# Thread: Rings -- real-valued functions

1. ## Rings -- real-valued functions

Let R be the ring of all real-valued functions defined on the real line.

(a) Find all the zero-divisors in R.
(b) Find all the units in R.

2. Hint: Let A be the set of zero divisors and B the set of units then $\displaystyle R\setminus A =B$

3. Of course, "units" are precisely those members that have multiplicative inverses and the "zero divisors" those that do not.

Of course, the real valued functions form a ring with two different kinds of "multiplication"- f(x)g(x) and f(g(x)). Which do you mean?

4. Originally Posted by HallsofIvy
Of course, "units" are precisely those members that have multiplicative inverses and the "zero divisors" those that do not.

Of course, the real valued functions form a ring with two different kinds of "multiplication"- f(x)g(x) and f(g(x)). Which do you mean?

Most probably he means the first one since the second one isn't, in a strict fashion, a ring

since functions composition may not yield a function defined in the WHOLE real numbers.

For example, the functions $\displaystyle f(x):=x^2\,,\,\,g(x)=x$ are defined on the real line, but $\displaystyle g\circ f$ is only defined

in the half line $\displaystyle [0,\infty)$ .

Tonio

5. Originally Posted by HallsofIvy
Of course, "units" are precisely those members that have multiplicative inverses and the "zero divisors" those that do not.
Not really, in $\displaystyle \mathbb{Z}$ 2 is not a unit or a zero divisor, which is why I gave the fact that in this case it is true as a hint.

For example, the functions are defined on the real line, but is only defined

in the half line .
Isn't one of them the identity? I don't see how the composition is only defined there. Besides it's easier to see that composition fails to distribute over the sum when the functions involved are non-linear.

6. Originally Posted by Jose27
Not really, in $\displaystyle \mathbb{Z}$ 2 is not a unit or a zero divisor, which is why I gave the fact that in this case it is true as a hint.

Isn't one of them the identity? I don't see how the composition is only defined there. Besides it's easier to see that composition fails to distribute over the sum when the functions involved are non-linear.

The composition is only defined there since the image of $\displaystyle f(x):=x^2$ is the non-negative reals...

Tonio

7. Originally Posted by tonio
The composition is only defined there since the image of $\displaystyle f(x):=x^2$ is the non-negative reals...

Tonio
What? The composition, by definition, has the same domain as the "initial" function (the fact that when we apply g the image will be restricted is another matter entirely).

8. for any $\displaystyle f \in R$ let $\displaystyle Z(f)=\{a \in \mathbb{R}: \ f(a)=0 \}.$ then $\displaystyle \{ f \in R: \ Z(f) \neq \emptyset \}$ is the set of zero divisors of $\displaystyle R$ and $\displaystyle \{f \in R: \ Z(f) = \emptyset \}$ is the set of units of $\displaystyle R$.

9. Originally Posted by Jose27
What? The composition, by definition, has the same domain as the "initial" function (the fact that when we apply g the image will be restricted is another matter entirely).

You're right, I miswrote the example. Thanx.

Tonio