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?

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