Question

Decide whether the set of all real numbers of the form m+nroot(2) where m is in Z and n is E (E is the ring of all even numbers) is a integral domain (or a field) with respect to the usual operations of addition and multiplication. If it is not an integral domain (or a field), state at least one condition that fails to hold.

Answer 1

Z + 2n sqrt(2)

hmmm, what are the conditions that it needs to satisfy?

Answer 2

additive associativity and commutativity dont seem to be an issue

0+0.rt2 is fine for an additive identity

z + 2n.rt2
-z-2n.rt2 ; additive inverse works
----------
0 + 0.rt2

Answer 3

(1+0.rt2) would work as a mult identity

Answer 4

(a+b.rt2)(x+y.rt2) = ax + 4by + (ay+bx)rt2

is there an ax+by, that is not even?

ax + 2(2kn) .... if ax is odd we might have a fail on this since its not closed

Answer 5

Yes those conditions. Also Identity element

Answer 6

(a+b.rt2)(x+y.rt2) = ax + 4by + (ay+bx)rt2

is there an ay+bx, that is not even?

b and y are always even, so yeah, thats fine

Answer 7

do we have an inverse for multiplication?

Answer 8

ax + 4by + (ay+bx)rt2 = 1 + 0.rt2

Answer 9

ay + bx = 0; ay = -bx is fine since even numbers have inverses

ax + 4by = 1, hmmm

Answer 10

4by is even, is there an ax that is odd?

Answer 11

or at least with the conbditions that ay = -bx that is

Answer 12

0s tend to cause troubles, so we should prolly work those in

Answer 13

Okay Im reading them.

Answer 14

(a+0.rt)(x+0.rt) = 1

ax + 0 + (0+0)rt2 = 1

we would need a fraction, wouldnt we

Answer 15

a in Z
x in Z

ax = 1 is not possible

Answer 16

yes, so basically it is not an integer because it will not equal 1.

Answer 17

(3+0.rt2)(x+0.rt2) = (1+0.rt2) has no \(x\in Z\) solution; x=1/3

Answer 18

ok got it

Answer 19

that does not state which element does not hold.

Answer 20

it shows that there is no inverse for at least 1 element of the group.

the element: (3+0.rt2) is of the form m=3 is an integer, n=0 is an even number

the multiplicative identity is the element: (1+0.rt)

(a+b.rt)(1+0.rt) = (a+b.rt)(1) = (a+b.rt)

therefore the inverse of multiplcation would have to satisfy

(a+b.rt) (x+y.rt) = (1+0.rt2) for every element of the stated structure

since: (3+0.rt2) (x+y.rt2) = (1 + 0.rt2) cannot be satisfied ... there is at least one element that has no inverse in multiplication

Answer 21

(3+0.rt2) (x+y.rt2) = (1 + 0.rt2)

(3x + 3y.rt2) = (1 + 0.rt2)

when 3x = 1, and when 3y = 0

since y=0 is fine and dandy, we would need to find an integer x, such that 3x = 1

Answer 22

if we leave 3 just a random integer; then for any a,x in Z; ax = 1 would have to be satisfied

Answer 23

ok

Answer 24

Is this a ring?

Answer 25

a ring stops at distrbutive property and ignores the mult id and inv right?

Answer 26

a ring is:

+retrice
+id
+inv
*distr over +

Answer 27

+assoc...

Answer 28

some texts say: *assoc..

Answer 29

either way, it only fails for a *inv, so i would say it satisfies a ring