algebraic structures, Blogs, Proofs

A Ring of Mappings

Last time, in Composition of Mappings on a Set, we saw that the set of all mappings from a set to itself forms a monoid under composition. In this post we will work with the set of all mappings from \(\mathbb{R}\) to itself, but with different operations. In particular, we will look at what properties are held by the usual sum and product of mapping of real numbers. We will then show that the set of all mapping on \(\mathbb{R}\), with both of these properties, forms a commutative ring with identity.

Addition of Mappings

Let \(M(\mathbb{R})\) be the set of all mappings from \(\mathbb{R}\) to itself. Furthermore, define \((f+g)(x)\) as \(f(x)+g(x)\). Then show that \(\langle M(\mathbb{R}), +\rangle\) is a group.

Closure

We will again work through this problem looking at the each of the properties that must be maintained one at a time. The first property will look at is that the set \(M(\mathbb{R})\) is closed under addition, that is, addition of mappings is an operation on \(\mathbb{R}\). In order to show this, we need to ensure that the sum of any two mappings on \(\mathbb{R}\) is again a mapping on \(\mathbb{R}\).

Now, let \(f,g \in \mathbb{R}\). In order to show that \(f+g\) is a mapping, we need to show that \((f+g)(x)\) is well-defined for all \(x \in \mathbb{R}\). Then, let \(x \in \mathbb{R}\). Here, we see that
\begin{align*}
(f+g)(x)&=f(x)+g(x) \\
&=y+z,
\end{align*}
for some \(y\) and \(z\) in \(\mathbb{R}\) since \(f\) and \(g\) are mappings on \(\mathbb{R}\). Then, since \(\mathbb{R}\) is closed for addition, we get that \(y+z \in \mathbb{R}\), so \((f+g)(x) \in \mathbb{R}\) for all \(x \in \mathbb{R}\) and \(M(\mathbb{R})\) is closed under addition. We can now give a formal proof.

Proof of Closure

Let \(f,g \in \mathbb{R}\) with \(x \in \mathbb{R}\). Then
\begin{align*}
(f+g)(x)&=f(x)+g(x) \in \mathbb{R}
\end{align*}
since \(f(x), g(x) \in \mathbb{R}\). Hence, \(M(\mathbb{R})\) is closed for addition.

Associativity

For associativity, we need to show that for any \(f,g,h \in M(\mathbb{R})\), we have that \((f+g)+h=f+(g+h)\). That is, for all \(x \in \mathbb{R}\), we have that \(((f+g)+h)(x)=(f+(g+h))(x)\). Note that \(((f+g)+h))(x)=f(x)+g(x)+h(x)=(f+(g+h))(x)\), therefore, we will have associativity.

Proof of Associativity

Let \(f,g,h \in M(\mathbb{R})\) and \(x \in \mathbb{R}\). Then
\begin{align*}
((f+g)+h))(x)&=(f+g)(x)+h(x) \\
&=f(x)+g(x)+h(x) \\
&=f(x)+(g+h)(x) \\
&=(f+(g+h))(x).
\end{align*}

Hence, \(+\) is associative on \(M(\mathbb{R})\).

Identity

Here, we just need to note that the identity for addition will be the \(0\) mapping. That is, we will need \(e(x)=0\) for all \(x \in \mathbb{R}\).

Proof of Identity

Let \(g \in M(\mathbb{R})\). Define \(e: \mathbb{R} \to \mathbb{R}\) as \(e(x)=0\) for all \(x \in \mathbb{R}\). Then note that \(e\) is well-defined. Furthermore, for \(y \in \mathbb{R}\), we have that
\begin{align*}
(g+e)(y)&=g(y)+e(y)=g(y)+0=g(y) \\
(e+g)(y)&=e(y)+g(y)=0+g(y)=g(y).
\end{align*}
Hence, \(e\) is the identity on \(M(\mathbb{R})\).

Inverses

Next, let \(f \in M(\mathbb{R})\). We will now need to find the inverse of this mapping. That is, we need to find \(g(x) \in \mathbb{R}\) such that \(f+g=0\), where \(0\) is the identity mapping we just found. Note that, in this case, we will need
\begin{align*}
f(x)+g(x)&=0 \\
g(x)&=-f(x).
\end{align*}
for all \(x \in \mathbb{R}\). That is, the inverse mapping will just be the mapping that sends \(x\) to \(-f(x)\). We can now give our proof.

Proof of Inverses

Let \(f \in M(\mathbb{R})\). Then defined \(g(x)\) as \(g(x)=-f(x)\). Note that \(f(x)\) is well-defined for all \(x \in \mathbb{R}\) so \(-f(x)\) is well-defined for all \(x \in \mathbb{R}\). Hence \(g \in M(\mathbb{R})\). Now,
\begin{align*}
(f+g)(x)&=f(x)+g(x) \\
&=f(x)+(-f(x)) \\
&=0 \\
&=(-f(x))+f(x) \\
&=g(x)+f(x).
\end{align*}

Hence \(g\) is the additive inverse of \(f\). Therefore, there exists an inverse of \(f\) for all \(f \in M(\mathbb{R})\).

Commutativity

While commutativity is not needed for a group, we may as well show that \(+\) satisfies commutativity on \(M(\mathbb{R})\). In order to do this, we need to show that \((f+g)=(g+f)\). That is, for all \(x \in \mathbb{R}\), \((f+g)(x)=(g+f)(x)\). However, we note that \(f(x)+g(x)=g(x)+f(x)\) since the real numbers are commutative, so commutativity of mappings follows.

Proof of Commutativity

Let \(f,g \in M(\mathbb{R})\) and \(x \in \mathbb{R}\). Then, note that
\begin{align*}
(f+g)(x)&=f(x)+g(x) \\
&=g(x)+f(x) \\
&=(g+f)(x).
\end{align*}
Hence, \(+\) is a commutative.

Group

We have shown that on \(M(\mathbb{R})\), \(+\) is an associative and commutative operation with an identity and inverses. Hence, \(\langle M(\mathbb{R}),+\rangle\) is an Abelian (commutative) group.

Product of Mappings

We have now seen that \(M(\mathbb{R})\) forms a group for \(+\), but we can also determine what properties holds under the product of functions. That is, for \(f,g \in M(\mathbb{R})\) let \((f \cdot g)(x)=f(x) \cdot g(x)\). Show that \(\langle M(\mathbb{R}), \cdot \rangle\) is a commutative monoid.

As we look at this, we should note that the proofs will be very similar to what we had for \(+\).

Closure

In order to show that \(\cdot\) is an operation on \(M(\mathbb{R})\), we need to show that \(f \cdot g \in M(\mathbb{R})\) for all \(f,g \in M(\mathbb{R})\). That is, we need to show that \((f \cdot g)(x)\) is well-defined in \(\mathbb{R}\) for all \(x \in \mathbb{R}\). However, since \((f \cdot g)(x)=f(x) \cdot g(x)\) is the product of real numbers, it must again be a real number. Therefore, we can give our proof.

Proof of Closure

Let \(f,g \in M(\mathbb{R})\) and \(x \in \mathbb{R}\). Then
\begin{align*}
(f \cdot g)(x) &=f(x) \cdot g(x)
\end{align*}
is well-defined in \(\mathbb{R}\) since \(\mathbb{R}\) is closed under multiplication.

Associativity

As was the case with addition, associativity of mapping products will follow from the associativity of products of real numbers.

Proof of Associativity

Let \(f,g,h \in M(\mathbb{R})\) and \(x \in \mathbb{R}\). We then have that
\begin{align*}
(f \cdot (g \cdot h))(x)&=f(x) \cdot (g \cdot h)(x) \\
&=f(x) \cdot g(x) \cdot h(x) \\
&=(f \cdot g)(x) \cdot h(x) \\
&=((f \cdot g) \cdot h)(x).
\end{align*}

Hence, \(\cdot\) is associative on \(M(\mathbb{R})\).

Identity

When we looked at addition, we noted that the identity was the \(0\) function. Therefore, it would make sense that the \(1\) function would be the identity for multiplication. That is, we should define \(e(x)=1\) for all \(x \in \mathbb{R}\).

Proof of Identity

Let \(f \in M(\mathbb{R})\). Then define \(e(x)=1\) for all \(x \in \mathbb{R}\). Since \(e(x)\) is well-defined for all \(x \in \mathbb{R}\), we have that \(e(x) \in M(\mathbb{R})\). Furthermore,
\begin{align*}
(f \cdot e)(x) &=f(x) \cdot e(x) \\
&=f(x) \cdot 1 \\
&=f(x) \\
&=1 \cdot f(x) \\
&=e(x) \cdot f(x) \\
&=(e \cdot f)(x).
\end{align*}

Hence, \(e(x)\) is the identity for \(\cdot\) on \(M(\mathbb{R})\).

Inverse

We note here that the inverse of a mapping \(f\) with respect to \(\cdot\) would be the mapping \(\frac{1}{f}\). However, if \(f(x)=0\) for any \(x \in \mathbb{R}\), then this would not be a well-defined mapping on \(\mathbb{R}\). Hence, any mapping that sends anything to \(0\) would not have an inverse. Two such examples would be \(f(x)=0\) or \(g(x)=x\). Since we were not asked to show that we did not have inverses, we will not include a formal proof of this.

Combining This Together

Now that we have finished determining which properties hold for sums and product of mappings for the set of mapping on \(\mathbb{R}\), we need to look at the interaction between these operations. In particular, we need to determine if the distributive properties hold.

Distributivity

For distributivity, we will look at three mappings. Therefore, let \(f,g,h \in M(\mathbb{R})\). We will then need to show that \(f\cdot(g+h)=f\cdot g+f\cdot h\). In order to do this, we let \(x \in \mathbb{R}\) and we need to show that both mappings send any such \(x\) to the same place. Therefore, we find that
\begin{align*}
[f\cdot (g +h)](x)&=f(x) \cdot (g+h)(x) \\
&=f(x) \cdot (g(x)+h(x)).
\end{align*}
Noting that the real numbers satisfy distributivity for multiplication and addition, we then find that
\begin{align*}
[f\cdot (g +h)](x)&=f(x) \cdot (g+h)(x) \\
&=f(x) \cdot (g(x)+h(x)) \\
&=f(x)\cdot g(x) +f(x) \cdot h(x) \\
&=(f \cdot g)(x)+(f \cdot h)(x).
\end{align*}
Hence, we will have left distributivity. Furthermore, since we have shown that the product is commutative, we will get right distributivity as well.

Proof of Distributivity

Let \(f,g,h \in M(\mathbb{R})\) and \(x \in \mathbb{R}\). We then note that
\begin{align*}
[f\cdot (g +h)](x)&=f(x) \cdot (g+h)(x) \\
&=f(x) \cdot (g(x)+h(x)) \\
&=f(x)\cdot g(x) +f(x) \cdot h(x) \\
&=(f \cdot g)(x)+(f \cdot h)(x).
\end{align*}

Hence, left distributivity holds. Furthermore
\begin{align*}
(g +h) \cdot f &=f \cdot (g+h) \\
&=(f \cdot g)+(f \cdot h) \\
&=(g \cdot f)+ (h \cdot f).
\end{align*}
Therefore, right distributivity holds.

Determining the type of Algebra

We are now ready to look at all properties that are held by these operations in order to determine what type of algebra we have. We note the following.

  • \(\langle M(\mathbb{R}), + \rangle\) is an Abelian group.
  • We also have that \(\cdot\) is an associative and commutative operation with an identity on \(M(\mathbb{R})\). Hence, \(\langle M(\mathbb{R}),\cdot\rangle\) is a commutative monoid.
  • Left and right distributivity are maintained.
  • This then give us that \(\langle M(\mathbb{R}), +, \cdot \rangle \) is a commutative ring with identity.

Conclusion

Thank you for reading along as we work on more algebra. I hope you found the step-by-step process with helpful as you study more about groups, monoids or rings. If you have more questions, please let me know by leaving a comment below or by contacting me. If you did find this helpful, make sure to share it with others that would also find it useful. Also, be sure to check out our YouTube channel for helpful videos.

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