of the set of real numbers R, Z[√3] is closed under operations of addition, subtraction, and multiplication. In terms of the two components associated with every number in Z[√3], the arithmetic operations are expressed as following.
 Addition:

(a_{1} + b_{1}√3) + (a_{2} + b_{2}√3) = (a_{1} + a_{2}) + (b_{1} + b_{2})√3.
 Subtraction:

(a_{1} + b_{1}√3)  (a_{2}  b_{2}√3) = (a_{1}  a_{2}) + (b_{1} + b_{2})√3.
 Multiplication:

(a_{1} + b_{1}√3)·(a_{2} + b_{2}√3) = (a_{1}·a_{2} + 3b_{1}·b_{2}) + (a_{1}b_{2} + a_{2}b_{1})√3.
Both addition and multiplication are commutative. 0 and 1 are elements of Z[√3] (0 = 0 + 0√3 and 1 = 1 + 0√3) Since the operations are actually the usual arithmetic operations on real numbers, we also have the distributive law.This makes Z[√3] a commutative ring just like Z. In Z, from ab = 0 we can conclude that either a = 0 or b = 0. In Algebra, nonzero elements for which ab = 0 are known as divisors of zero. So no element of Z is a divisor of zero. Rings with this property are called integral domains. Z is one example of integral domain. Z[√3] is another. To show this, it is convenient to introduce a couple of notations.
For any A = a + b√3 define its conjugate A' = a  b√3. The conjugate has several important properties. For example,
 (A + B)' = A' + B', and
 (A · B)' = A' · B'.
Lemma
Z[√3] is an integral domain.
Proof
Note that N(AB) = N(A)N(B). Indeed, N(AB) = (AB)·(AB)' = ABA'B' = AA'BB' = N(A)N(B).
Assume AB = 0. Then also N(AB) = 0. From the preceding sentence,
(Write down the identity N(AB) = N(A)N(B) explicitly. You may be surprised to discover what actually has been proven so easily in an abstract form.)
Z and Z[√3] have other features in common. Methods that work in Z may be useful for solving problems in Z[√3]. Here is a problem that has at least two solutions, one borrowed from Z, another that makes use of the peculiar structure of Z[√3].
Problem
Is 99999 + 222222√3 a square of a number from Z[√3]?
Solution 1
Is integer 3726125 a square of an integer? This is the kind of questions to answer which K.F.Gauss (17771855) in 1801 invented Modular Arithmetic. Consider operations modulo 9. Look at the main diagonal of the multiplication table. Unless modulo 9 a number is one of 0, 1, 4, 7, it can't be a square of an integer. 3726125 = 8 (mod 9). (Casting out 9s is easy; for
Let's apply this approach to solving our problem.
(a + b√3)² = (a² + 3b²) + 2ab√3
If 99999 = a² + 3b² for a couple of integers a and b, then a² is divisible by 3. Then by Euclid's Proposition VII.30 a is divisible by 3. Thus a² = 0 (mod 9), and necessarily 3b² = 0 (mod 9). Which implies that b = 0 (mod 3) also. We finally get 2ab = 0 (mod 9). However, 222222 = 3 (mod 9). The answer to our problem is negative, 99999 + 222222√3 can't be a square of a number from Z[√3].
Solution 2
Assume that 99999 + 222222√3 = (a + b√3)² for some integer a and b. Then it must be that also
In Z[√3], the two methods lead to different generalizations of the problem.
 From Solution 1, if integers a and b are such that a = 0 (mod 9) while b is not then a + b√3 is not a square of a number from Z[√3].
 From Solution 2, if A' < 0, then a + b√3 is not a square of a number from Z[√3]. (This is of course true also if A < 0.)
Both admit further generalizations. For example, we can define Z[√m] where m is not a square of an integer. #1 extends easily to the case of an odd m. For #2, consider
Remark
To establish that Q[√m] is a field, one has to verify that it is closed under division. I.e., the result of division of two numbers from Q[√m] is itself an element of Q[√m]. (This has to do with removing irrationality from the denominator.)
Problem
Show that the set Z[√3] is dense on the real line.
Solution
We have to show that for any real number r and any positive ε there exists a number
r  (a + b√3) < ε.
In other words, we have to show that, by a proper choice of integers a and b,
First of all, note that we can easily achieve this goal for
(*)  k < ε, k = a + b√3. 
Assume k > 0. (If it's not, take k instead.)
For a given real number r
[r]≤ r < [r] + 1,
where [r] is the whole part of r. Consider the sequence
[r], [r] + k, [r] + 2k, [r] + 3k, ...
Since k is not 0, r will fall between a pair of successive terms of the sequence. It's distance to either of the terms would not exceed k and, from (*), ε.
And we are done.
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