hcju/proofwikiqa huggingface dataset:
This proof assumes that $R$ is a field, which makes the triangulation process slightly quicker. By this assumptions, all elements of $\mathbf A$ have multiplicative inverses. Let $\mathbf A$ be a square matrix of order $n$. We proceed by induction on $n$, the number of rows of $\mathbf A$. === Basis for the Induction === For $n = 1$, we have a matrix of just one element, which is trivially diagonal, hence both upper and lower triangular. This is the basis for the induction. === Induction Hypothesis === Fix $n \in \N$, and assume all $n \times n$-matrices can be upper triangularised by elementary row operations.  If $R$ is a field, assume all $n \times n$-matrices can be upper triangularised by elementary row operations of type 2. This forms our induction hypothesis. === Induction Step === Let $\mathbf A = \left[{a}\right]_{n + 1}$ be a square matrix of order $n + 1$. When the first column of $\mathbf A$ contains only zeroes, it is upper triangularisable {{iff}} the submatrix $\mathbf A \left({1; 1}\right)$ is. Each elementary row operation used in triangularisation process of the submatrix $\mathbf A \left({1; 1}\right)$ will not change the zeros of the first column of $\mathbf A$. So when $\mathbf A \left({1; 1}\right)$ is upper triangularised, then $\mathbf A$ will also be upper triangularised. From the induction hypothesis, we conclude that $\mathbf A$ can be upper triangularised by elementary row operations. Now suppose that its first column contains a non-zero value. Suppose that $a_{11} = 0$. Let $j$ be the smallest row index such that $a_{j1} \ne 0$, and note that $j$ exists by assumption. Now apply the following operation of type 2: :$r_1 \to r_1 + r_j$ As $a_{j1} \ne 0$, this enforces $a_{11} \ne 0$, and we continue as in the case below. Suppose $a_{11} \ne 0$.  We use the following operations of type 2: :$\forall j \in \left\{{2, \ldots, n+1}\right\}: r_j \to r_j - \dfrac {a_{j1}} {a_{11}} r_1$ This will put the first column to zero (except for the first element, $a_{11}$). It follows that $\mathbf A$ can be upper triangularised precisely when the submatrix $\mathbf A \left({1; 1}\right)$ can. Again, the induction hypothesis renders $\mathbf A$ upper triangularisable by elementary row operations. This completes the case distinction, and hence the result follows by induction. To put the matrix $\mathbf A$ into lower triangular form, just do the same thing, but start with the last column and the last diagonal element $a_{n + 1 \; n + 1}$. {{qed}}
From the definition of matrix product, we have: :$\displaystyle \forall i, j \in \closedint 1 n: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} b_{k j}$ Now when $i = j$ (as on the main diagonal): :$\displaystyle c_{j j} = \sum_{k \mathop = 1}^n a_{j k} b_{k j}$ Now both $\mathbf A$ and $\mathbf B$ are upper triangular. Thus: :if $k > j$, then $b_{k j} = 0$ and thus $a_{j k} b_{k j} = 0$ :if $k < j$, then $a_{j k} = 0$ and thus $a_{j k} b_{k j} = 0$. So $a_{j k} b_{k j} \ne 0$ only when $j = k$. So: :$\displaystyle c_{j j} = \sum_{k \mathop = 1}^n a_{j k} b_{k j} = a_{j j} b_{j j}$ Now if $i > j$, it follows that either $a_{i k}$ or $b_{k j}$ is zero for all $k$, and thus $c_{i j} = 0$. Thus $\mathbf C$ is upper triangular. The same argument can be used for when $\mathbf A$ and $\mathbf B$ are both lower triangular matrices. {{Qed}} Category:Triangular Matrices Category:Conventional Matrix Multiplication kfypsdf63kqm9gggh85dbj560bfaa6i
From Elementary Row Operations as Matrix Multiplications, an elementary row operation on $\mathbf A$ is equivalent to matrix multiplication by the elementary row matrices corresponding to the elementary row operations. From Determinant of Elementary Row Matrix, the determinants of those elementary row matrices are as follows: === Scale Row === {{:Determinant of Elementary Row Matrix/Scale Row}} === Add Scalar Product of Row to Another === {{:Determinant of Elementary Row Matrix/Scale Row and Add}} === Exchange Rows === {{:Determinant of Elementary Row Matrix/Exchange Rows}} Hence the result. {{qed}}
Let $z = a + i b$. Then: {{begin-eqn}} {{eqn | l = z \overline z       | r = a^2 + b^2       | c = Product of Complex Number with Conjugate }} {{eqn | r = \cmod z^2       | c = {{Defof|Complex Modulus}} }} {{end-eqn}} {{qed}}
The result follows from Existence of Positive Root of Positive Real Number and Uniqueness of Positive Root of Positive Real Number. {{qed}}
Let $n \in \Z$ be an even integer. Then $n = 2 k$ for some $k \in \Z$. Thus: :$\forall x \in \R: x^n = x^{2 k} = \paren {x^k}^2$ But from Square of Real Number is Non-Negative: :$\forall x \in \R: \paren {x^k}^2 \ge 0$ and so there is no real number whose square is negative. The result follows from Solution to Quadratic Equation. {{Qed}}
If $n$ is an odd integer, then $n = 2 k + 1$ for some $k \in \N$. Thus $x^n = x \cdot x^{2 k}$. But $x^{2 k} \ge 0$ from Even Power is Non-Negative. The result follows. {{qed}}
Let $P$ be the (strict) positivity property on $D$. Let $<$ be the (strict) total ordering defined on $D$ as: :$a < b \iff a \le b \land a \ne b$ Let $N$ be the strict negativity property on $D$. We consider all possibilities in turn. $(1): \quad a = 0_D$ or $b = 0_D$ In this case, both the {{LHS}} $\size a \times \size b$ and the {{RHS}} are equal to zero. So: :$\size a \times \size b = \size {a \times b}$ $(2): \quad \map P a, \map P b$ First: {{begin-eqn}} {{eqn | l = \map P a, \map P b       | o = \leadsto       | r = \size a = a, \size b = b       | c = {{Defof|Absolute Value on Ordered Integral Domain}} }} {{eqn | o = \leadsto       | r = \size a \times \size b = a \times b       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map P a, \map P b       | o = \leadsto       | r = \map P {a \times b}       | c = Strict Positivity Property: $(P \, 2)$ }} {{eqn | o = \leadsto       | r = \size {a \times b} = a \times b       | c = {{Defof|Absolute Value on Ordered Integral Domain}} }} {{end-eqn}} So: :$\size a \times \size b = \size {a \times b}$   $(3): \quad \map P a, \map N b$ First: {{begin-eqn}} {{eqn | l = \map P a, \map N b       | o = \leadsto       | r = \size a = a, \size b = -b       | c = {{Defof|Absolute Value on Ordered Integral Domain}} }} {{eqn | o = \leadsto       | r = \size a \times \size b = -a \times b       | c = Product with Ring Negative }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map P a, \map N b       | o = \leadsto       | r = \map N {a \times b}       | c = Properties of Strict Negativity: $(5)$ }} {{eqn | o = \leadsto       | r = \map P {-a \times b}       | c = {{Defof|Strict Negativity Property}} }} {{eqn | o = \leadsto       | r = \size {a \times b} = -a \times b       | c = {{Defof|Absolute Value on Ordered Integral Domain}} }} {{end-eqn}} So: :$\size a \times \size b = \size {a \times b}$ Similarly $\map N a, \map P b$. $(4): \quad \map N a, \map N b$ First: {{begin-eqn}} {{eqn | l = \map N a, \map N b       | o = \leadsto       | r = \size a = -a, \size b = -b       | c = {{Defof|Absolute Value on Ordered Integral Domain}} }} {{eqn | o = \leadsto       | r = \size a \times \size b = a \times b       | c = Product of Ring Negatives }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map N a, \map N b       | o = \leadsto       | r = \map P {a \times b}       | c = Properties of Strict Negativity: $(4)$ }} {{eqn | o = \leadsto       | r = \map P {a \times b}       | c = {{Defof|Strict Negativity Property}} }} {{eqn | o = \leadsto       | r = \size {a \times b} = a \times b       | c = {{Defof|Absolute Value on Ordered Integral Domain}} }} {{end-eqn}} So: :$\size a \times \size b = \size {a \times b}$ In all cases the result holds. {{qed}}
Assume $x \le y$. As $\le$ is compatible with the ring structure of $\struct {R, +, \circ, \le}$, we have: :$x \ge 0 \implies x \circ x \le x \circ y$ :$y \ge 0 \implies x \circ y \le y \circ y$ and thus as $\le$ is transitive, it follows that $x \circ x \le y \circ y$. {{qed}}
Let $S$ be bounded above. Let $L$ be the set of real numbers defined as: :$\alpha \in L \iff \exists x \in S: \alpha < x$ Let $R := \relcomp \R L$, where $\complement_\R$ denotes complement in $\R$. By construction of $L$, every element of $L$ is less than some element of $S$. Hence no element of $L$ is an upper bound of $S$. By construction of $R$, for every element $x$ of $R$, there exists no element of $S$ which is greater than $x$. Hence every element of $R$ is an upper bound of $S$. So, to prove the existence of $\sup S$, it is sufficient to demonstrate that $R$ contains a smallest number. We verify that $L$ and $R$ fulfil the conditions for Dedekind's Theorem to hold. We confirm that $\tuple {L, R}$ is a Dedekind cut of $\R$: :$(1): \quad \set {L, R}$ is a partition of $\R$ :$(2): \quad L$ does not have a greatest element :$(3): \quad \forall x \in L: \forall y \in R: x < y$ By Union with Relative Complement: :$L \cup R = \R$ By corollary $2$ to Set Difference and Intersection form Partition, $\tuple {L, R}$ forms a partition of $\R$. So $(1)$ holds immediately. Let $\alpha \in L$. Then there exists $x \in S$ such that $\alpha < x$. Let $\alpha'$ be such that $\alpha < \alpha' < x$. Then $\alpha' \in L$ So whatever $\alpha \in L$ is, it cannot be the greatest element of $L$. Thus $(2)$ holds. Let $\alpha \in L$. Let $\beta \in R$. Then there exists $x \in S$ such that $\alpha < x$. By construction of $R$, $x \le \beta$. Thus $\alpha < \beta$ for all $\alpha \in L, \beta \in R$. Thus $(3)$ holds. By the corollary to Dedekind's Theorem, either $L$ contains a greatest element or $R$ contains a smallest element. We have shown that $L$ does not contain a greatest element. Hence $R$ contains a smallest element. Hence if $S$ is bounded above, it has a supremum. Thus $\R$ is Dedekind complete by definition. Now let $S$ be bounded below. By Dedekind Completeness is Self-Dual, it follows that $S$ admits an infimum. {{qed}}
From the informal definition of complex numbers, we define the following: : $z = x_1 + i y_1$ : $w = x_2 + i y_2$ where $i = \sqrt {-1}$ and $x_1, x_2, y_1, y_2$. Then from the definition of complex multiplication: : $z w = \left({x_1 x_2 - y_1 y_2}\right) + i \left({x_1 y_2 + x_2 y_1}\right)$ From Real Numbers form Field: : $x_1 x_2 - y_1 y_2 \in \R$ and: : $x_1 y_2 + x_2 y_1 \in \R$ Hence the result. {{qed}}
From the formal definition of complex numbers, we define the following: : $z = \left({x_1, y_1}\right)$ : $w = \left({x_2, y_2}\right)$ Then from the definition of complex multiplication: : $z w = \left({x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1}\right)$ From Real Numbers form Field: : $x_1 x_2 - y_1 y_2 \in \R$ and: : $x_1 y_2 + x_2 y_1 \in \R$ Hence the result. {{qed}}
From the definition of complex numbers, we define the following: {{begin-eqn}} {{eqn | l = z_1       | o = :=       | r = \tuple {x_1, y_1} }} {{eqn | l = z_2       | o = :=       | r = \tuple {x_2, y_2} }} {{eqn | l = z_3       | o = :=       | r = \tuple {x_3, y_3} }} {{end-eqn}} where $x_1, x_2, x_3, y_1, y_2, y_3 \in \R$. Thus: {{begin-eqn}} {{eqn | r = z_1 \left({z_2 z_3}\right)       | o =  }} {{eqn | r = \tuple {x_1, y_1} \paren {\tuple {x_2, y_2} \tuple {x_3, y_3} }       | c = {{Defof|Complex Number|index = 2}} }} {{eqn | r = \tuple {x_1, y_1} \tuple {x_2 x_3 - y_2 y_3, x_2 y_3 + y_2 x_3}       | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}} }} {{eqn | r = \tuple {x_1 \paren {x_2 x_3 - y_2 y_3} - y_1 \paren {x_2 y_3 + y_2 x_3}, y_1 \paren {x_2 x_3 - y_2 y_3} + x_1 \paren {x_2 y_3 + y_2 x_3} }       | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}} }} {{eqn | r = \tuple {x_1 x_2 x_3 - x_1 y_2 y_3 - y_1 x_2 y_3 - y_1 y_2 x_3, y_1 x_2 x_3 - y_1 y_2 y_3 + x_1 x_2 y_3 + x_1 y_2 x_3}       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \tuple {\paren {x_1 x_2 x_3 - y_1 y_2 x_3} - \paren {x_1 y_2 y_3 + y_1 x_2 y_3}, \paren {x_1 x_2 y_3 - y_1 y_2 y_3} + \paren {y_1 x_2 x_3 + x_1 y_2 x_3} }       | c = Real Multiplication is Commutative }} {{eqn | r = \tuple {\paren {x_1 x_2 - y_1 y_2} x_3 - \paren {x_1 y_2 + y_1 x_2} y_3, \paren {x_1 x_2 - y_1 y_2} y_3 + \paren {x_1 y_2 + y_1 x_2} x_3}       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2} \tuple {x_3, y_3}       | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}} }} {{eqn | r = \paren {\tuple {x_1, y_1} \tuple {x_2, y_2} } \tuple {x_3, y_3}       | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}} }} {{eqn | r = \paren {z_1 z_2} z_3       | c = {{Defof|Complex Number|index = 2}} }} {{end-eqn}} {{qed}}
$A_n$ is defined as: :$\displaystyle A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n x_k}$ $H_n$ is defined as: :$\displaystyle \frac 1 H_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {x_k} }$ We have that: :$\forall k \in \closedint 1 n: x_k > 0$ From Positive Real has Real Square Root, we can express each $x_k$ as a square: :$\forall k \in \closedint 1 n: x_k = y_k^2$ without affecting the result. Thus we have: :$\displaystyle A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n y_k^2}$ :$\displaystyle \frac 1 {H_n} = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {y_k^2} }$ Multiplying $A_n$ by $\dfrac 1 {H_n}$: {{begin-eqn}} {{eqn | l = \frac {A_n} {H_n}       | r = \frac 1 n \paren {\sum_{k \mathop = 1}^n y_k^2} \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {y_k^2} }       | c =  }} {{eqn | o = \ge       | r = \frac 1 {n^2} \paren {\sum_{k \mathop = 1}^n \frac {y_k} {y_k} }^2       | c = Cauchy's Inequality }} {{eqn | r = \frac 1 {n^2} \paren {\sum_{k \mathop = 1}^n 1}^2       | c =  }} {{eqn | r = \frac {n^2} {n^2} = 1       | c =  }} {{end-eqn}} So: :$\dfrac {A_n} {H_n} \ge 1$ and so from Real Number Axioms: $\R \text O 2$: compatible with multiplication: :$A_n \ge H_n$ {{qed}}
Let $B = \map \sup S$. Then by definition, $B$ is the smallest number such that $x \in S \implies x \le B$. Let $T = \set {z x: x \in S}$. Because $z > 0$, it follows that: :$\forall x \in S: z x \le z B$ So $T$ is bounded above by $z B$. By the Continuum Property, $T$ has a supremum which we will call $C$. We need to show that $C = z B$. Since $z B$ is ''an'' upper bound for $T$, and $C$ is the ''smallest'' upper bound for $T$, it follows that $C \le z B$. Now as $z > 0$ and is a real number: :$\exists z^{-1} \in \R: z^{-1} > 0$ So we can reverse the roles of $S$ and $T$: :$S = \set {z^{-1} y: y \in T}$ We know that $C$ is the smallest number such that: :$\forall y \in T: y \le C$ So it follows that: :$\forall y \in T: z^{-1} y \le z^{-1} C$ So $z^{-1} C$ is ''an'' upper bound for $S$. But $B$ is the ''smallest'' upper bound for $S$. So: :$B \le z^{-1} C \implies z B \le C$ So we have shown that: :$z B \le C$ and $C \le z B$ Hence the result. {{qed}}
Let: :$\map f x = \ln x$ for $x > 0$. With a view to apply Jensen's Inequality: Real Analysis: Corollary, we can show that $f$ is concave on $\openint 0 \infty$.  By Second Derivative of Concave Real Function is Non-Positive, it is sufficient to show that $\map {f''} x \le 0$ for all $x > 0$.  We have, by Derivative of Natural Logarithm: :$\map {f'} x = \dfrac 1 x$ We then have, by Derivative of Power: :$\map {f''} x = -\dfrac 1 {x^2}$  As $x^2 > 0$ for all $x > 0$, we have: :$\dfrac 1 {x^2} > 0$ Therefore:  :$-\dfrac 1 {x^2} = \map {f''} x < 0$ so $f$ is indeed concave on $\openint 0 \infty$. As $x_1, x_2, \ldots, x_n$ are all positive, they all lie in the interval $\openint 0 \infty$.  We therefore have, by Jensen's Inequality: Real Analysis: Corollary:  :$\displaystyle \map \ln {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \ln {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$ for real $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$, with at least one of which being non-zero.  As $n$ is a positive integer, we have:  :$\dfrac 1 n > 0$ We can therefore set:  :$\lambda_i = \dfrac 1 n$ for $1 \le i \le n$. This gives:  {{begin-eqn}} {{eqn	| l = \map \ln {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } 	| r = \map \ln {\frac {\sum_{k \mathop = 1}^n x_k} {n \sum_{k \mathop = 1}^n \frac 1 n} } }} {{eqn	| r = \map \ln {\frac {\sum_{k \mathop = 1}^n x_k} n} }} {{eqn	| r = \map \ln {A_n} 	| c = {{Defof|Arithmetic Mean}} }} {{end-eqn}} and: {{begin-eqn}} {{eqn	| l = \frac {\sum_{k \mathop = 1}^n \lambda_k \map \ln {x_k} } {\sum_{k \mathop = 1}^n \lambda_k} 	| r = \frac {\sum_{k \mathop = 1}^n \map \ln {x_k} } {n \sum_{k \mathop = 1}^n \frac 1 n} }} {{eqn	| r = \frac {\sum_{k \mathop = 1}^n \map \ln {x_k} } n }} {{eqn	| r = \frac 1 n \map \ln {\prod_{k \mathop = 1}^n x_k} 	| c = Sum of Logarithms }} {{eqn	| r = \map \ln {\paren {\prod_{k \mathop = 1}^n x_k}^{1/n} } 	| c = Logarithm of Power }} {{eqn	| r = \map \ln {G_n} 	| c = {{Defof|Geometric Mean}} }} {{end-eqn}} We therefore have:  :$\map \ln {A_n} \ge \map \ln {G_n}$ Note that for $x > 0$:  :$\dfrac 1 x = \map {f'} x > 0$ Therefore, by Derivative of Monotone Function, $f$ is increasing on $\openint 0 \infty$.  We therefore have:  :$A_n \ge G_n$ {{qed}}
The arithmetic mean of $x_1, x_2, \ldots, x_n$ is defined as: :$\displaystyle A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n x_k}$ The geometric mean of $x_1, x_2, \ldots, x_n$ is defined as: :$\displaystyle G_n = \paren {\prod_{k \mathop = 1}^n x_k}^{1/n}$ We prove the result by induction: For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :For all positive real numbers $x_1, x_2, \ldots, x_n: A_n \ge G_n$. $\map P 1$ is true, as this just says: :$\dfrac {x_1} 1 \ge x_1^{1/1}$ which is trivially true. === Basis for the Induction === $\map P 2$ is the case: :$\dfrac {x_1 + x_2} 2 \ge \sqrt {x_1 x_2}$ As $x_1, x_2 > 0$ we can take their square roots and do the following: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = \paren {\sqrt {x_1} - \sqrt {x_2} }^2       | c =  }} {{eqn | r = x_1 - 2\sqrt {x_1 x_2} + x_2       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {x_1 x_2}       | o = \le       | r = \frac {x_1 + x_2} 2       | c =  }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we show that: :$(1): \quad$ If $\map P {2^k}$ is true, where $k \ge 1$, then it logically follows that $\map P {2^{k + 1} }$ is true :$(2): \quad$ If $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k - 1}$ is true. The result will follow by Backwards Induction. This is our first induction hypothesis: :$A_{2^k} \ge G_{2^k}$ Then we need to show: :$A_{2^{k + 1} } \ge G_{2^{k + 1} }$ === Induction Step === This is our induction step: Let $m = 2^k$. Then $2^{k + 1} = 2 m$. Because $\map P m$ is true: :$\paren {x_1 x_2 \dotsm x_m}^{1/m} \le \dfrac 1 m \paren {x_1 + x_2 + \dotsb + x_m}$ Also: :$\paren {x_{m + 1} x_{m + 2} \dotsm x_{2 m} }^{1/m} \le \dfrac 1 m \paren {x_{m + 1} + x_{m + 2} + \dotsb + x_{2 m} }$ But we have $\map P 2$, so: :$\paren {\paren {x_1 x_2 \dotsm x_m}^{1/m} \paren {x_{m + 1} x_{m + 2} \dotsm x_{2 m} }^{1/m} }^{1/2} \le \dfrac 1 2 \paren {\dfrac {x_1 + x_2 + \cdots + x_m} m + \dfrac {x_{m + 1} + x_{m + 2} + \dotsb + x_{2 m} } m}$ So: :$\paren {x_1 x_2 \dotsm x_{2 m} }^{1/2m} \le \dfrac {x_1 + x_2 + \dotsb + x_{2 m} } {2 m}$ So $\map P {2 m} = \map P {2^{k + 1} }$ holds. So $\map P {2^n}$ holds for all $n$ by induction. Now suppose $\map P k$ holds. Then: {{begin-eqn}} {{eqn | l = \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1/k}       | o = \le       | r = \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k       | c =  }} {{eqn | ll= \leadsto       | l = \paren {G_{k - 1}^{k - 1} G_{k - 1} }^{1/k}       | o = \le       | r = \dfrac {\paren {k - 1} A_{k - 1} + G_{k - 1} } k       | c =  }} {{eqn | ll= \leadsto       | l = k G_{k - 1}       | o = \le       | r = \paren {k - 1} A_{k - 1} + G_{k - 1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {k - 1} G_{k - 1}       | o = \le       | r = \paren {k - 1} A_{k - 1}       | c =  }} {{eqn | ll= \leadsto       | l = G_{k - 1}       | o = \le       | r = A_{k - 1}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k - 1}$ and the result follows by Backwards Induction. Therefore $A_n \ge G_n$ for all $n$. {{qed}}
We check the metric space axioms in turn. === Axiom $(\text M 1)$ === The statement of this axiom is: :$(\text M 1): \forall x \in X: \size {x - x} = 0$ This follows from the definition of absolute value. {{qed|lemma}} === Axiom $(\text M 2)$ === The statement of this axiom is: :$(\text M 2): \forall x, y, z \in X: \size {x - y} + \size {y - z} \ge \size {x - z}$ We have: :$\paren {x - y} + \paren {y - z} = \paren {x - z}$ The result follows from the Triangle Inequality for Real Numbers. {{qed|lemma}} === Axiom $(\text M 3)$ === The statement of this axiom is: :$(\text M 3): \forall x, y \in X: \size {x - y} = \size {y - x}$ As $x - y = -\paren {y - x}$, it follows from the definition of absolute value that $\size {x - y} = \size {y - x}$. {{qed|lemma}} === Axiom $(\text M 4)$ === The statement of this axiom is: :$(\text M 4): \forall x, y \in X: x \ne y \implies \size {x - y} > 0$ This follows from the definition of absolute value. {{qed|lemma}} Having verified all the axioms, we conclude $d$ is a metric. {{qed}}
Let $M = \struct {A, d}$ be a metric space. Let $\sequence {x_n}$ be a sequence in $M$ which is convergent, and so $x_n \to l$ as $n \to \infty$. From the definition, in order to prove boundedness, all we need to do is find $K \in \R$ such that $\forall n \in \N: \map d {x_n, l} \le K$. Since $\sequence {x_n}$ converges, it is true that: :$\forall \epsilon > 0: \exists N: n > N \implies \map d {x_n, l} < \epsilon$ In particular, this is true when $\epsilon = 1$, for example. That is: :$\exists N_1: \forall n > N_1: \map d {x_n, l} < 1$ So now we set: :$K = \max \set {\map d {x_1, l}, \map d {x_2, l}, \ldots, \map d {x_{N_1}, l}, 1}$ The result follows. {{qed}}
We note that all of $\Q, \R, \C$ can be considered as metric spaces. Then under the usual metric: : $d \left({x_n, l}\right) = \left|{x_n - l}\right|$. The result follows from the definition of metric: $d \left({x_n, l}\right) = 0 \iff x_n = l$. {{qed}}
First we show that $\sequence {x_n}$ is increasing. Let $a_1 = a_2 = \cdots = a_{n - 1} = 1 + \dfrac 1 {n - 1}$. Let $a_n = 1$. Let: :$A_n$ be the arithmetic mean of $a_1 \ldots a_n$ :$G_n$ be the geometric mean of $a_1 \ldots a_n$ Thus: :$A_n = \dfrac {\paren {n - 1} \paren {1 + \dfrac 1 {n - 1} } + 1} n = \dfrac {n + 1} n = 1 + \dfrac 1 n$ :$G_n = \paren {1 + \dfrac 1 {n - 1} }^{\dfrac {n - 1} n}$ By Cauchy's Mean Theorem: : $G_n \le A_n$ Thus: :$\paren {1 + \dfrac 1 {n - 1} }^{\frac {n - 1} n} \le 1 + \dfrac 1 n$ and so: :$x_{n - 1} = \paren {1 + \dfrac 1 {n - 1} }^{n - 1} \le \paren {1 + \dfrac 1 n}^n = x_n$ Hence $\sequence {x_n}$ is increasing. Next, we show that $\sequence {x_n}$ is bounded above. Using the Binomial Theorem: {{begin-eqn}} {{eqn | l = \paren {1 + \frac 1 n}^n       | r = 1 + n \paren {\frac 1 n} + \frac {n \paren {n - 1} } 2 \paren {\frac 1 n}^2 + \cdots + \paren  {\frac 1 n}^n }} {{eqn | r = 1 + 1 + \paren {1 - \frac 1 n} \frac 1 {2!} + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \frac 1 {3!} + \cdots + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {n - 1} n} \frac 1 {n!} }} {{eqn | o = \le       | r = 1 + 1 + \frac 1 {2!} + \frac 1 {3!} + \cdots + \frac 1 {n!} }} {{eqn | o = \le       | r = 1 + 1 + \frac 1 2 + \frac 1 {2^2} + \cdots + \frac 1 {2^n}       | c = (because $2^{n - 1} \le n!$) }} {{eqn | r = 1 + \frac {1 - \paren {\frac 1 2}^n} {1 - \frac 1 2} }} {{eqn | r = 1 + 2 \paren {1 - \paren {\frac 1 2}^n} }} {{eqn | o = <       | r = 3 }} {{end-eqn}} So $\sequence {x_n}$ is bounded above by $3$. From the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x_n}$ converges to a limit.