Let F(R) denote the set of all functions from R to R. Define addition and multip
ID: 2977437 • Letter: L
Question
Let F(R) denote the set of all functions from R to R. Define addition and multiplication on F(R) as follows: For all f. g e F(R), (f + g): R rightarrow R is the function defined by for all x e R. For all is the function defined by for all x e R. Prove that F(R) is an Abelian group under addition. Does F(R) have an identity element for multiplication? Find an element in F(R) that does not have a multiplicative inverse in F(R). Explain how this shows F(R) is not a group under multiplication. Find necessary and sufficient conditions for an element in F(R) to be a unit in F(R). State your result in a lemma of the form "The function f e F(R) is a unit in F(R) if and only if"... Your lemma must say something more than just a rehash of the definition of a unit: rather, it must actually characterize the functions that are invertible under multiplication in F(R).Explanation / Answer
1) If f and g belong to F, they are functions from R to R. So h(x)=(f+g)(x)=f(x)+g(x) is also a function from R to R. So h also belongs to R.
For f,g,h belongs to F, (f+g)(x) + h(x)= f(x)+g(x)+h(x)=f(x)+(g+h)(x). So addition is associative.
Say, h(x)=0 for all x belong to R. So for any f belong to F, f+h=f. So identity alement exists.
If f belongs to R, -1*falso belongs to R since -f is also function from R to R. f+(-f)=0. So inverse also exists.
Since addition is abelian, F is an abelian group under addition.
2) Consider a function f(x)=1,for all x belong to R, so f also belongs toF, for any function g belong to F, fg(x)=f(x)*g(x)=1*g(x)=g(x). So a multiplicative identity in F exists.
3) Suppose f(x)= sin x, this is a function from R to R, so it belongs to F(R). But 1/sin(x) is not defined on points such as 0,and n. So 1/sin x does not belong to F(R). So F(R) cannot be a group since the multiplicative inverse does not exist for all f belong to F.
4) From the above example we can see that f has to be a non zero function for it to have a multiplicative inverse. Since the multiplicative inverse which is 1/f will not exist at points at which f is 0.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.