Let Zsquareroot 2 a b squareroot 2 a b epsilon Z Its not t
     Let Z|squareroot 2| = {a + b squareroot 2| a, b epsilon  Z}. It\'s not too hard to show (and you don\'t have to) that Z[squareroot 2] is a subring of R. Observe that the element 1 + squareroot 2 is a unit in Z[squareroot 2], because (1 + squareroot 2)(-1 + squareroot 2) = 1  Find ten other units in Z[squareroot 2], not counting 1 + squareroot 2, -1 + squareroot 2, 1, or -1.![Let Z|squareroot 2| = {a + b squareroot 2| a, b epsilon Z}. It\'s not too hard to show (and you don\'t have to) that Z[squareroot 2] is a subring of R. Observe  Let Z|squareroot 2| = {a + b squareroot 2| a, b epsilon Z}. It\'s not too hard to show (and you don\'t have to) that Z[squareroot 2] is a subring of R. Observe](/WebImages/30/let-zsquareroot-2-a-b-squareroot-2-a-b-epsilon-z-its-not-t-1084350-1761569912-0.webp) 
  
  Solution
Other units is Z[2] = a+b2 such that a and b are integers are
 2+5 (multiply with -2+ 5 and you get = 1)
 3+10 (multiply with -3+ 10 and you get=1) and so on for otehr units
 4+17
 5+26
 6+37
 7+ 50
 8+65
 9+82
 10+101
 and 11+122
![Let Z|squareroot 2| = {a + b squareroot 2| a, b epsilon Z}. It\'s not too hard to show (and you don\'t have to) that Z[squareroot 2] is a subring of R. Observe  Let Z|squareroot 2| = {a + b squareroot 2| a, b epsilon Z}. It\'s not too hard to show (and you don\'t have to) that Z[squareroot 2] is a subring of R. Observe](/WebImages/30/let-zsquareroot-2-a-b-squareroot-2-a-b-epsilon-z-its-not-t-1084350-1761569912-0.webp)
