Let f: (R,+) → (R,, x) is defined s f(x) = ex for all x in R, where R→ set of real numbers ond R. → set of positive real numbers. Prove that ƒ is a homomorphism. Is f an isomorphism?

To prove that f: (R,+) → (R*,x) defined as f(x) = e^x for all x in R is a homomorphism, we need to show that it preserves the group structure.

Let’s consider two elements a, b ∈ R. Then we have:

f(a+b) = e^(a+b) = e^a * e^b = f(a) * f(b)

where the second equality follows from the properties of exponentiation. Therefore, f preserves the group structure, and is a homomorphism.

To prove that f is an isomorphism, we need to show that it is both injective and surjective.

First, let’s show that f is injective. Suppose that f(a) = f(b) for some a, b ∈ R. Then we have:

e^a = e^b

Taking the natural logarithm of both sides, we get:

a = b

Therefore, f is injective.

Next, let’s show that f is surjective. Given any y ∈ R*, we want to find an x ∈ R such that f(x) = y. Taking the natural logarithm of both sides of the equation f(x) = e^x = y, we get:

x = ln(y)

Since ln(y) is defined for all y ∈ R*, we can always find an x such that f(x) = y. Therefore, f is surjective.

Since f is both injective and surjective, it is bijective. Moreover, since f preserves the group structure, it is an isomorphism.