Answer the following completely:
1. Determine whether each is a subset of the complex numbers is a subgroup of group c of complex numbers under ADDITION
c. The set iR of pure imaginary numbers including 0.
1 Answer | Add Yours
To show that an infinite subset H of a group G forms a subgroup under the operation + we need only show that `a,b in H => a+b, a+(-b) in H`
(1) Determine if `Q^(+)` is a subgroup of `CC` :
Let `a,b in Q^(+)` ( e.g. a and b are positive rational numbers.)
Suppose `b>a` . Then `a+b in Q^(+)` but `a+(-b) notin Q^(+)` since `a+(-b)<0` .
So `Q^(+)` is not a subgroup of `CC` under addition.
** Another way to look at this is that every element in `Q^(+)` does not have an additive inverse. **
(2) Determine if `7ZZ` is a subgroup of `CC` under addition:
Let `a,b in 7ZZ` (e.g. a and b are intger multiples of 7).
(a) `a+b in 7ZZ` : let a=7m, b=7n with `m,n in ZZ` . Then
a+b=7m+7n=7(m+n) and `7(m+n) in 7ZZ`
(b) `a+(-b) in 7ZZ` : again let a=7m, b=7n with m,n integers. Then a+(-b)=7m-7n=7(m-n) and `7(m-n) in 7ZZ`
Therefore `7ZZ` is a subgroup of `CC` under addition.
(3) Determine if `iRR` is a subset of `CC` under addition:
Let `a,b in iRR` (e.g. a and b are pure imaginary numbers).
(a) `a+b in iRR` : Let a=ci and b=di where `i=sqrt(-1);c,d in RR` .
a+b=ci+di=(c+d)i and `(c+d)i in iRR`
(b) `a+(-b) in iRR` : Again let a=ci,b=di with i the imaginary number and c,d real numbers. Then
a+(-b)=ci-di=(c-d)i and `(c-d)i in iRR`
Therefore `iRR` is a subset of `CC` under addition.
We’ve answered 287,822 questions. We can answer yours, too.Ask a question