hcju/proofwikiqa huggingface dataset:
Suppose that $a \ge 0$. As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$. Thus: :$\dfrac 1 {b - a} \in \R$ By the Archimedean Principle: :$\exists n \in \N: n > \dfrac 1 {b - a}$ Let $M := \set {x \in \N: \dfrac x n > a}$. By the Well-Ordering Principle, there exists $m \in \N$ such that $m$ is the smallest element of $M$. That is: :$m > a n$ and, by definition of smallest element: :$m - 1 \le a n$ As $n > \dfrac 1 {b - a}$, it follows from Ordering of Reciprocals that: :$\dfrac 1 n < b - a$ Thus: {{begin-eqn}} {{eqn | l = m - 1       | o = \le       | r = a n       | c =  }} {{eqn | ll= \leadsto       | l = m       | o = \le       | r = a n + 1       | c =  }} {{eqn | ll= \leadsto       | l = \frac m n       | o = \le       | r = a + \frac 1 n       | c =  }} {{eqn | o = <       | r = a + \paren {b - a}       | c =  }} {{eqn | r = b       | c =  }} {{end-eqn}} Thus we have shown that $a < \dfrac m n < b$. That is: : $\exists r \in \Q: a < r < b$ such that $r = \dfrac m n$. Now suppose $a < 0$. If $b > 0$ then $0 = r$ is a rational number such that $a < r < b$. Otherwise we have $a < b \le 0$. Then $0 \le -b < -a$ and there exists $r \in \Q$ such that: :$-b < r < -a$ where $r$ can be found as above. That is: :$a < -r < b$ All cases have been covered, and the result follows. {{qed}}
As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$. Thus: :$\dfrac 1 {b - a} \in \R$ By the Archimedean Principle: :$\exists n \in \N: n > \dfrac 1 {b - a}$ Let $M := \set {x \in \Z: x > a n}$. By Set of Integers Bounded Below has Smallest Element, there exists $m \in \Z$ such that $m$ is the smallest element of $M$. That is: :$m > a n$ and, by definition of smallest element: :$m - 1 \le a n$ As $n > \dfrac 1 {b - a}$, it follows from Ordering of Reciprocals that: :$\dfrac 1 n < b - a$ Thus: {{begin-eqn}} {{eqn | l = m - 1       | o = \le       | r = a n       | c =  }} {{eqn | ll= \leadsto       | l = m       | o = \le       | r = a n + 1       | c =  }} {{eqn | ll= \leadsto       | l = \frac m n       | o = \le       | r = a + \frac 1 n       | c =  }} {{eqn | o = <       | r = a + \paren {b - a}       | c =  }} {{eqn | r = b       | c =  }} {{end-eqn}} Thus we have shown that $a < \dfrac m n < b$. That is: :$\exists r \in \Q: a < r < b$ such that $r = \dfrac m n$. {{qed}}
We need to show that $x_n \to 0$ as $n \to \infty$. Let $N \in \N$ be the smallest natural number which satisfies $N > x$. (From the Archimedean Principle, such an $N$ always exists.) First we show that: : $\forall n > N: \dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{n - N + 1}$ Note that as $N > x$, $\dfrac x N < 1$. Also: :$m > n \implies \dfrac x m < \dfrac x n$ Thus: {{begin-eqn}} {{eqn | l = \frac {x^n} {n!}       | r = \frac x 1 \frac x 2 \cdots \frac x {N - 1} \frac x N \frac x {N + 1} \cdots \frac x n       | c =  }} {{eqn | r = \frac {x^{N - 1} } {\paren {N - 1}!} \frac x N \frac x {N + 1} \cdots \frac x n       | c =  }} {{eqn | o = \le       | r = \frac {x^{N - 1} } {\paren {N - 1}!} \paren {\frac x N}^{n - N + 1}       | c = as $\dfrac x {N + 1}, \dfrac x {N + 2}, \ldots, \dfrac x n < \dfrac x N$ }} {{eqn | r = \frac {x^{N - 1} } {\paren {N - 1}!} \paren {\frac x N}^{1 - N} \paren {\frac x N}^n       | c =  }} {{end-eqn}} As $\dfrac x N < 1$, it follows from Sequence of Powers of Number less than One that $\paren {\dfrac x N}^n \to 0$ as $n \to \infty$. For a given $x$ and $N$, $\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N}$ is constant. Thus by the Multiple Rule for Real Sequences: :$\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n \to 0$ as $n \to \infty$ As (from above): :$\dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n$ the result follows from the Squeeze Theorem for Real Sequences. {{qed}}
Let $\epsilon > 0$. We need to show that: :$\exists N \in \N: n > N \implies \size {\dfrac 1 {n^r} } < \epsilon$ That is, that $n^r > 1 / \epsilon$. Let us choose $N = \ceiling {\paren {1 / \epsilon}^{1/r} }$. By Reciprocal of Strictly Positive Real Number is Strictly Positive and power of positive real number is positive, it follows that: :$\paren {\dfrac 1 \epsilon}^{1/r} \gt 0$ Then by Positive Power Function on Non-negative Reals is Strictly Increasing: :$\forall n > N: n^r > N^r \ge 1 / \epsilon$ {{qed}}
We expand $\left({1 + \dfrac 1 n}\right)^n$ by the Binomial Theorem, using that $\dfrac {n - k} n = 1 - \dfrac k n$: {{begin-eqn}} {{eqn | l = \left({1 + \frac 1 n}\right)^n       | r = 1 + n \left({\frac 1 n}\right) + \frac {n \left({n - 1}\right)} 2 \left({\frac 1 n}\right)^2 + \cdots + \left({\frac 1 n}\right)^n       | c =  }} {{eqn | r = \frac 1 {0!} + \frac 1 {1!} + \left({1 - \frac 1 n}\right) \frac 1 {2!} + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right)\frac 1 {3!} + \cdots + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right) \cdots \left({1 - \frac {n - 1} n}\right) \frac 1 {n!}       | c =  }} {{end-eqn}} Take one of the terms in the above: :$x = \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \cdots \left({1 - \dfrac {k - 1} n}\right) \dfrac 1 {k!}$ From Sequence of Powers of Reciprocals is Null Sequence, $\dfrac 1 n \to 0$ as $n \to \infty$. From the Combination Theorem for Sequences: :$\forall \lambda \in \R: \dfrac \lambda n \to 0$ as $n \to \infty$ :$\forall \lambda \in \R: 1 - \dfrac \lambda n \to 1$ as $n \to \infty$ :$x = \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \cdots \left({1 - \dfrac {k - 1} n}\right) \dfrac 1 {k!} \to \dfrac 1 {k!}$ as $n \to \infty$ Hence: :$\displaystyle \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n = \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \cdots = \sum_{k \mathop = 0}^\infty \frac 1 {k!}$ {{qed}}
First, note that $E / F$ is a field extension as $F \subseteq K \subseteq E$. Suppose that $\index E K = m$ and $\index K F = n$. Let $\alpha = \set {a_1, \ldots, a_m}$ be a basis of $E / K$, and $\beta = \set {b_1, \ldots, b_n}$ be a basis of $K / F$. We wish to prove that the set: :$\gamma = \set {a_i b_j: 1 \le i \le m, 1 \le j \le n}$ is a basis of $E / F$. As $\alpha$ is a basis of $E / K$, we have, for all $c \in E$: :$\displaystyle c = \sum_{i \mathop = 1}^m c_i a_i$, for some $c_i \in K$. Define $\displaystyle b := \sum_{j \mathop = 1}^n b_i$ and $d_i := \dfrac {c_i} b$. Note $b \ne 0$ since $\beta$ is linearly independent over $F$, and $d_i \in K$ since $b, c_i \in K$. Now we have: :$\displaystyle c = \sum_{i \mathop = 1}^m \frac {c_i} b \cdot b \cdot a_i = \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n d_i a_i b_j$ Thus $\gamma$ is seen to be a spanning set of $E / F$. To show $\gamma$ is linearly independent, suppose that for some $c_{i j} \in F$: :$\displaystyle \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n c_{i j} a_i b_j = 0$ Then we have (as fields are commutative rings): {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n c_{i j} a_i b_j       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \forall i: \sum_{j \mathop = 1}^n c_{i j} b_j       | r = 0       | c = by the linear independence of $\alpha$ over $K$ }} {{eqn | ll= \leadstoandfrom       | l = \forall i,j: c_{i j}       | r = 0       | c = by the linear independence of $\beta$ over $F$ }} {{end-eqn}} Hence $\gamma$ is a linearly independent spanning set; thus it is a basis. Recalling the definition of $\gamma$ as $\set {a_i b_j: 1 \le i \le m, 1 \le j \le n}$, we have: :$\size {\gamma} = m n = \index E K \index K F$ as was to be shown. {{qed}}
Suppose $\displaystyle \lim_{n \mathop \to \infty} x_n = l$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = m$. Let $\epsilon > 0$. Then, provided $n$ is sufficiently large: {{begin-eqn}} {{eqn | l = \map d {l, m}       | o = \le       | r = \map d {l, x_n} + \map d {x_n, m}       | c = Triangle Inequality }} {{eqn | o = <       | r = \epsilon + \epsilon       | c = {{Defof|Limit of Sequence (Metric Space)}} }} {{eqn | r = 2 \epsilon }} {{end-eqn}} So $0 \le \dfrac {\map d {l, m} } 2 < \epsilon$. This holds for ''any'' value of $\epsilon > 0$. Thus from Real Plus Epsilon it follows that $\dfrac {\map d {l, m} } 2 = 0$, that is, that $l = m$. {{qed}}
We have that a Metric Space is Hausdorff. The result then follows from Convergent Sequence in Hausdorff Space has Unique Limit. {{qed}}
$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$: Let $\epsilon > 0$. Then: :$\exists N \in \N: n > N \implies \size {x_n - l} < \epsilon$ So from Negative of Absolute Value: :$l - \epsilon < x_n < l + \epsilon$ But $x_n \ge a$, so: :$a \le x_n < l + \epsilon$ Thus, for ''any'' $\epsilon > 0$: :$a < l + \epsilon$ From Real Plus Epsilon it follows that $a \le l$. {{qed|lemma}} $(2): \quad \forall n \in \N: x_n \le b \implies l \le b$: If $x_n \le b$ it follows that $-x_n \ge -b$ and the above result can be used. {{qed}}
Let $\sequence {x_n}$ be the sequence in $\R$ which forms the basis of Grandi's series, defined as: :$x_n = \paren {-1}^n$ It is clear that $\sequence {x_n}$ is bounded: above by $1$ and below by $-1$. {{AimForCont}} $x_n \to l$ as $n \to \infty$. Let $\epsilon > 0$. Then $\exists N \in \R: \forall n > N: \size {\paren {-1}^n - l} < \epsilon$. But there are values of $n > N$ for which $\paren {-1}^n = \pm 1$. It follows that $\size {1 - l} < \epsilon$ and $\size {-1 - l} < \epsilon$. From the Triangle Inequality for Real Numbers, we have: {{begin-eqn}} {{eqn | l = 2       | r = \size {1 - \paren {-1} }       | c =  }} {{eqn | o = \le       | r = \size {1 - l} + \size {l - \paren {-1} }       | c =  }} {{eqn | o = <       | r = 2 \epsilon       | c =  }} {{end-eqn}} This is a contradiction whenever $\epsilon < 1$. Thus $\sequence {x_n}$ has no limit and, while definitely bounded, is unmistakably divergent. {{qed}}
Let $\sequence {x_n}$ be the sequence in $\R$ which forms the basis of Grandi's series,  defined as: :$x_n = \paren {-1}^n$ It is clear that $\sequence {x_n}$ is bounded: above by $1$ and below by $-1$. Note the following subsequences of $\sequence {x_n}$: :$(1): \quad \sequence {x_{n_r} }$ where $\sequence {n_r}$ is the integer sequence defined as $n_r = 2 r$ :$(2): \quad \sequence  {x_{n_s} }$ where $\sequence {n_s}$ is the integer sequence defined as $n_s = 2 s + 1$. We have that: :$\sequence {x_{n_r} }$ is the sequence $1, 1, 1, 1, \ldots$ :$\sequence {x_{n_s} }$ is the sequence $-1, -1, -1, -1, \ldots$ So $\sequence {x_n}$ has two subsequences with different limits. From Limit of Subsequence equals Limit of Real Sequence, that means $\sequence {x_n}$ can not be convergent. {{qed}}
We have that: :The kernel of a homomorphism is an ideal of $R$ :the only ideals of a division ring are trivial. So $\map \ker \phi = \set {0_R}$ or $R$.   If $\map \ker \phi = \set {0_R}$, then $\phi$ is injective by Kernel is Trivial iff Monomorphism. If $\map \ker \phi = R$, $\phi$ is the zero homomorphism by definition. {{qed}}
From Surjection by Restriction of Codomain, we can restrict the codomain of $\phi$ and consider the mapping $\phi': R \to \operatorname {Im} \left({R}\right)$ As $\phi'$ is now a surjective homomorphism, it is by definition an epimorphism. Then we invoke the result that an epimorphism from a division ring to a ring is either null or an isomorphism. As an isomorphism is by definition injective, the result follows. {{qed}}
=== Existence of Kernel === By Kernel of Ring Homomorphism is Ideal: :The kernel of $\phi$ is an ideal of $R_1$. {{qed|lemma}} === Uniqueness of Quotient Mapping === By Quotient Ring of Kernel of Ring Epimorphism: :there exists a unique ring isomorphism $g: R_1 / K \to R_2$ such that $g \circ q_K = \phi$ :$\phi$ is a ring isomorphism {{iff}} $K = \set {0_{R_1} }$. {{qed}}
Suppose $x_n \to 0$ as $n \to \infty$. Let $H > 0$. So $H^{-1} > 0$. Since $x_n \to 0$ as $n \to \infty$: :$\exists N: \forall n > N: \size {x_n} < H^{-1}$ That is: :$\size {\dfrac 1 {x_n} } > H$. So: :$\exists N: \forall n > N: \size {\dfrac 1 {x_n} } > H$ and thus: :$\sequence {\size {\dfrac 1 {x_n} } }$ diverges to $+\infty$. {{qed|lemma}} Suppose $\size {\dfrac 1 {x_n} } \to \infty$ as $n \to \infty$. By reversing the argument above, we see that $x_n \to 0$ as $n \to \infty$. {{qed}}
By the Triangle Inequality, we have: :$\cmod {\cmod {x_n} - \cmod l} \le \cmod {x_n - l}$ Hence by the Squeeze Theorem and Convergent Sequence Minus Limit, $\cmod {x_n} \to \cmod l$ as $n \to \infty$. {{Qed}}
This is to be proved by induction on $r$. For all $r \in \N_{>0}$, let $P \left({r}\right)$ be the proposition $n_r \ge r$. === Basis for the Induction === When $r = 1$, it follows that $n_1 \ge 1$ as $\N_{>0}$ is bounded below by $1$. Thus $P(1)$ is true. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So our induction hypothesis is that $n_k \ge k$. Then we need to show that $n_{k+1} \ge k+1$. === Induction Step === This is our induction step: Suppose that $n_k \ge k$. Then as $\left \langle {n_r} \right \rangle$ is strictly increasing: :$n_{k+1} > n_k \ge k$ From Sum with One is Immediate Successor in Naturally Ordered Semigroup, we have: :$n_{k+1} > k \implies n_{k+1} \ge k+1$ So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore $\forall r \in \N_{>0}: n_r \ge r$. {{qed}} {{improve|Better link instead of Sum with One is Immediate Successor in Naturally Ordered Semigroup}} Category:Sequences tl74r1b0764lp29mrj86082k7u4u23s
Let $U \in \tau$ be an open set such that $l \in U$. By definition of convergence, we have: :$\exists N \in \N: \forall n > N: x_n \in U$. When $r > N$, we have $n_r > n_N > N$ by Strictly Increasing Sequence of Natural Numbers. It follows that: :$\exists N \in \N: \forall r > N: x_{n_r} \in U$. Therefore, as $U$ was arbitrary, we have established $\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$, by definition of convergence. {{qed}} Category:Convergence Category:Limits of Sequences 1c9kwdnmw5pp6isssb8tcvj4v1sitqm
Let $y = x^{1/n}$. From the definition of the $n$th root of $x$, it follows that $x = y^n$. We will show by induction that $\forall n \in \N^*: y^n \ge 1 \implies y \ge 1$. For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition: : $y^n \ge 1 \implies y \ge 1$ ==== Basis for the Induction ==== By definition, $y^1 = y$. Thus $P(1)$ is true, as this just says $y \ge 1 \implies y \ge 1$. This is our basis for the induction. ==== Induction Hypothesis ==== * Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So our induction hypothesis is that: : $y^k \ge 1 \implies y \ge 1$ Now we need to show that: : $y^{k+1} \ge 1 \implies y \ge 1$. ==== Induction Step ==== This is our induction step: By definition, $y^{k+1} = y \cdot y^k$. Suppose $y^k \ge 1$. From the induction hypothesis it follows that $y > 1$. As $y \ge 1$ it follows that $y > 0$. Let $y^{k+1} = y \cdot y^k \ge 1$. Then $y \cdot y^k$ By Real Number Ordering is Compatible with Multiplication, $y \cdot y^k \ge y \times 1$ and hence the result. So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore $\forall n \in \N^*: y^n \ge 1 \implies y \ge 1$. As $y^n = x$ and $y = x^{1/n}$, the result follows. {{qed}} Category:Analysis Category:Inequalities jz0ugo79w1zoig24q41ykvss46x96vx
Let us define $a_1 = a_2 = \cdots = a_{n-1} = 1$ and $a_n = x$. Let $G_n$ be the geometric mean of $a_1, \ldots, a_n$. Let $A_n$ be the arithmetic mean of $a_1, \ldots, a_n$. From their definitions: :$G_n = x^{1/n}$ and: :$A_n = \dfrac {n - 1 + x} n = 1 + \dfrac{x - 1} n$ From Arithmetic Mean Never Less than Geometric Mean: :$x^{1/n} \le 1 + \dfrac{x - 1} n$ That is: :$x^{1/n} - 1 \le \dfrac{x - 1} n$ There are two cases to consider: $x \ge 1$ and $0 < x < 1$. Let $x \ge 1$. From Root of Number Greater than One, it follows that: :$x^{1/n} \ge 1$ Thus: :$0 \le x^{1/n} - 1 \le \dfrac 1 n \paren {x - 1}$ But from Sequence of Powers of Reciprocals is Null Sequence: :$\dfrac 1 n \to 0$ as $n \to \infty$ From the Combination Theorem for Sequences: :$\dfrac 1 n \paren {x - 1} \to 0$ as $n \to \infty$ Thus by the Squeeze Theorem: :$x^{1/n} - 1 \to 0$ as $n \to \infty$ Hence, again from the Combination Theorem for Sequences: :$x^{1/n} \to 1$ as $n \to \infty$ Now let $0 < x < 1$. Then $x = \dfrac 1 y$ where $y > 1$. But from the above: :$y^{1/n} \to 1$ as $n \to \infty$ Hence by the Combination Theorem for Sequences: :$x^{1/n} = \dfrac 1 {y^{1/n} } \to \dfrac 1 1 = 1$ as $n \to \infty$ {{qed}}
We consider the case where $x \ge 1$; when $0 < x < 1$ the proof can be completed as for proof 1. From Root of Number Greater than One: :$x^{1/n} \ge 1$ Hence $\sequence {x^{1/n} }$ is bounded below by $1$. Now consider $x^{1/n} / x^{1 / \paren {n + 1} }$: {{begin-eqn}} {{eqn | l = \frac {x^{1/n} } {x^{\frac 1 {n + 1} } }       | r = x^{\frac 1 n - \frac 1 {n + 1} }       | c =  }} {{eqn | r = x^{\frac {n + 1 - n} {n \paren {n + 1} } }       | c =  }} {{eqn | r = x^{\frac 1 {n \paren {n + 1} } }       | c =  }} {{eqn | o = >       | r = 1       | c = Root of Number Greater than One }} {{end-eqn}} So: :$x^{1/n} > x^{\frac 1 {n + 1} }$ and so $\sequence {x^{1 / n} }$ is strictly decreasing. Hence from the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x^{1 / n} }$ converges to a limit $l$ and that $l \ge 1$. Now, since we know that $\sequence {x^{1 / n} }$ is convergent, we can apply Limit of Subsequence equals Limit of Real Sequence. That is, any subsequence of $\sequence {x^{1 / n} }$ must also converge to $l$. So we take the subsequence: :$\sequence {x^{1 / {2 n} } }$ From what has just been shown: :$x^{1 / {2 n} } \to l$ as $n \to \infty$ Using the Combination Theorem for Sequences, we have: :$x^{1 / n} = x^{1 / {2 n} } \cdot x^{1 / {2 n} } \to l \cdot l = l^2$ as $n \to \infty$ But a Convergent Real Sequence has Unique Limit, so $l^2 = l$ and so $l = 0$ or $l = 1$. But $l \ge 1$ and so $l = 1$. Hence the result. {{qed}}
Consider $x_n - x_{n + 1}$. {{begin-eqn}} {{eqn | l = x_n - x_{n + 1}       | r = x_n - \frac {x_n + \dfrac a {x_n} } 2       | c =  }} {{eqn | r = \frac 1 {2 x_n} \paren {x_n^2 - a}       | c =  }} {{eqn | o = \ge       | r = 0       | c = for $n \ge 2$       | cc= Lemma 2 }} {{end-eqn}} So, providing we ignore the first term (about which we can state nothing), the sequence $\sequence {x_n}$ is decreasing and bounded below by $\sqrt a$. Thus by the Monotone Convergence Theorem (Real Analysis), $x_n \to l$ as $n \to \infty$, where $l \ge \sqrt a$. Now we want to find exactly what that value of $l$ actually is. By Limit of Subsequence equals Limit of Real Sequence we also have $x_{n + 1} \to l$ as $n \to \infty$. But $x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$. Because $l \ge \sqrt a$ it follows that $l \ne 0$. So by the Combination Theorem for Sequences: :$x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2 \to \dfrac {l + \dfrac a l} 2$ as $n \to \infty$ Since a Convergent Real Sequence has Unique Limit, that means: :$l = \dfrac {l + \dfrac a l} 2$ and so (after some straightforward algebra): :$l^2 = a$ Thus: :$l = \pm \sqrt a$ and as $l \ge +\sqrt a$ it follows that: :$l = +\sqrt a$ Hence the result. {{Qed}}
Let $a > 0$. We make no statement about $x_1$. We specify that: :$x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$ Now: {{begin-eqn}} {{eqn | l = x_{n + 1} - \sqrt a       | r = \frac {x_n + \dfrac a {x_n} } 2 - \sqrt a       | c =  }} {{eqn | r = \frac 1 {2 x_n} \paren {x_n^2 - 2 x_n \sqrt a + a}       | c =  }} {{eqn | r = \frac 1 {2 x_n} \paren {x_n - \sqrt a}^2       | c =  }} {{eqn | r = \frac 1 {2 x_n} \paren {\dfrac {\paren {x_{n - 1} - \sqrt a}^2} {2 x_{n - 1} } }^2       | c =  }} {{eqn | r = \frac 1 {2 x_n} \frac 1 {\paren {2 x_{n - 1} }^2} \paren {x_{n - 1} - \sqrt a}^4       | c =  }} {{eqn | r = \frac 1 {2 x_n} \frac 1 {\paren {2 x_{n - 1} }^2} \frac 1 {\paren {2 x_{n - 2} }^4} \paren {x_{n - 2} - \sqrt a}^8       | c =  }} {{eqn | r = \frac 1 {2 x_n} \frac 1 {\paren {2 x_{n - 1} }^2} \cdots \frac 1 {\paren {2 x_1}^{2 n - 1} } \paren {x_1 - \sqrt a}^{2 n}       | c =  }} {{end-eqn}} If we now assume that $x_1 \ge \sqrt a$, then it follows from Lemma 2 that $x_n \ge \sqrt a$. So: {{begin-eqn}} {{eqn | l = \size {x_{n + 1} - \sqrt a}       | o = \le       | r = \paren {\frac 1 {2 \sqrt a} }^{1 + 2 + 2^2 + \cdots + 2^{n - 1} } \paren {x_1 - \sqrt a}^{2 n}       | c =  }} {{eqn | r = \paren {\frac 1 {2 \sqrt a} }^{\dfrac {2^n - 1} {2 - 1} } \paren {x_1 - \sqrt a}^{2 n}       | c =  }} {{eqn | r = 2 \sqrt a \paren {\frac {x_1 - \sqrt a} {2 \sqrt a} }^{2 n}       | c =  }} {{end-eqn}} If $\size y < 1$, then $y^n \to 0$ as $n \to \infty$ from Sequence of Powers of Number less than One. So, by Limit of Subsequence equals Limit of Real Sequence: :$y^{2^n} \to 0$ as $n \to \infty$ Thus we see that: :$x_n \to \sqrt a$ as $n \to \infty$ provided that: :$\dfrac {x_1 - \sqrt a} {2 \sqrt a} < 1$ that is, that: :$\sqrt a \le x_1 < 3 \sqrt a$ We assumed above that $x_1 \ge \sqrt a$. Now we have shown that $x_n \to \sqrt a$ as $n \to \infty$ provided that $\sqrt a \le x_1 < 3 \sqrt a$. However, we have already shown that $x_n \to \sqrt a$ as long as $x_1 \ge 0$. The advantage to this analysis is that this gives us an opportunity to determine how close $x_n$ gets to $\sqrt a$. {{Qed}}
From Number to Reciprocal Power is Decreasing we have that the real sequence $\sequence {n^{1/n} }$ is decreasing for $n \ge 3$. Now, as $n^{1 / n} > 0$ for all positive $n$, it follows that $\sequence {n^{1 / n} }$ is bounded below (by $0$, for a start). Thus the subsequence of $\sequence {n^{1 / n} }$ consisting of all the terms of $\sequence {n^{1 / n} }$ where $n \ge 3$ is convergent by the Monotone Convergence Theorem (Real Analysis). Now we need to demonstrate that this limit is in fact $1$. Let $n^{1 / n} \to l$ as $n \to \infty$. Having established this, we can investigate the subsequence $\sequence {\paren {2 n}^{1 / {2 n} } }$. By Limit of Subsequence equals Limit of Real Sequence, this will converge to $l$ also. From Limit of Root of Positive Real Number, we have that $2^{1 / {2 n} } \to 1$ as $n \to \infty$. So $n^{1 / {2 n} } \to l$ as $n \to \infty$ by the Combination Theorem for Sequences. Thus: :$n^{1 / n} = n^{1 / {2 n} } \cdot n^{1 / {2 n} } \to l \cdot l = l^2$ as $n \to \infty$. So $l^2 = l$, and as $l \ge 1$ the result follows. {{qed}}
We have the definition of the power to a real number: :$\displaystyle n^{1/n} = \map \exp {\frac 1 n \ln n}$ From Powers Drown Logarithms, we have that: :$\displaystyle \lim_{n \mathop \to \infty} \frac 1 n \ln n = 0$ Hence: :$\displaystyle \lim_{n \mathop \to \infty} n^{1/n} = \exp 0 = 1$ and the result. {{qed}}
We have: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = \sqrt {n + 1} - \sqrt n       | c =  }} {{eqn | r = \frac {\paren {\sqrt {n + 1} - \sqrt n} \paren {\sqrt {n + 1} + \sqrt n} } {\sqrt {n + 1} + \sqrt n}       | c = multiplying top and bottom by $\sqrt {n + 1} - \sqrt n$ }} {{eqn | r = \frac {n + 1 - n} {\sqrt {n + 1} + \sqrt n}       | c = Difference of Two Squares }} {{eqn | r = \frac 1 {\sqrt {n + 1} + \sqrt n}       | c =  }} {{eqn | o = <       | r = \frac 1 {\sqrt n}       | c = as $\sqrt {n + 1} + \sqrt n > \sqrt n$ }} {{end-eqn}} But from Sequence of Powers of Reciprocals is Null Sequence, $\dfrac 1 {\sqrt n} \to 0$ as $n \to \infty$. The result follows by the Squeeze Theorem. {{Qed}}