Question

Let G= {a+b*2^(1/2) | a,b Element of Q} Q is the set of rational numbers. Prove that G is a group under addition.

Answer 1

ok we need to show G is closed, associative, has identity, and has inverse.
*CLOSURE!
so we need to show when we take to elements of set G then we still get an element in set G.

so Let a,b,c,d be rational.
so we have
that
a+b*2^(1/2) is element of G
and
c+d*2^(1/2) is element of G

so
when we add:

(a+b*2^(1/2))+(c+d*2^(1/2))
=(a+c)+(b+d)*2(1/2)

so we need a+c to be rational and b+d to be rational

if you take a rational number and add it to another rational number you still have a rational number so G is closed!
now step 2
show that G is associative

Answer 2

Ok, I think I am starting to get it.

Answer 3

*Associative
let a,b,c,d,e,f be rational.
so we have
that
a+b*2^(1/2) is element of G
c+d*2^(1/2) is element of G
e+f*2^(1/2) is element of G

so we need to show

[(a+b*2^(1/2))+(c+d*2^(1/2))]+(e+f*2^(1/2))=
(a+b*2^(1/2))+[(c+d*2^(1/2))+(e+f*2^(1/2))]
this step is not hard to do

Answer 4

we have the top part =
(a+c)+(b+d)*2^(1/2)+(e+f*2^(1/2))
=(a+c+e)+(b+d+e)*2^(1/2)
bottom part=
(a+b*2^(1/2))+(c+e)+(d+f)*2^(1/2)
=(a+c+e)+(e+d+f)*2^(1/2)
since addition is commutative and associative then G is associative

Answer 5

*Identity
let a,b rational
then a+b*2^(1/2) is element of G

so we have
a+b*2^(1/2)+?=a+b*2^(1/2)

if ? is an element of G, then G is the identity property

Answer 6

Thank you so much for this.

Answer 7

if ?=0, then the equation holds but is 0 and element of G

0 and 0 are rational numbers

so 0+0*2^(1/2) is element of G
but 0+0*2^(1/2)=0
so this must mean 0 is element of G
and therefore has the identity property

Answer 8

last step is Inverse

Answer 9

*Inverse
again let a,b rational
then a+b*2^(1/2) is element of G
now we
want to find some element in G,?, such that

(a+b*2^(1/2))+?=0

if we can ? is in G, then we have the inverse property

additive inverse of a is -a

additive inverse of b*2^(1/2) is -b*2^(1/2)

-a and -b are both rational
so therefore

-a-b*2^(1/2) is element of G

and therefore has inverse property

Answer 10

*in conclusion
since we have all four of the above, we have that G is a group under addition.

Answer 11

was that fine? :)

Answer 12

oh yeah. thank you.

Answer 13

np :)

Answer 14

you could have also use the subgroup test to show that G is a subgroup of the group
\[(\mathbb{R},+)\]

subgroups are groups

Answer 15

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e6576330b8b1f45b4b49133