Question

T is the set of real numbers of the form a+b√2, where a,b € Q and a,b are not simultaneously zero. Show that (T,•) is a group where "•" is an ordinary multiplication?

Answer 1

For `" "(T,*) " "`to be a group it has to satisfy Four axioms

(1) Closure

(2) Asssociativity

(3) An identity element exists

(3) Inverse elements exist for each member of the set.

Each axiom will be explained and demonstrated in turn

Explanation:

we have `" "T={a+bsqrt2:a,b inQQ}`

and `a, b` are not simultaneously zero.

To show `" " (T,*)" "` is a group

(1) Closure:

`" i.e. " AA x, yinT" " x*yin T`

`(a+bsqrt2)*(c+dsqrt2)" " a,b,c,d inQQ`

`=(ac+adsqrt2+bcsqrt2+2bd)`

`=(ac+2bd)+(ad+bc)sqrt2`

since `*` is closed in `QQ`

`(ac+2bd) nn (ad+bc)in T`

`:. (T,*) " is closed"`

(2) associative property

`" i.e. " x,y,z,inT" "x*(y*z)=(x*y)*z`

in this case we have to prove

`(a+bsqrt2)* [(c+dsqrt2)*(e+fsqrt2)]`

`= [(a+bsqrt2)* (c+dsqrt2)]*(e+fsqrt2)`

This is a tedious bit of algebra and it is left as an exercise for the reader to do. But it does work!

(3) identity element exists

`" i.e. "AA x inT" " EE e:" "x*e=e*x=x`

`because (a+bsqrt2) * 1 = 1* (a+bsqrt2)=a+bsqrt2`

so an identity element exists, and is it `" "e= 1`.

(4) Inverse elements exist

`" i.e. " AAx in T, EEx^(-1):" "x*x^(-1)=x^(-1)*x=e`

where `" "e" "` is the identity element

In this case `e=1" "`

take `x*x^(-1)=1`

`x*x^(-1)=(a+bsqrt2)=1`

`x^(-1) = 1/(a+bsqrt2) =(a+bsqrt2)/(a^2-2b^2)`

now since `a` and `b` are not simultaneously zero this inverse exists

similarly for `x^(-1)*x=1" "`

so inverses exist

Since all four axioms are satisfied

`(T,*)` is a group