avewright/proofwiki-math huggingface dataset:
Theorem: Definition of Polynomial from Polynomial Ring over Sequences

Let $\struct {R, +, \circ}$ be a ring with unity.
Let $\struct {P \sqbrk R, \oplus, \odot}$ be the polynomial ring over the set of all sequences in $R$:
:$P \sqbrk R = \set {\sequence {r_0, r_1, r_2, \ldots} }$
where the operations $\oplus$ and $\odot$ on $P \sqbrk R$ be defined as:
{{:Definition:Operations on Polynomial Ring of Sequences}}
Let $\struct {R \sqbrk X, +, \circ}$ be the ring of polynomials over $R$ in $X$.
{{explain|Strictly speaking the definition in terms of Definition:Polynomial Form is needed here, with $X$ specifically being an Definition:Indeterminate (Polynomial Theory) or Definition:Transcendental over Integral Domain.}}
Then $\struct {R \sqbrk X, +, \circ}$ and $\struct {P \sqbrk R, \oplus, \odot}$ are isomorphic.

Proof:
Let $P \sqbrk R$ be the polynomial ring over $R$.
Consider the injection $\phi: R \to P \sqbrk R$ defined as:
:$\forall r \in R: \map \phi r = \sequence {r, 0, 0, \ldots}$
It is easily checked that $\phi$ is a ring monomorphism.
So the set $\set {\sequence {r, 0, 0, \ldots}: r \in R}$ is a subring of $P \sqbrk R$ which is isomorphic to $R$.
So we ''identify'' $r \in R$ with the sequence $\sequence {r, 0, 0, \ldots}$.
Next we note that $P \sqbrk R$ contains the element $\sequence {0, 1, 0, \ldots}$ which we can call $x$.
From the definition of ring product on the polynomial ring over $R$, we have that:
:$x^2 = \sequence {0, 1, 0, \ldots} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, 0, 1, 0, 0, \ldots}$
:$x^3 = \sequence {0, 0, 1, 0, 0, \ldots} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, 0, 0, 1, 0, \ldots}$
and in general:
:$x^n = \sequence {0, 1, 0, \ldots}^{n - 1} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, \ldots \paren n \ldots, 0, 1, 0, \ldots}$
for all $n \ge 1$.
Hence we see that:
{{begin-eqn}}
{{eqn | l = \sequence {r_0, r_1, \ldots, r_n, 0, \ldots \ldots}
      | r = \sequence {r_0, 0, 0, \ldots \ldots} \odot \sequence {1, 0, 0, \ldots}
      | c = 
}}
{{eqn | ro= \oplus
      | o = 
      | r = \sequence {r_1, 0, 0, \ldots \ldots} \odot \sequence {0, 1, 0, \ldots}
      | c = 
}}
{{eqn | ro= \oplus
      | o = 
      | r = \cdots
      | c = 
}}
{{eqn | ro= \oplus
      | o = 
      | r = \sequence {r_n, 0, 0, \ldots \ldots} \odot \sequence {0, \ldots \paren n \ldots, 0, 1, 0, \ldots}
      | c = 
}}
{{eqn | r = r_0 \oplus r_1 \circ x \oplus r_2 \circ x^2 \oplus \ldots \oplus r_n \circ x^n
      | c = 
}}
{{end-eqn}}
So by construction, $R \sqbrk X$ is seen to be equivalent to $P \sqbrk R$.
{{qed}}
{{handwaving|The entirety of the above argument needs to be made properly rigorous.}}
It can also be shown that this proof works for the general ring whether it be a ring with unity or not.
Theorem: Integers under Multiplication form Countably Infinite Semigroup

The set of integers under multiplication $\struct {\Z, \times}$ is a countably infinite semigroup.

Proof:
From Integers under Multiplication form Semigroup, $\struct {\Z, \times}$ is a countably infinite semigroup.
Then we have that the Integers are Countably Infinite.
The criteria for $\struct {\Z, \times}$ to be a countably infinite semigroup are seen to be satisfied. 
{{Qed}}
Category:Integer Multiplication
Category:Examples of Semigroups
ofhk6z2f8lxht1xp6vezhh6swr7te1j
Theorem: Integer Addition is Closed

The set of integers is closed under addition:
:$\forall a, b \in \Z: a + b \in \Z$

Proof:
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested.
Integer addition is defined as:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} + \eqclass {c, d} {} = \eqclass {a + c, b + d} {}$
We have that:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \in \Z, \eqclass {c, d} {} \in \Z$
Also:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} + \eqclass {c, d} {} = \eqclass {a + c, b + d} {}$
But:
:$a + c \in \N, b + d \in \N$
So:
:$\forall a, b, c, d \in \N: \eqclass {a + c, b + d} {} \in \Z$
Therefore integer addition is closed.
{{qed}}
Theorem: Integer Addition Identity is Zero

The identity of integer addition is $0$:
:$\exists 0 \in \Z: \forall a \in \Z: a + 0 = a = 0 + a$

Proof:
Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$.
$\boxtimes$ is the congruence relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, as suggested.
From the method of construction, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the isomorphic copy of $0 \in \N$.
So, we need to show that:
:$\forall a, b, c \in \N: \eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$
Thus:
{{begin-eqn}}
{{eqn | l = \eqclass {a, b} {} + \eqclass {c, c} {}
      | r = \eqclass {a + c, b + c} {}
      | c = 
}}
{{eqn | r = \eqclass {a, b} {}
      | c = Construction of Inverse Completion: Members of Equivalence Classes
}}
{{end-eqn}}
So:
:$\eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {}$
The identity $\eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$ is demonstrated similarly.
{{qed}}
Theorem: Inverse for Integer Addition

Each element $x$ of the set of integers $\Z$ has an inverse element $-x$ under the operation of integer addition:
:$\forall x \in \Z: \exists -x \in \Z: x + \paren {-x} = 0 = \paren {-x} + x$

Proof:
Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$.
$\boxtimes$ is the congruence relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, as suggested.
From the method of construction, the element $\eqclass {a, a + x} {}$ has an inverse $\eqclass {a + x, a} {}$ where $a$ and $x$ are elements of the natural numbers $\N$.
Thus:
{{begin-eqn}}
{{eqn | l = \eqclass {a, a + x} {} + \eqclass {a + x, a} {}
      | r = \eqclass {a + a + x, a + x + a} {}
      | c = 
}}
{{eqn | r = \eqclass {a, a} {}
      | c = Construction of Inverse Completion: Members of Equivalence Classes
}}
{{eqn | r = \eqclass {a + x + a , a + a + x} {}
      | c = 
}}
{{eqn | r = \eqclass {a + x, a} {} + \eqclass {a, a + x} {}
      | c = 
}}
{{end-eqn}}
So $\eqclass {a, a + x} {}$ has the inverse $\eqclass {a + x, a} {}$.
{{explain|Needs more detail explaining the connection to the natural numbers. The Inverse Completion page is okay, but the link to what's going on here is hard to follow.}}
{{qed}}
Theorem: Rational Numbers form Ring

The set of rational numbers $\Q$ forms a ring under addition and multiplication: $\struct {\Q, +, \times}$.

Proof:
Recall that $\struct {\Q, +, \times}$ is a field.
As a field is also by definition a division ring, which is an example of a ring, the result follows.
{{qed}}
Theorem: Real Numbers form Ring

The set of real numbers $\R$ forms a ring under addition and multiplication: $\struct {\R, +, \times}$.

Proof:
From Real Numbers under Addition form Infinite Abelian Group, $\struct {\R, +}$ is an abelian group.
We also have that:
:Real Multiplication is Closed:
::$\forall x, y \in \R: x \times y \in \R$
:Real Multiplication is Associative:
::$\forall x, y, z \in \R: x \times \paren {y \times z} = \paren {x \times y} \times z$
Thus $\struct {\R, +}$ is a semigroup.
Finally we have that Real Multiplication Distributes over Addition:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in \R
      | l = x \times \paren {y + z}
      | r = x \times y + x \times z
}}
{{eqn | l = \paren {y + z} \times x
      | r = y \times x + z \times x
}}
{{end-eqn}}
Hence the result, by definition of ring.
{{qed}}
Theorem: Complex Numbers form Ring

The set of complex numbers $\C$ forms a ring under addition and multiplication: $\struct {\C, +, \times}$.

Proof:
From Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is an abelian group.
We also have that:
:Complex Multiplication is Closed:
::$\forall x, y \in \C: x \times y \in \C$
:Complex Multiplication is Associative:
::$\forall x, y, z \in \C: x \times \paren {y \times z} = \paren {x \times y} \times z$
Thus $\struct{\C, +}$ is a semigroup.
Finally we have that Complex Multiplication Distributes over Addition:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in \C
      | l = x \times \paren {y + z}
      | r = x \times y + x \times z
}}
{{eqn | l = \paren {y + z} \times x
      | r = y \times x + z \times x
}}
{{end-eqn}}
Hence the result, by definition of ring.
{{qed}}
Theorem: Complex Addition Identity is Zero

Let $\C$ be the set of complex numbers.
The identity element of $\struct {\C, +}$ is the complex number $0 + 0 i$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \paren {x + i y} + \paren {0 + 0 i}
      | r = \paren {x + 0} + i \paren {y + 0} = x + i y
}}
{{eqn | l = \paren {0 + 0 i} + \paren {x + i y}
      | r = \paren {0 + x} + i \paren {0 + y} = x + i y
}}
{{end-eqn}}
{{qed}}
Theorem: Rational Addition Identity is Zero

The identity of rational number addition is $0$:
:$\exists 0 \in \Q: \forall a \in \Q: a + 0 = a = 0 + a$

Proof:
From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
From Zero of Inverse Completion of Integral Domain, for any $k \in \Z^*$, the element $\dfrac {0_D} k$ of $\Q$ serves as the zero of $\struct {\Q, +, \times}$.
Hence $\dfrac 0 k$ is the identity for $\left({\Q, +}\right)$:
{{begin-eqn}}
{{eqn | l = \frac a b + \frac 0 k
      | r = \frac {a \times k + b \times 0} {b \times k}
      | c = 
}}
{{eqn | r = \frac {a \times k} {b \times k}
      | c = 
}}
{{eqn | r = \frac a b
      | c = Equal Elements of Field of Quotients
}}
{{end-eqn}}
Similarly for $\dfrac 0 k + \dfrac a b$.
Next we note that it ''is'' a zero:
{{begin-eqn}}
{{eqn | l = \frac a b \times \frac 0 k
      | r = \frac {a \times 0} {b \times k}
      | c = 
}}
{{eqn | r = \frac 0 {b \times k}
      | c = 
}}
{{eqn | r = \frac {0 \times b} {k \times b}
      | c = 
}}
{{eqn | r = \frac 0 k
      | c = Equal Elements of Field of Quotients
}}
{{end-eqn}}
Hence we ''define'' the zero of $\struct {\Q, +, \times}$ as $0$ and identify it with the set of all elements of $\Q$ of the form $\dfrac 0 k$ where $k \in \Z^*$.
{{qed}}
Category:Rational Addition
7xbsf9voemuqibzxl3t8s0tzo1ramkn
Theorem: Inverse for Complex Addition

Let $z = x + i y \in \C$ be a complex number.
Let $-z = -x - i y \in \C$ be the negative of $z$.
Then $-z$ is the inverse element of $z$ under the operation of complex addition:
:$\forall z \in \C: \exists -z \in \C: z + \paren {-z} = 0 = \paren {-z} + z$

Proof:
From Complex Addition Identity is Zero, the identity element for $\struct {\C, +}$ is $0 + 0 i$.
Then:
{{begin-eqn}}
{{eqn | o =  
      | r = \paren {x + i y} + \paren {-x - i y}
      | c = 
}}
{{eqn | r = \paren {x - x} + i \paren {y - y}
      | c = 
}}
{{eqn | r = 0 + 0 i
      | c = 
}}
{{end-eqn}}
Similarly for $\paren {-x - i y} + \paren {x + i y}$.
{{qed}}
Theorem: Equal Elements of Field of Quotients

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $\struct {K, +, \circ}$ be the field of quotients of $\struct {D, +, \circ}$.
Let $x = \dfrac p q \in K$.
Then:
:$\forall k \in D^*: x = \dfrac {p \circ k} {q \circ k}$
where:
:$D^* := D \setminus \set {0_D}$
that is, $D$ with its zero removed.

Proof:
We have that the field of quotients $\struct {K, +, \circ}$ of an integral domain is its inverse completion.
Thus we have:
:$\forall x_1, x_2 \in D, y_1, y_2 \in D^*: \dfrac {x_1} {y_1} = \dfrac {x_2} {y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
So:
{{begin-eqn}}
{{eqn | l = \paren {p \circ k} \circ q
      | r = p \circ \paren {q \circ k}
      | c = as $D$ is an integral domain: $\circ$ is commutative and associative
}}
{{eqn | ll= \leadsto
      | l = \frac p q
      | r = \frac {p \circ k} {q \circ k}
      | c = 
}}
{{end-eqn}}
Hence the result.
Note that in order for $\dfrac {p \circ k} {q \circ k}$ to be defined, $q \circ k \ne 0_D$, that is, $k \ne 0_D$.
{{qed}}
Category:Integral Domains
Category:Inverse Completions
Category:Fields of Quotients
8z2cxquhurho9p83ll897w71ug6ast7
Theorem: Rational Addition is Closed

The operation of addition on the set of rational numbers $\Q$ is well-defined and closed:
:$\forall x, y \in \Q: x + y \in \Q$

Proof:
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $+$ is well-defined and closed on $\Q$.
{{qed}}
Theorem: Inverse for Rational Addition

Each element $x$ of the set of rational numbers $\Q$ has an inverse element $-x$ under the operation of rational number addition:
:$\forall x \in \Q: \exists -x \in \Q: x + \paren {-x} = 0 = \paren {-x} + x$

Proof:
Let $x = \dfrac a b$ where $b \ne 0$.
We take the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
From Existence of Field of Quotients, we have that the inverse of $\dfrac a b$ for $+$ is $\dfrac {-a} b$:
{{begin-eqn}}
{{eqn | l = \frac a b + \frac {-a} b
      | r = \frac {a \times b + b \times \paren {-a} } {b \times b}
      | c = 
}}
{{eqn | r = \frac {b \times \paren {a + \paren {-a} } } {b \times b}
      | c = 
}}
{{eqn | r = \frac {b \times 0} {b \times b}
      | c = 
}}
{{eqn | r = \frac 0 {b \times b}
      | c = 
}}
{{eqn | r = 0
      | c = Rational Addition Identity is Zero
}}
{{end-eqn}}
From Negative of Division Product, we have that:
:$-\dfrac a b = \dfrac {-a} b = \dfrac a {-b}$
So $\dfrac a b$ has a unique and unambiguous inverse for $+$.
{{qed}}
Category:Rational Addition
Category:Examples of Inverse Elements
3i3498hj1qxrki0s59psrhf0mbl3s6h
Theorem: Rational Multiplication is Closed

The operation of multiplication on the set of rational numbers $\Q$ is well-defined and closed:
:$\forall x, y \in \Q: x \times y \in \Q$

Proof:
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore {{afortiori}} $\times$ is well-defined and closed on $\Q$.
{{qed}}
Theorem: Rational Multiplication is Closed

The operation of multiplication on the set of rational numbers $\Q$ is well-defined and closed:
:$\forall x, y \in \Q: x \times y \in \Q$

Proof:
From the definition of rational numbers, there exists four integers $p$, $q$, $r$, $s$, where:
:$q \ne 0$
:$s \ne 0$
:$\dfrac p q = x$
:$\dfrac r s = y$
We have that:
:$p \times r \in \Z$
:$q \times s \in \Z$
Since $q \ne 0$ and $s \ne 0$, we have that:
:$q \times s \ne 0$
Therefore, by the definition of rational numbers:
:$x \times y = \dfrac {p \times r} {q \times s} \in \Q$
Hence the result.
{{qed}}
Theorem: Rational Multiplication Identity is One

The identity of rational number multiplication is $1$:
:$\exists 1 \in \Q: \forall a \in \Q: a \times 1 = a = 1 \times a$

Proof:
From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
From the properties of the quotient structure, elements of $\Q$ of the form $\dfrac p p$ where $p \ne 0$ act as the identity for multiplication.
From Equal Elements of Field of Quotients, we have that:
:$\dfrac p p = \dfrac {1 \times p} {1 \times p} = \dfrac 1 1$
Hence $\dfrac p p$ is the identity for $\struct {\Q, \times}$:
{{begin-eqn}}
{{eqn | l = \frac a b \times \frac p p
      | r = \frac {a \times p} {b \times p}
      | c = 
}}
{{eqn | r = \frac a b
      | c = Equal Elements of Field of Quotients
}}
{{end-eqn}}
Similarly for $\dfrac p p \times \dfrac a b$.
Hence we ''define'' the unity of $\struct {\Q, +, \times}$ as $1$ and identify it with the set of all elements of $\Q$ of the form $\dfrac p p$ where $ \in \Z^*$.
{{qed}}
Category:Rational Multiplication
scokc9618ph3cghsfsoigfgrdh39k01
Theorem: Inverse for Rational Multiplication

Each element $x$ of the set of non-zero rational numbers $\Q_{\ne 0}$ has an inverse element $\dfrac 1 x$ under the operation of rational number multiplication:
:$\forall x \in \Q_{\ne 0}: \exists \dfrac 1 x \in \Q_{\ne 0}: x \times \dfrac 1 x = 1 = \dfrac 1 x \times x$

Proof:
From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
From Rational Multiplication Identity is One, the identity for $\struct {\Q, \times}$ is $1 = \dfrac 1 1 = \dfrac p p$ where $p \in \Z$ and $p \ne 0$.
From the definition of product inverses in field of quotients, we have that:
:$\paren {\dfrac p q}^{-1} = \dfrac q p$
which can be demonstrated directly:
{{begin-eqn}}
{{eqn | l = \frac p q \times \frac q p
      | r = \frac {p \times q} {q \times p}
      | c = 
}}
{{eqn | r = \frac {p \times q} {p \times q}
      | c = 
}}
{{eqn | r = \frac 1 1
      | c = Equal Elements of Field of Quotients
}}
{{end-eqn}}
Note that this is defined only when $p \ne 0$ and $q \ne 0$.
Now let $x = \dfrac p q \in \Q$.
We define $\dfrac 1 x$ as $1 \times \paren {\dfrac p q}^{-1}$.
Hence:
{{begin-eqn}}
{{eqn | l = \frac 1 x
      | r = \frac 1 1 \times \paren {\frac p q}^{-1}
      | c = 
}}
{{eqn | r = \frac 1 1 \times \frac q p
      | c = 
}}
{{eqn | r = \frac {1 \times q} {1 \times p}
      | c = 
}}
{{eqn | r = \frac q p
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Real Addition Identity is Zero

The identity of real number addition is $0$:
:$\exists 0 \in \R: \forall x \in \R: x + 0 = x = 0 + x$

Proof:
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.
Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.
From the definition of real addition, $x + y$ is defined as $\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$.
Let $\sequence {0_n}$ be such that $\forall i: 0_n = 0$.
Then we have:
{{begin-eqn}}
{{eqn | l = \eqclass {\sequence {0_n} } {} + \eqclass {\sequence {x_n} } {}
      | r = \eqclass {\sequence {0_n + x_n} } {}
      | c = 
}}
{{eqn | r = \eqclass {\sequence {0 + x_n} } {}
      | c = 
}}
{{eqn | r = \eqclass {\sequence {x_n} } {}
      | c = 
}}
{{end-eqn}}
Similarly for $\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {0_n} } {}$.
Thus the identity element of $\struct {\R, +}$ is the real number $0$.
{{qed}}
Theorem: Inverse for Real Addition

Each element $x$ of the set of real numbers $\R$ has an inverse element $-x$ under the operation of real number addition:
:$\forall x \in \R: \exists -x \in \R: x + \paren {-x} = 0 = \paren {-x} + x$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \eqclass {\sequence {x_n} } {} + \paren {-\eqclass {\sequence {x_n} } {} }
      | r = \eqclass {\sequence {x_n - x_n} } {}
      | c = 
}}
{{eqn | r = \eqclass {\sequence {0_n} } {}
      | c = 
}}
{{end-eqn}}
Similarly for $\paren {-\eqclass {\sequence {x_n} } {} } + \eqclass {\sequence {x_n} } {}$.
Thus the inverse of $x \in \struct {\R, +}$ is $-x$.
{{qed}}
Category:Real Addition
Category:Examples of Inverse Elements
21b4dil1gjwlplla2a923jj0elvyzhj
Theorem: Real Multiplication Identity is One

The identity element of real number multiplication is the real number $1$:
:$\exists 1 \in \R: \forall a \in \R_{\ne 0}: a \times 1 = a = 1 \times a$

Proof:
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.
Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.
From the definition of real multiplication, $x \times y$ is defined as:
:$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$
Let $\sequence {1_n}$ be such that $\forall i: 1_n = 1$.
Then we have:
{{begin-eqn}}
{{eqn | l = \eqclass {\sequence {1_n} } {} \times \eqclass {\sequence {x_n} } {}
      | r = \eqclass {\sequence {1_n \times x_n} } {}
      | c = 
}}
{{eqn | r = \eqclass {\sequence {1 \times x_n} } {}
      | c = 
}}
{{eqn | r = \eqclass {\sequence {x_n} } {}
      | c = 
}}
{{end-eqn}}
Similarly for $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {1_n} } {}$.
So the identity element of $\struct {\R_{\ne 0}, \times}$ is the real number $1$.
{{qed}}
Theorem: Inverse for Real Multiplication

Each element $x$ of the set of non-zero real numbers $\R_{\ne 0}$ has an inverse element $\dfrac 1 x$ under the operation of real number multiplication:
:$\forall x \in \R_{\ne 0}: \exists \dfrac 1 x \in \R_{\ne 0}: x \times \dfrac 1 x = 1 = \dfrac 1 x \times x$

Proof:
By the definition of real number:
:$\forall \epsilon \in \R_{>0}: \exists t \in \N: \forall i > t: \size {x_i - x} < \epsilon$
{{explain|In order for this line to make sense, it needs the definition (or at least the relevant part) extracted and posted here, so the context of the convergence condition is clear.}}
Let $\epsilon = \size x$.
This is possible because $x \ne 0$, since it is required that $\epsilon > 0$.
{{begin-eqn}}
{{eqn | q = \exists t \in \N: \forall i > t
      | l = \size {x_i - x}
      | o = <
      | r = \size x
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists t \in \N: \forall i > t
      | l = \size {x_i - x}
      | o = <
      | r = \size {x_i - x} + \size {x_i}
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | ll= \leadsto
      | q = \exists t \in \N : \forall i > t
      | l = \size {x_i}
      | o = >
      | r = 0
      | c = 
}}
{{end-eqn}}
Construct a sequence $\sequence {y_n}$ as follows:
:$y_n = \begin {cases} \dfrac 1 {x_n} & n > t \\ 0 & n \le t \end {cases}$
Let $u_n$ denote the Heaviside step function with parameter $t$.
We have:
{{begin-eqn}}
{{eqn | l = \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {}
      | r = \eqclass {\sequence {x_n y_n} } {}
      | c = 
}}
{{eqn | r = \eqclass {\sequence {u_n} } {}
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Similarly for $\eqclass {\sequence {y_n} } {} \times \eqclass {\sequence {x_n} } {}$.
So the inverse of $x \in \struct {\R_{\ne 0}, \times}$ is $x^{-1} = \dfrac 1 x = y$.
{{qed}}
Theorem: Complex Multiplication Identity is One

Let $\C_{\ne 0}$ be the set of complex numbers without zero.
The identity element of $\struct {\C_{\ne 0}, \times}$ is the complex number $1 + 0 i$.

Proof:
{{begin-eqn}}
{{eqn | l = \paren {x + i y} \paren {1 + 0 i}
      | r = \paren {x \cdot 1 - y \cdot 0} + i \paren {x \cdot 0 + y \cdot 1}
      | c = 
}}
{{eqn | r = \paren {x + i y}
      | c = 
}}
{{end-eqn}}
and similarly:
{{begin-eqn}}
{{eqn | l = \paren {1 + 0 i} \paren {x + i y}
      | r = \paren {1 \cdot x - 0 \cdot y} + i \paren {0 \cdot x + 1 \cdot y}
      | c = 
}}
{{eqn | r = \paren {x + i y}
      | c = 
}}
{{end-eqn}}
{{qed}}