Define a function f:\mathbb R \to \mathbb R using the following rule:

"For any x, y \in \mathbb R, f(x + y) = f(x)f(y)"

Show that, if f is non-zero:

a) f(0) = 1

b) f(x) = 0 has no solutions

c) f(-x) = \dfrac 1 {f(x)}

Define a function f:\mathbb R \to \mathbb R using the following rule:

"For any x, y \in \mathbb R, f(x + y) = f(x)f(y)"

Show that, if f is non-zero:

a) f(0) = 1

b) f(x) = 0 has no solutions

c) f(-x) = \dfrac 1 {f(x)}

Better Latex:

Define a function f:\mathbb R \to \mathbb R using the following rule:

"For any x, y \in \mathbb R, f(x + y) = f(x)f(y)"

Show that, if f is non-zero:

(a) f(0) = 1

(b) f(x) = 0 has no solutions

© f(-x) = \dfrac 1 {f(x)}

proof next post.

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(a)

We have f(x) = f(x + 0) = f(x)f(0) for all x \in \mathbb R. Since f is not identically zero we must therefore have that f(0) = 1.

(b)

Suppose that \alpha is such that f(\alpha) = 0. Then f(x) = f((x - \alpha) + \alpha) = f(x - \alpha)f(\alpha) = 0 for all x \in \mathbb R. But f is not identically zero, so there exists no such \alpha.

©

We have:

f(x)f(-x) = f(x - x) = f(0) = 1

for all x \in \mathbb R, so:

\displaystyle f(-x) = \frac 1 {f(x)}

NB: I assume that by f being non-zero, they mean f is not identically zero, otherwise (b) is trivial from construction of f

1 Like

Tidied the latex