nyuuzyou/wikis huggingface dataset:
Squares of form 2 n^2 - 1

== Theorem ==
The sequence of integers $\left\langle{n}\right\rangle$ such that $2 n^2 - 1$ is square begins:
:$1, 5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, \ldots$
== Proof ==
== Sources ==
* : $29$
* : $29$
Category:Square Numbers
Category:Pythagorean Triangles
Squares which are 4 Less than Cubes

== Theorem ==
The only two square numbers which are $4$ less than a cube are:
:$2^2 + 4 = 2^3$
:$11^2 + 4 = 5^3$
== Proof ==
== Historical Note ==
== Sources ==
* : $121$
* : $121$
Category:4
Category:121
Category:Square Numbers
Category:Cube Numbers
Squares which are Sum of Two Cubes

== Theorem ==
Solutions of the equation
:$A^3 + B^3 = C^2$
are given by:
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
:$\begin{array} {c|c|rclcc}
\text n & \text k & A^3 & + & B^3 & = & C^2 \\
\hline
\frac 1 4 & 8 & \paren {11}^3 & + & \paren{37}^3 & = & \paren{228}^2 \\
\frac 1 2 & 4 & \paren {22}^3 & + & \paren{26}^3 & = & \paren{168}^2 \\
1 & \frac 1 2 & \paren {2}^3 & + & \paren{1}^3 & = & \paren{3}^2 \\
2 & \frac 1 2 & \paren {\frac {23} 2}^3 & + & \paren{\frac 1 2}^3 & = & \paren{39}^2 \\
3 & \frac 1 2 & \paren {33}^3 & + & \paren{-6}^3 & = & \paren{189}^2 \\
4 & \frac 1 2 & \paren {71}^3 & + & \paren{-23}^3 & = & \paren{588}^2 \\
\end{array}$
Category:Sums of Cubes
Squares which are Sum of Two Cubes/Formulation 1

== Theorem ==
Solutions of the equation
:$A^3 + B^3 = C^2$
are given by:
where $n \in \C$.
== Proof ==
Assume that:
:$A + B = 12 n^2$
Thus, factoring $A^3 + B^3$, we find that:
Therefore:
Next, assume that:
:$C = n^2 \paren {\alpha n^2 + \beta}$
for some $\alpha, \beta \in \Z$
Furthermore, we would like to write:
:$\paren {A - B}^2 = \dfrac 1 9 n^2 \paren {\alpha n^2 - \beta}^2$
Letting $\alpha = 18$, then $\beta = 6$ and $C = n^2 \paren {18 n^2 + 6} = 6 n^2 \paren {3 n^2 + 1}$
== Sources ==
* : Chapter $\text {16}$. $q$-Series and Theta-Functions
Category:Sums of Cubes
Squares whose Digits can be Separated into 2 other Squares/Mistake

== Source Work ==
:
:The Dictionary
::$1444$
== Mistake ==
:$1444$ is also the $4$th square whose digits form two other squares juxtaposed, $1444 = 144 : 4$.
== Correction ==
It is actually the $5$th, after $49$, $169$, $361$ and $1225$.
This mistake is corrected in of $1997$, which explicitly lists the squares in question.
== Sources ==
* : $1444$
Category:Mistakes/Curious and Interesting Numbers
Squares whose Digits form Consecutive Decreasing Integers

== Theorem ==
The sequence of integers whose squares have a decimal representation consisting of the concatenation of $2$ consecutive decreasing integers begins:
:$91, 9079, 9901, 733 \, 674, 999 \, 001, 88 \, 225 \, 295, 99 \, 990 \, 001, \ldots$
== Proof ==
We have:
They can be determined by inspection.
== Also see ==
* Squares whose Digits form Consecutive Integers
* Squares whose Digits form Consecutive Increasing Integers
Category:Square Numbers
Category:Recreational Mathematics
Squares whose Digits form Consecutive Increasing Integers

== Theorem ==
The sequence of integers whose squares have a decimal representation consisting of the concatenation of $2$ consecutive increasing integers begins:
:$428, 573, 727, 846, 7810, 36 \, 365, 63 \, 636, 326 \, 734, \ldots$
== Proof ==
We have:
They can be determined by inspection.
== Also see ==
* Squares whose Digits form Consecutive Integers
* Squares whose Digits form Consecutive Decreasing Integers
== Sources ==
* : $183,184$
* : $183,184$
Category:Square Numbers
Category:Recreational Mathematics
Squares whose Digits form Consecutive Integers

== Theorem ==
The sequence of integers whose squares have a decimal representation consisting of the concatenation of $2$ consecutive integers, either increasing or decreasing begins:
:$91, 428, 573, 727, 846, 7810, 9079, 9901, 36 \, 365, 63 \, 636, 326 \, 734, 673 \, 267, 733 \, 674, \ldots$
This sequence can be divided into two subsequences: 
Those where the consecutive integers are increasing:
:$428, 573, 727, 846, 7810, 36 \, 365, 63 \, 636, 326 \, 734, 673 \, 267, \ldots$
Those where the consecutive integers are decreasing:
:$91, 9079, 9901, 733 \, 674, 999 \, 001, 88 \, 225 \, 295, \ldots$
== Proof ==
We have:
They can be determined by inspection.
== Also see ==
* Squares whose Digits form Consecutive Increasing Integers
* Squares whose Digits form Consecutive Decreasing Integers
== Sources ==
* 
* : $60,996,100$
* : $60,996,100$
Category:Square Numbers
Category:Recreational Mathematics
Squares with All Odd Digits

== Theorem ==
The only squares whose digits (when written in base $10$ notation) are all odd are $1$ and $9$.
== Proof ==
If $n$ is even, then at least the last digit of $n^2$ is even.
So for $n^2$ to have its digits all odd, $n$ itself must be odd.
We can see immediately that $1 = 1^2$ and $9 = 3^2$ fit the criterion.
Of the other $1$-digit odd integers, we have $5^2 = 25, 7^2 = 49, 9^2 = 81$, all of which have an even digit.
Now, let $n > 10$ be an odd integer.
There are $5$ cases to consider:
* $n = 10 p + 1$: we have $\paren {10 p + 1}^2 = 100 p^2 + 20 p + 1 = 10 \paren {10 p^2 + 2 p} + 1$.
* $n = 10 p + 3$: we have $\paren {10 p + 3}^2 = 100 p^2 + 60 p + 9 = 10 \paren {10 p^2 + 6 p} + 9$.
* $n = 10 p + 5$: we have $\paren {10 p + 5}^2 = 100 p^2 + 100 p + 25 = 10 \paren {10 p^2 + 10 p + 2} + 5$.
* $n = 10 p + 7$: we have $\paren {10 p + 7}^2 = 100 p^2 + 140 p + 49 = 10 \paren {10 p^2 + 14 p + 4} + 9$.
* $n = 10 p + 9$: we have $\paren {10 p + 9}^2 = 100 p^2 + 180 p + 81 = 10 \paren {10 p^2 + 18 p + 8} + 1$.
It is clear that in all cases the $10$s digit is even.
So the square of every odd integer greater than $3$ always has at least one even digit.
Hence the result.
Category:Square Numbers
Squares with No More than 2 Distinct Digits

== Sequence ==
The sequence of square numbers which have no more than $2$ distinct digits and not ending in $0$ begins:
:$1, 4, 9, 16, 25, 36, 49, 64, 81, 121, 144, 225, 441, 484, 676, 1444, 7744, 11 \, 881, 29 \, 929, 44 \, 944, 55 \, 225, 69 \, 696, 9 \, 696 \, 996, 6 \, 661 \, 661 \, 161, \ldots$
It is not known whether there are any more.
== Historical Note ==
== Sources ==
* : $6,661,661,161$
Category:Square Numbers
Category:Recreational Mathematics
Squares with No More than 2 Distinct Digits/Historical Note

== Historical Note on Squares with No More than 2 Distinct Digits ==
Writing in $1997$, reports in his attributes the largest known square numbers which has no more than $2$ distinct (non-zero) digits, $6\,661\,661\,161$, to Yoshigahara.
It is assumed that this is the renowned puzzler .
== Sources ==
* : $6,661,661,161$
Category:Historical Notes
Squaring the Circle

== Classic Problem ==
Squaring the Circle is creating a square with the same area as that of a given circle.
== Solution ==
The problem of squaring the circle reduces to:
:starting with a line segment of given length ($1$, say)
:constructing a line segment of length $\sqrt \pi$ of the first.
== Fallacious Proofs ==
== Also see ==
* Squaring the Circle by Archimedean Spiral
* Squaring the Circle by Compass and Straightedge Construction is Impossible
* Trisecting the Angle
* Doubling the Cube
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* 
* 
* 
* 
* 
* 
* 
Category:Squaring the Circle
Category:Classic Problems
Category:Euclidean Geometry
Category:Circles
Squaring the Circle/Historical Note

== Historical Note on Squaring the Circle ==
The problem of squaring the circle using a compass and straightedge construction dates from the time of , who considered the question while in prison for political reasons.
It subsequently became an exercise that the ancient Greeks tried but failed to succeed in.
This was one of three such problems: the other two being Trisecting the Angle and Doubling the Cube.
There are several techniques available that use other tools, but these were considered unacceptably vulgar to the followers of .
The problem remained unsolved until its impossibility was proved in $1882$ by .
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* 
* 
* 
Category:Historical Notes
Squaring the Circle by Archimedean Spiral

== Theorem ==
Let $C$ be a circle.
It is possible to construct a square of the same area as $C$ using an Archimedean spiral.
== Construction ==
== Historical Note ==
== Also see ==
* Trisection of Angle by Archimedean Spiral
Category:Squaring the Circle
Category:Archimedean Spiral
Squaring the Circle by Compass and Straightedge Construction is Impossible

== Theorem ==
There is no compass and straightedge construction to allow a square to be constructed whose area is equal to that of a given circle.
== Proof ==
Squaring the Circle consists of constructing a line segment of length $\sqrt \pi$ of another.
From Constructible Length with Compass and Straightedge, any such line segment has a length which is an algebraic number of degree $2$.
But $\pi$ is transcendental.
Hence $\pi$ and therefore $\sqrt \pi$ is not such an algebraic number.
Therefore any attempt at such a construction will fail.
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* : Chapter $2$: The Logic of Shape: Problems for the Greeks
Category:Compass and Straightedge Constructions
Category:Squaring the Circle
Squaring the Circle by Compass and Straightedge Construction is Impossible/Historical Note

== Historical Note on Squaring the Circle by Compass and Straightedge Construction is Impossible ==
The crucial final step in the proof of the age-old classic problem of Squaring the Circle was made by , who proved in $1882$ that $\pi$ is transcendental.
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Historical Notes
Squeeze Theorem

== Theorem ==
== Sequences ==
== Functions ==
== Also known as ==
== Also see ==
* Sandwich Principle
== Comment ==
A useful tool to determine the limit of a sequence or function which is difficult to calculate or analyze.
If you can prove it is always between two sequences, both converging to the same limit, whose behavior is considerably more tractable, you can save yourself the trouble of working on that awkward case.
Category:Named Theorems
Category:Limits of Sequences
Category:Squeeze Theorem
Squeeze Theorem/Also known as

== Squeeze Theorem: Also known as ==
This result is also known, in the UK in particular, as the sandwich theorem or the sandwich rule.
In that culture, the word traditionally means specifically enclosing food between two slices of bread, as opposed to the looser usage of the , where the there is only one such slice.
Hence, in idiomatic British English, one can refer to the (often uncomfortable) situation of being between two entities as being sandwiched between them.
As the idiom is not universal globally, the term squeeze theorem is preferred on , for greatest comprehension.
Category:Squeeze Theorem
Squeeze Theorem/Functions

== Theorem ==
Let $a$ be a point on an open real interval $I$.
Let $f$, $g$ and $h$ be real functions defined at all points of $I$ except for possibly at point $a$.
Suppose that:
:$\forall x \ne a \in I: \map g x \le \map f x \le \map h x$
:$\ds \lim_{x \mathop \to a} \map g x = \lim_{x \mathop \to a} \map h x = L$
Then:
:$\ds \lim_{x \mathop \to a} \ \map f x = L$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
== Sources ==
* : $\S 8.14$
Category:Squeeze Theorem for Functions
Category:Limits of Real Functions
Category:Squeeze Theorem
Squeeze Theorem/Functions/Proof 1

== Theorem ==
== Proof ==
We start by proving the special case where $\forall x: \map g x = 0$ and $L = 0$, in which case:
:$\ds \lim_{x \mathop \to a} \map h x = 0$
Let $\epsilon > 0$ be a positive real number.
Then by the definition of the limit of a function:
:$\exists \delta > 0: 0 < \size {x - a} < \delta \implies \size {\map h x} < \epsilon$
Now:
:$\forall x \ne a: 0 = \map g x \le \map f x \le \map h x$
so that:
:$\size {\map f x} \le \size {\map h x}$
Thus:
:$0 < \size {x - a} < \delta \implies \size {\map f x} \le \size {\map h x} < \epsilon$
By the transitive property of $\le$, this proves that:
:$\ds \lim_{x \mathop \to a} \map f x = 0 = L$
We now move on to the general case, with $\map g x$ and $L$ arbitrary.
For $x \ne a$, we have:
:$\map g x \le \map f x \le \map h x$
By subtracting $\map g x$ from all expressions, we have:
:$0 \le \map f x - \map g x \le \map h x - \map g x$
Since as $x \to a, \map h x \to L$ and $\map g x \to L$, we have:
:$\map h x - \map g x \to L - L = 0$
From the special case, we now have:
:$\map f x - \map g x \to 0$
We conclude that:
:$\map f x = \paren {\map f x - \map g x} + \map g x \to 0 + L = L$
Category:Squeeze Theorem for Functions
Squeeze Theorem/Functions/Proof 2

== Theorem ==
== Proof ==
Let $f, g, h$ be real functions defined on an open interval $\openint a b$, except possibly at the point $c \in \openint a b$.
Let:
:$\ds \lim_{x \mathop \to c} \map g x = L$
:$\ds \lim_{x \mathop \to c} \map h x = L$
:$\map g x \le \map f x \le \map h x$ except perhaps at $x = c$.
Let $\sequence {x_n}$ be a sequence of points of $\openint a b$ such that:
:$\forall n \in \N_{>0}: x_n \ne c$
and:
:$\ds \lim_{n \mathop \to \infty} \ x_n = c$
By Limit of Function by Convergent Sequences:
:$\ds \lim_{n \mathop \to \infty} \map g {x_n} = L$
and:
:$\ds \lim_{n \mathop \to \infty} \map h {x_n} = L$
Since:
:$\map g {x_n} \le \map f {x_n} \le \map h {x_n}$
it follows from the Squeeze Theorem for Real Sequences that:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = L$
The result follows from Limit of Function by Convergent Sequences.
== Sources ==
* : Appendix: $\S 18.6$: Limits of functions
Category:Squeeze Theorem for Functions
Squeeze Theorem/Functions/Proof 3

== Theorem ==
== Proof ==
By the definition of the limit of a real function, we have to prove that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \paren {\size {x - a} < \delta \implies \size {\map f x - L} < \epsilon}$
Let $\epsilon \in \R_{>0}$ be given.
We have:
:$\ds \lim_{x \mathop \to a} \map g x = \lim_{x \mathop \to a} \map h x$
Hence by Sum Rule for Limits of Real Functions:
:$\ds \lim_{x \mathop \to a} \paren {\map h x - \map g x} = 0$
By the definition of the limit of a real function:
:$(1): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \paren {\size {x - a} < \delta \implies \size {\map h x - L} < \epsilon'}$
:$(2): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \paren {\size {x - a} < \delta \implies \size {\map g x - L} < \epsilon'}$
:$(3): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \paren {\size {x - a} < \delta \implies \size {\map h x - \map g x} < \epsilon'}$
Take $\epsilon' = \dfrac {\epsilon} 3$ in $(1)$, $(2)$, $(3)$.
Then there exists $\delta_1, \delta_2, \delta_3$ that satisfies $(1)$, $(2)$, $(3)$ with $\epsilon' = \dfrac \epsilon 3$. 
Take $\delta = \min \set {\delta_1, \delta_2, \delta_3}$.
Then:
So, if $\size {x - a} < \delta$:
Category:Squeeze Theorem for Functions
Squeeze Theorem/Sequences

== Theorem ==
There are two versions of this result:
* one for sequences in the set of complex numbers $\C$, and more generally for sequences in a metric space.
* one for sequences in the set of real numbers $\R$, and more generally in linearly ordered spaces (which is stronger). 
=== Sequences of Real Numbers ===
=== Sequences of Complex Numbers ===
=== Sequences in a Linearly Ordered Space ===
=== Sequences in a Metric Space ===
== Also known as ==
This result is also known, in the UK in particular, as the sandwich theorem.
Category:Limits of Sequences
Squeeze Theorem/Sequences/Complex Numbers

== Theorem ==
Let $\sequence {a_n}$ be a null sequence in $\R$, that is:
:$a_n \to 0$ as $n \to \infty$
Let $\sequence {z_n}$ be a sequence in $\C$.
Suppose $\sequence {a_n}$ dominates $\sequence {z_n}$.
That is:
: $\forall n \in \N: \cmod {z_n} \le a_n$
Then $\sequence {z_n}$ is a null sequence.
== Proof ==
Thus $\sequence {z_n}$ is a null sequence.
== Also known as ==
Category:Limits of Sequences
Squeeze Theorem/Sequences/Linearly Ordered Space

== Theorem ==
Let $\struct {S, \le, \tau}$ be a linearly ordered space.
Let $\sequence {x_n}$, $\sequence {y_n}$, and $\sequence {z_n}$ be sequences in $S$.
Let $p \in S$.
Let $\sequence {x_n}$ and $\sequence {z_n}$ both converge to $p$.
Let $\forall n \in \N: x_n \le y_n \le z_n$.
Then $\sequence {y_n}$ converges to $p$.
== Proof ==
Let $m \in S$ and $m < p$.
Then $\sequence {x_n}$ eventually succeeds $m$.
Thus by Extended Transitivity, $\sequence {y_n}$ eventually succeeds $m$.
A similar argument using $\sequence {z_n}$ proves the dual statement.
Thus $\sequence {y_n}$ is eventually in each ray containing $p$, so it converges to $p$.
Category:Order Topology
Category:Limits of Sequences
Squeeze Theorem/Sequences/Metric Spaces

== Theorem ==
Let $M = \struct {S, d}$ be a metric space or pseudometric space.
Let $p \in S$.
Let $\sequence {r_n}$ be a null sequence in $\R$.
Let $\sequence {x_n}$ be a sequence in $S$ such that:
:$\forall n \in \N: \map d {p, x_n} \le r_n$.
Then $\sequence {x_n}$ converges to $p$.
== Proof ==
Thus $\sequence {x_n}$ converges to $p$.
Category:Metric Spaces
Category:Limits of Sequences
Squeeze Theorem/Sequences/Real Numbers

== Theorem ==
Let $\sequence {x_n}$, $\sequence {y_n}$ and $\sequence {z_n}$ be sequences in $\R$.
Let $\sequence {y_n}$ and $\sequence {z_n}$ both be convergent to the following limit:
:$\ds \lim_{n \mathop \to \infty} y_n = l, \lim_{n \mathop \to \infty} z_n = l$
Suppose that:
:$\forall n \in \N: y_n \le x_n \le z_n$
Then:
:$x_n \to l$ as $n \to \infty$
that is:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Thus, if $\sequence {x_n}$ is always between two other sequences that both converge to the same limit, $\sequence {x_n} $ is said to be sandwiched or squeezed between those two sequences and itself must therefore converge to that same limit.
=== Corollary ===
== Proof ==
From Negative of Absolute Value: Corollary 1:
:$\size {x - l} < \epsilon \iff l - \epsilon < x < l + \epsilon$
Let $\epsilon > 0$.
We need to prove that:
:$\exists N: \forall n > N: \size {x_n - l} < \epsilon$
As $\ds \lim_{n \mathop \to \infty} y_n = l$ we know that:
:$\exists N_1: \forall n > N_1: \size {y_n - l} < \epsilon$
As $\ds \lim_{n \mathop \to \infty} z_n = l$ we know that:
:$\exists N_2: \forall n > N_2: \size {z_n - l} < \epsilon$
Let $N = \max \set {N_1, N_2}$.
Then if $n > N$, it follows that $n > N_1$ and $n > N_2$.
So:
:$\forall n > N: l - \epsilon < y_n < l + \epsilon$
:$\forall n > N: l - \epsilon < z_n < l + \epsilon$
But:
:$\forall n \in \N: y_n \le x_n \le z_n$
So:
:$\forall n > N: l - \epsilon < y_n \le x_n \le z_n < l + \epsilon$
and so:
:$\forall n > N: l - \epsilon < x_n < l + \epsilon$
So:
:$\forall n > N: \size {x_n - l} < \epsilon$
Hence the result.
== Also known as ==
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.10$: Theorem (the sandwich theorem)
Category:Real Analysis
Category:Squeeze Theorem
Squeeze Theorem for Absolutely Convergent Series

== Theorem ==
Let $\ds \sum \size {a_n}$ be an absolutely convergent series in $\R$.
Suppose that:
:$\ds -\sum \size {a_n} = \sum \size {a_n}$
Then $\ds \sum a_n$ equals the above two series.
== Proof ==
From Absolutely Convergent Real Series is Convergent, the convergence of:
:$\ds \sum_{n \mathop = 1}^\infty \size {a_n}$
implies that of:
:$\ds \sum_{n \mathop = 1}^\infty a_n$
By Negative of Absolute Value:
:$\ds -\size {\sum_{n \mathop = 1}^j a_n} \le \sum_{n \mathop = 1}^j a_n \le \size {\sum_{n \mathop = 1}^j a_n}$
By repeated application of Triangle Inequality:
:$\ds -\sum_{n \mathop = 1}^j \size {a_n} \le \sum_{n \mathop = 1}^j a_n \le \sum_{n \mathop = 1}^j \size {a_n}$
We have that the leftmost and rightmost terms converge as $j \to +\infty$.
Hence the result, from Squeeze Theorem for Real Sequences.
Category:Series
Squeeze Theorem for Filter Bases

== Theorem ==
Let $\struct {S, \le, \tau}$ be a linearly ordered space.
Let $F_1$, $F_2$, and $F_3$ be filter bases in $S$.
Let:
:$\forall T \in F_1: \exists M \in F_2: \forall x \in M: \exists y \in T: y \le x$
That is:
:for each $T \in F_1$, $F_2$ has an element $M$ such that all elements of $M$ succeed some element of $T$.
Similarly, let:
:$\forall U \in F_3: \exists N \in F_2: \forall x \in N: \exists y \in U: x \le y$
That is:
:for each $U \in F_3$, $F_2$ has an element $N$ such that all elements of $N$ precede some element of $U$.
Let $F_1$ and $F_3$ each converge to a point $p \in S$.
Then $F_2$ converges to $p$.
== Proof ==
Let $q \in S$ such that $q < p$.
We will show that $F_2$ has an element which is a subset of $q^\ge$.
Since $F_1$ converges to $p$, it has an element:
:$A \subseteq q^\ge$.
Thus there is an element $k$ in $A$ and an element $M$ in $F_2$ such that all elements of $M$ succeed $k$.
Then by Extended Transitivity:
:$M \subseteq q^\ge$
A similar argument using $F_3$ proves the dual statement.
Thus $F_2$ converges to $p$.
Category:Filter Theory
Category:Order Topology
Category:Convergence
Squeeze Theorem for Real Sequences/Corollary

== Corollary to Squeeze Theorem for Real Sequences ==
Let $\sequence {y_n}$ be a sequence in $\R$ which is null, that is:
:$y_n \to 0$ as $n \to \infty$
Let:
:$\forall n \in \N: \size {x_n - l} \le y_n$
Then $x_n \to l$ as $n \to \infty$.
== Proof ==
From :
:$\size {x_n - l} \le y_n \iff l - y_n \le x_n \le l + y_n$
From the Difference Rule for Real Sequences:
:$l - y_n \to l$ as $n \to \infty$
and from the Sum Rule for Real Sequences:
:$l + y_n \to l$ as $n \to \infty$
So by the Squeeze Theorem for Real Sequences, $x_n \to l$ as $n \to \infty$.
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.11$: Corollary
Category:Squeeze Theorem
St. Ives Problem

== Classic Problem ==
:As I was going to 
:I met a man with seven wives
:Every wife had seven sacks
:Every sack had seven cats
:Every cat had seven kits
:Kits, cats, sacks and wives
:How many were going to St. Ives?
=== Rhind Papyrus Variant ===
=== Fibonacci Variant ===
== Solution 1 ==
== Solution 2 ==
== Historical Note ==
== Linguistic Note ==
== Sources ==
* : $7$
* : $7$
* 
Category:St. Ives Problem
Category:Trick Puzzles
Category:Classic Problems
Category:7
St. Ives Problem/Fibonacci Variant

== Classic Problem ==
:$7$ old women are travelling to Rome.
:Each old woman has $7$ mules.
:On each mule are $7$ sacks.
:In each sack are $7$ loaves of bread.
:In each loaf are $7$ knives.
:Each knife has $7$ sheaths.
What is the total of all of these?
== Solution ==
Thus we have:
As with the St. Ives Problem, the total can be calculated using the Sum of Geometric Sequence:
== Sources ==
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* : The World's Oldest Puzzle: $2$
Category:St. Ives Problem