Rings -- real-valued functions

**liedora** · February 4th 2011, 03:06 PM

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.

Can someone please help me out a bit, thanks.

**Jose27** · February 4th 2011, 06:33 PM

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

**HallsofIvy** · February 5th 2011, 07:03 AM

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?

**tonio** · February 5th 2011, 07:45 AM

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

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

Tonio

**Jose27** · February 5th 2011, 07:52 AM

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 $\mathbb{Z}$ 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.

**tonio** · February 5th 2011, 09:46 AM

Originally Posted by Jose27

Not really, in $\mathbb{Z}$ 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 $f(x):=x^2$ is the non-negative reals...

Tonio

**Jose27** · February 5th 2011, 12:02 PM

Originally Posted by tonio

The composition is only defined there since the image of $f(x):=x^2$ 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).

**NonCommAlg** · February 5th 2011, 05:32 PM

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

**tonio** · February 5th 2011, 06:44 PM

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

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