hcju/proofwikiqa huggingface dataset:
=== Lemma === {{:Numbers with Square-Free Binomial Coefficients/Lemma}}{{qed|lemma}} === Case 1: $n \ge 25$ === Consider the case $n \ge 25$.  {{AimForCont}} for all $0 \le m \le n$, $\dbinom n m$ is square-free. Then for any prime $p$: :$p^2 \nmid \dbinom n m$ for all $0 \le m \le n$. In particular we have: :$2^2 \nmid \dbinom n m$ and: :$5^2 \nmid \dbinom n m$ There exists an integer $r$ where $2^r \le n < 2^{r + 1}$. There also exists an integer $s$ where $5^s \le n < 5^{s + 1}$. We have $r \ge 4$ and $s \ge 2$ from $n \ge 25 = 5^2 \ge 16 = 2^4$. By the Lemma, we have: :$2^{r - 1} \divides \paren {n + 1}$ and: :$5^{s - 1} \divides \paren {n + 1}$ Because $2^{r - 1}$ and $5^{s - 1}$ are coprime, we have by Product of Coprime Factors: :$2^{r - 1} 5^{s - 1} \divides \paren {n + 1}$ Thus: {{begin-eqn}} {{eqn | l = n + 1       | o = \ge       | r = 2^{r - 1} 5^{s - 1}       | c = from $2^{r - 1} 5^{s - 1} \divides \paren {n + 1}$ }} {{eqn | o = \ge       | r = 2^{r - 1} 5       | c = $s - 1 \ge 1$ }} {{eqn | r = \dfrac {2^{r + 1} } 4 \times 5 }} {{eqn | o = >       | r = \dfrac n 4 \times 5       | c = from $2^{r + 1} > n$ }} {{eqn | r = n + \dfrac n 4 }} {{eqn | o = \ge       | r = n + \dfrac {25} 4       | c = from $n \ge 25$ }} {{end-eqn}} which is a contradiction. Therefore $\dbinom n m$ is not square-free for any $0 \le m \le n$. {{qed|lemma}} === Case 2: $9 \le n \le 24$ === Consider the case $9 \le n \le 24$. Suppose for all $0 \le m \le n$, $\dbinom n m$ is square-free. Then for any prime $p$: :$p^2 \nmid \dbinom n m$ for all $0 \le m \le n$. In particular we have: :$2^2 \nmid \dbinom n m$ and: :$3^2 \nmid \dbinom n m$ There exists an integer $r$ where $2^r \le n < 2^{r + 1}$. There also exists an integer $s$ where $3^s \le n < 3^{s + 1}$. From: :$n \ge 9 = 3^2 \ge 8 = 2^3$ we have $r \ge 3$ and $s \ge 2$ By the Lemma, we have: :$2^{r - 1} \divides \paren {n + 1}$ and: :$3^{s - 1} \divides \paren {n + 1}$ So we have: :$2^2 \divides \paren {n + 1}$ and: :$3^1 \divides \paren {n + 1}$ Since $4$ and $3$ are coprime, we have by Product of Coprime Factors: :$12 \divides \paren {n + 1}$ In the range $9 \le n \le 24$, only $11$ and $23$ satisfy this condition. {{qed|lemma}} === Case 3: $4 \le n \le 8$ === Consider the case $4 \le n \le 8$. Suppose for all $0 \le m \le n$, $\dbinom n m$ is square-free. Then for any prime $p$: :$p^2 \nmid \dbinom n m$ for all $0 \le m \le n$. In particular we have: :$2^2 \nmid \dbinom n m$ There exists an integer $r$ where $2^r \le n < 2^{r + 1}$. From $n \ge 4 = 2^2$: :$r \ge 2$ By the Lemma, we have: :$2^{r - 1} \divides \paren {n + 1}$ So we have: :$2^1 \divides \paren {n + 1}$ In the range $4 \le n \le 8$, only $5$ and $7$ satisfy this condition. {{qed|lemma}} === Manual Checking === We have narrowed our list of possible candidates to: :$1, 2, 3, 5, 7, 11, 23$ Manual checking each $n$ in the list shows that $\dbinom n m$ is square-free for each $0 \le m \le n$. {{qed}}
thumb Consider an isosceles triangle $\triangle ABC$ with base $BC$ and apex $\angle BAC = 2 \alpha$. Construct the angle bisector to $\angle BAC$ and name it $AH$: :$\angle BAH = \angle CAH = \alpha$ From Bisector of Apex of Isosceles Triangle is Perpendicular to Base: :$AH \perp BC$ From Area of Triangle in Terms of Two Sides and Angle: {{begin-eqn}} {{eqn | l = \map \Area {\triangle BAH}       | r = \dfrac {BA \cdot AH \sin \alpha} 2 }} {{eqn | l = \map \Area {\triangle CAH}       | r = \dfrac {CA \cdot AH \sin \alpha} 2 }} {{end-eqn}} By definition of sine: {{begin-eqn}} {{eqn | l = AH       | r = CA \cos \alpha }} {{eqn | l = AH       | r = BA \cos \alpha }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \map \Area {\triangle BAH}       | r = \dfrac {BA \cdot CA \cos \alpha \sin \alpha} 2 }} {{eqn | l = \map \Area {\triangle CAH}       | r = \dfrac {CA \cdot BA \cos \alpha \sin \alpha} 2 }} {{eqn | l = \map \Area {\triangle ABC}       | r = \map \Area {\triangle BAH} + \map \Area {\triangle CAH} }} {{eqn | r = \frac {BA \cdot CA \cos \alpha \sin \alpha} 2 + \frac {CA \cdot BA \cos \alpha \sin \alpha} 2 }} {{eqn | r = BA \cdot CA \cos \alpha \sin \alpha }} {{eqn | r = \frac {BA \cdot CA \sin 2 \alpha} 2       | c = Area of Triangle in Terms of Two Sides and Angle ($\triangle ABC$) }} {{eqn | ll= \leadsto       | l = \sin 2 \alpha       | r = 2 \cos \alpha \sin \alpha       | c = dividing both sides by $\dfrac {BA \cdot CA} 2$ }} {{end-eqn}} {{qed}}
Let $A$ be subset of $T$. By definition of complete lattice: :$A$ admits an infimum in $L$. Thus by definition of infima inheriting: :$A$ admits an infimum in $S$. Hence by dual of Lattice is Complete iff it Admits All Suprema: :$S$ is a complete lattice. {{qed}}
For each $k \in \N$, there are $9^k$ $k$-digit numbers containing no instances of the digit $0$. Each of these numbers is at least $10^{k - 1}$. Hence the reciprocals of each of these numbers is at most $\dfrac 1 {10^{k - 1}}$. Thus: {{begin-eqn}} {{eqn | l = \sum_{\map P n} \frac 1 n       | o = <       | r = \sum_{k \mathop = 1}^\infty \frac {9^k} {10^{k - 1} } }} {{eqn | r = \frac 9 {1 - \frac 9 {10} }       | c = Sum of Geometric Sequence }} {{eqn | r = 90 }} {{end-eqn}} showing that the sum converges. {{finish|Finer approximations can be obtained (e.g. in virtue of Closed Form for Triangular Numbers/Direct Proof), but due to the slow growth of the harmonic series, many numbers must to summed to obtain the approximation in the theorem. Case in point: $H_{5 \times 10^9} < 23$}}
We have: {{begin-eqn}} {{eqn | l = 1^2 + 2^2 + 3^2 + \cdots + 24^2       | r = \dfrac {24 \times \paren {24 + 1} \times \paren {2 \times 24 + 1} } 6       | c = Sum of Sequence of Squares }} {{eqn | r = \dfrac {24 \times 25 \times 49} 6       | c =  }} {{eqn | r = \dfrac {2^3 \times 3 \times 5^2 \times 7^2} {2 \times 3}       | c =  }} {{eqn | r = 2^2 \times 5^2 \times 7^2       | c =  }} {{eqn | r = \paren {2 \times 5 \times 7}^2       | c =  }} {{eqn | r = 70^2       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = 18^2 + 19^2 + \cdots + 28^2       | r = \dfrac {28 \times \paren {28 + 1} \times \paren {2 \times 28 + 1} } 6 - \dfrac {17 \times \paren {17 + 1} \times \paren {2 \times 17 + 1} } 6       | c = Sum of Sequence of Squares }} {{eqn | r = \dfrac {28 \times 29 \times 57 - 17 \times 18 \times 35} 6       | c =  }} {{eqn | r = \dfrac {\paren {2^2 \times 7} \times 29 \times \paren {3 \times 19} - 17 \times \paren {2 \times 3^2} \times \paren {5 \times 7} } {2 \times 3}       | c =  }} {{eqn | r = 2 \times 7 \times 29 \times 19 - 17 \times 3 \times 5 \times 7       | c =  }} {{eqn | r = 7 \times \paren {1102 - 255}       | c =  }} {{eqn | r = 7 \times 847       | c =  }} {{eqn | r = 7 \times 7 \times 11^2       | c =  }} {{eqn | r = 77^2       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = 25^2 + 26^2 + \cdots + 624^2       | r = \dfrac {624 \times \paren {624 + 1} \times \paren {2 \times 624 + 1} } 6 - \dfrac {24 \times \paren {24 + 1} \times \paren {2 \times 24 + 1} } 6       | c = Sum of Sequence of Squares }} {{eqn | r = \dfrac {624 \times 625 \times 1249 - 24 \times 25 \times 49} 6       | c =  }} {{eqn | r = \dfrac {\paren {2^4 \times 3 \times 13} \times 5^4 \times 1249 - \paren {2^3 \times 3} \times 5^2 \times 7^2} {2 \times 3}       | c =  }} {{eqn | r = 2^3 \times 5^4 \times 13 \times 1249 - 2^2 \times 5^2 \times 7^2       | c =  }} {{eqn | r = 2^2 \times 5^2 \times \paren {2 \times 5^2 \times 13 \times 1249 - 7^2}       | c =  }} {{eqn | r = 2^2 \times 5^2 \times \paren {811 \, 850 - 49}       | c =  }} {{eqn | r = 2^2 \times 5^2 \times \paren {811 \, 801}       | c =  }} {{eqn | r = 2^2 \times 5^2 \times \paren {17^2 \times 53^2}       | c =  }} {{eqn | r = 9010^2       | c =  }} {{end-eqn}} {{qed}}
The divisors of $24$ are: :$1, 2, 3, 4, 6, 8, 12, 24$ The divisors of $26$ are: :$1, 2, 13, 26$ Then we have: {{begin-eqn}} {{eqn | r = 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 8^2 + 12^2 + 24^2       | o =        | c =  }} {{eqn | r = 1 + 4 + 9 + 16 + 36 + 64 + 144 + 576       | c =  }} {{eqn | r = 850       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | r = 1^2 + 2^2 + 13^2 + 26^2       | o =        | c =  }} {{eqn | r = 1 + 4 + 169 + 676       | c =  }} {{eqn | r = 850       | c =  }} {{end-eqn}} {{qed}}
From Heron's Formula, the area $A$ of $\triangle ABC$ is given by: :$A = \sqrt{s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right)}$ where $s = \dfrac{a + b + c} 2$ is the semiperimeter of $\triangle ABC$. Here we have: {{begin-eqn}} {{eqn | l = s       | r = \dfrac{4 + 13 + 15} 2       | c =  }} {{eqn | r = 16       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = A       | r = \sqrt{16 \left({16 - 4}\right) \left({16 - 13}\right) \left({16 - 15}\right)}       | c =  }} {{eqn | r = \sqrt{16 \times 12 \times 3 \times 1}       | c =  }} {{eqn | r = \sqrt{2^4 \times 2^2 \times 3 \times 3 \times 1}       | c =  }} {{eqn | r = \sqrt{2^6 \times 3^2}       | c =  }} {{eqn | r = 2^3 \times 3       | c =  }} {{eqn | r = 24       | c =  }} {{end-eqn}} {{ProofWanted|It remains to be shown that it is the smallest. This can be done by exhaustion: the number of triples defining an obtuse scalene triangle are not that many.}}
thumb Consider an isosceles triangle $\triangle ABC$ with base $BC$, and head angle $\angle BAC = 2 \alpha$. Draw an angle bisector to $\angle BAC$ and name it $AH$. :$\angle BAH = \angle CAH = \alpha$ From Angle Bisector and Altitude Coincide iff Triangle is Isosceles: :$AH \perp BC$ From Law of Cosines: {{begin-eqn}} {{eqn | n = 1       | l = CB^2       | r = AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha }} {{end-eqn}} From Pythagoras's Theorem: {{begin-eqn}} {{eqn | l = AC ^ 2       | r = CH^2 + AH^2       | c = in triangle $\triangle AHC$ }} {{eqn | n = 2.1       | ll= \leadsto       | l = CH^2       | r = AC^2 - AH^2 }} {{eqn | o =  }} {{eqn | l = AB ^ 2       | r = BH^2 + AH^2       | c = in triangle $\triangle AHB$ }} {{eqn | n = 2.2       | ll= \leadsto       | l = BH^2       | r = BC^2 - AH^2 }} {{end-eqn}} By definition of sine: {{begin-eqn}} {{eqn | n = 3.1       | l = CH       | r = AC \sin \alpha }} {{eqn | n = 3.2       | l = BH       | r = AB \sin \alpha }} {{end-eqn}} By definition of cosine: :$AH = AB \cos \alpha = AC \cos \alpha$ So: {{begin-eqn}} {{eqn | n = 4       | l = AH^2       | r = AB \cdot AC \cdot \cos^2 \alpha }} {{eqn | o =  }} {{eqn | l = CH^2       | r = AC^2 - AH^2       | c = $(2.1)$ }} {{eqn | n = 5.1       | r = AC^2 - AB \cdot AC \cdot \cos^2 \alpha       | c = assigning $(4)$ }} {{eqn | o =  }} {{eqn | l = BH^2       | r = AB^2 - AH^2       | c = $(2.2)$ }} {{eqn | n = 5.2       | r = AB^2 - AB \cdot AC \cdot \cos^2 \alpha       | c = assigning $(4)$ }} {{end-eqn}} Now: {{begin-eqn}} {{eqn | l = CB^2       | r = (CH + BH)^2 }} {{eqn | r = CH^2 + BH^2 + 2 \cdot CH \cdot BH       | c = Square of Sum }} {{eqn | r = AC^2 - AB \cdot AC \cdot \cos^2 \alpha + AB^2 - AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot CH \cdot BH       | c = assigning $(5.1)$,$(5.2)$ }} {{eqn | r = AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot CH \cdot BH       | c = simplifying }} {{eqn | r = AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot AB \cdot AC \cdot \sin^2 \alpha       | c = assigning $(3.1)$,$(3.2)$ }} {{eqn | r = AC^2 + AB^2 - 2 \cdot AB \cdot AC \paren {\cos^2 \alpha - \sin^2 \alpha}       | c = simplifying }} {{eqn | r = AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha       | c = equating to $(1)$ }} {{end-eqn}} Hence we get the equation: {{begin-eqn}} {{eqn | l = AC^2 + AB^2 - 2 \cdot AB \cdot AC \paren {\cos^2 \alpha - \sin^2 \alpha}       | r = AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha }} {{eqn | ll= \leadsto       | l = \cos^2 \alpha - \sin^2 \alpha       | r = \cos 2 \alpha       | c = simplifying }} {{end-eqn}} {{qed}}
We will prove that :$K \left({\left({c \left[{S}\right], \precsim}\right)}\right) \subseteq c \left[{K \left({L}\right)}\right]$ By Image of Compact Subset under Directed Suprema Preserving Closure Operator is Subset of Compzct Subset: :$c \left[{K \left({L}\right)}\right] \subseteq K \left({\left({c \left[{S}\right], \precsim}\right)}\right)$ Thus the result by definition of set equality. {{qed}}
Let $S$ be the set of all positive integers which are divisible by both the sum and product of their digits. Trivially, the sum and product of the digits of a one-digit number $n$ are themselves $n$. Thus from Integer Divides Itself, the positive integers from $1$ to $9$ are in $S$. The product of any integer with a $0$ in it is $0$. From Zero Divides Zero $0$ is not a divisor of $n$ unless $n$ is itself $0$. So $10, 20, 30, \ldots$ are not in $S$. For all prime numbers $p$ with $2$ or more digits, the sum of its digits is greater than $1$ and less than $p$. Thus $p$ is not a multiple of the sum of its digits. {{ProofWanted|Having eliminated a fair number of potential candidates, the remaining ones can be found by trial and error.}}
As $p$ and $q$ are distinct prime numbers, it follows that $p$ and $q$ are coprime. Thus by Sigma Function is Multiplicative: :$\map \sigma n = \map \sigma p \map \sigma q$ From Sigma Function of Prime Number: :$\map \sigma p = \paren {p + 1}$ :$\map \sigma q = \paren {q + 1}$ Hence the result. {{qed}}
A semiprime with distinct prime factors is a square-free integer. By Sigma Function of Square-Free Integer: :$\ds \map \sigma n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$ Hence the result. {{qed}}
For a given $n$, to determine every $m$ such that $\sigma \left({m}\right) = n$ can be determined by evaluating the the $\sigma$ value of all integers up to $n - 1$. It is hence noted: {{begin-eqn}} {{eqn | l = 24       | r = \sigma \left({14}\right)       | c = {{SigmaLink|14}} }} {{eqn | r = \sigma \left({15}\right)       | c = {{SigmaLink|15}} }} {{eqn | r = \sigma \left({23}\right)       | c = $\sigma$ of Prime Number: $23$ is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 42       | r = \sigma \left({20}\right)       | c = {{SigmaLink|20}} }} {{eqn | r = \sigma \left({26}\right)       | c = {{SigmaLink|26}} }} {{eqn | r = \sigma \left({41}\right)       | c = $\sigma$ of Prime Number: $41$ is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 48       | r = \sigma \left({33}\right)       | c = {{SigmaLink|33}} }} {{eqn | r = \sigma \left({35}\right)       | c = {{SigmaLink|35}} }} {{eqn | r = \sigma \left({47}\right)       | c = $\sigma$ of Prime Number: $47$ is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 60       | r = \sigma \left({24}\right)       | c = {{SigmaLink|24}} }} {{eqn | r = \sigma \left({38}\right)       | c = {{SigmaLink|38}} }} {{eqn | r = \sigma \left({59}\right)       | c = $\sigma$ of Prime Number: $59$ is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 84       | r = \sigma \left({44}\right)       | c = {{SigmaLink|44}} }} {{eqn | r = \sigma \left({65}\right)       | c = {{SigmaLink|65}} }} {{eqn | r = \sigma \left({83}\right)       | c = $\sigma$ of Prime Number: $83$ is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 90       | r = \sigma \left({40}\right)       | c = {{SigmaLink|40}} }} {{eqn | r = \sigma \left({58}\right)       | c = {{SigmaLink|58}} }} {{eqn | r = \sigma \left({89}\right)       | c = $\sigma$ of Prime Number: $89$ is prime }} {{end-eqn}} {{qed}}
We have that the Sigma Function is Multiplicative. From the definition of prime number, each of the prime factors of $n$ is coprime to all other divisors of $n$. From Sigma Function of Prime Number, we have: :$\displaystyle \sigma \left({p_i}\right) = p_i + 1$ Hence the result. {{qed}}
From Sigma Function of Integer: :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where each of the $k_i$s are equal to $1$; Hence: :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^2 - 1} {p_i - 1}$ But from Difference of Two Squares: :$\dfrac {p_i^2 - 1} {p_i - 1} = \dfrac {\left({p_i + 1}\right) \left({p_i i 1}\right)} {p_i - 1} = p_i + 1$ Hence the result. {{qed}}
By definition: :$\forall x \in S: x^\preceq$ is a principal ideal. By Compact Element iff Principal Ideal: :$\forall x \in S: x^\preceq$ is a compact element in $I$. By definition of compact subset: :$\forall x \in S: x^\preceq \in K\left({I}\right)$ Then $f$ is well-defined. We will prove that :$f$ is an order embedding That means :$\forall x, y \in S: x \preceq y \iff f\left({x}\right) \precsim' f\left({y}\right)$ Let $x, y \in S$. We will prove as lemma that :$x^\preceq \subseteq y^\preceq \implies x \preceq y$ Assume that :$x^\preceq \subseteq y^\preceq$ By definition of reflexivity: :$x \preceq x$ By definition of lower closure of element: :$x \in x^\preceq$ By definition of subset: :$x \in y^\preceq$ Thus by definition of lower closure of element: :$x \preceq y$ {{qed|lemma}} :$x \preceq y$ {{iff}} :$x^\preceq \subseteq y^\preceq$ by lemma and Lower Closure is Increasing {{iff}} :$f\left({x}\right) \subseteq f\left({y}\right)$ by definition of $f$ {{iff}} :$f\left({x}\right) \precsim f\left({y}\right)$ by definition of $\precsim$ {{iff}} :$f\left({x}\right) \precsim' f\left({y}\right)$ by definition of ordered subset. {{qed|lemma}} We will prove that :$\forall a \in K\left({I}\right): \exists y \in S:a = f\left({y}\right)$ Let $a \in K\left({I}\right)$ By definition of compact subset: :$a$ is a compact element in $I$. by Compact Element iff Principal Ideal: :$a$ is a principal ideal. By definition of principal ideal: :$\exists y \in S: a = y^\preceq$ Thus by definition of $f$: :$a = f\left({y}\right)$ {{qed|lemma}} Then $f$ is a surjection. Hence $f$ is an order isomorphism. {{qed}}
We will prove that: :$\forall x, y \in \map K I: \exists z \in \map K I: z \precsim x \land z \precsim y \land \forall v \in \map K I: v \precsim x \land v \precsim y \implies v \precsim z$ Let $x, y \in \map K I$. By definition of compact subset: :$x$ and $y$ are compact elements in $I$. By Compact Element iff Principal Ideal and definition of principal ideal: :$\exists a \in S: x = a^\preceq$ and :$\exists b \in S: y = b^\preceq$ By Intersection of Semilattice Ideals is Ideal: :$x \cap y \in \mathit{Ids}\left({L}\right)$ Define $z = x \cap y$. By Meet in Set of Ideals: :$z = x \wedge_I y$ We will prove that :$\left({a \wedge b}\right)^\preceq \subseteq z$ Let $c \in \left({a \wedge b}\right)^\preceq$ By definition of lower closure of element: :$c \preceq a \wedge b$ By Meet Precedes Operands: :$a \wedge b \preceq a$ and $a \wedge b \preceq b$ By definition of transitivity: :$c \preceq a$ and $c \preceq b$ By definition of lower closure of element: :$c \in x$ and $c \in y$ Thus by definition of intersection: :$c \in z$ {{qed|lemma}} We will prove that :$z \subseteq \left({a \wedge b}\right)^\preceq$ Let $c \in z$. By definition of intersection: :$c \in x$ and $c \in y$ By definition of lower closure of element: :$c \preceq a$ and $c \preceq b$ By definitions of infimum and lower bound: :$c \preceq a \wedge b$ Thus by definition of lower closure of element: :$c \in \left({a \wedge b}\right)^\preceq$ {{qed|lemma}} By definition of set equality: :$z = \left({a \wedge b}\right)^\preceq$ By definition: :$z$ is a principal ideal. By Compact Element iff Principal Ideal: :$z$ is a compact element in $I$. Thus by definition of compact subset: :$z \in \map K I$ Thus by Meet Precedes Operands: :$z \precsim x$ and $z \precsim y$ Let $v \in \map K I$ such that :$v \precsim x$ and $v \precsim y$ By definition of $\precsim$: :$v \subseteq x$ and $v \subseteq y$ By Intersection is Largest Subset: :$v \subseteq z$ Thus by definition of $\precsim$: :$v \precsim z$ {{qed|lemma}} By definition: :$\map K I$ forms a meet semilattice. By Compact Subset is Join Subsemilattice: :$\map K I$ forms a join semilattice. By definition: :$\map K I$ forms a lattice. By Ideals form Algebraic Lattice: :$I$ is an algebraic lattice. Thus by Arithmetic iff Compact Subset form Lattice in Algebraic Lattice: :$I$ is an arithmetic lattice. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$ where $\paren {2 n + 1})^2 = 2 m + 1$. First it is worth rewriting this so as to eliminate $m$. {{begin-eqn}} {{eqn | l = \paren {2 n + 1}^2       | r = 4 n^2 + 4 n + 1       | c =  }} {{eqn | r = 2 \paren {2 n^2 + 2 n} + 1       | c =  }} {{end-eqn}} Thus the statement to be proved can be expressed: :$\displaystyle \sum_{j \mathop = 0}^n \paren {2 n^2 + 2 n - j}^2 = \sum_{j \mathop = 1}^n \paren {2 n^2 + 2 n + j}^2$ === Basis for the Induction === $\map P 1$ is the case: :$3^2 + 4^2 = 5^2$ which is the $3-4-5$ triangle. Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{j \mathop = 0}^k \paren {2 k^2 + 2 k - j}^2 = \sum_{j \mathop = 1}^k \paren {2 k^2 + 2 k + j}^2$ from which it is to be shown that: :$\displaystyle \sum_{j \mathop = 0}^{k + 1} \paren {2 \paren {k + 1}^2 + 2 \paren {k + 1} - j}^2 = \sum_{j \mathop = 1}^{k + 1} \paren {2 \paren {k + 1}^2 + 2 \paren {k + 1} + j}^2$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_{j \mathop = 0}^{k + 1} \paren {2 \paren {k + 1}^2 + 2 \paren {k + 1} - j}^2       | c =  }} {{eqn | r = \sum_{j \mathop = 0}^{k + 1} \paren {2 k^2 + 6 k + 4 - j}^2       | c =  }} {{end-eqn}} {{finish|hard algebra}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{>0}: \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$ where $\paren {2 n + 1}^2 = 2 m + 1$. {{qed}}
By definition of subset: :$\mathit{Ids}\left({L}\right) \subseteq \mathcal P\left({S}\right)$ where $\mathcal P\left({S}\right)$ denotes the power set of $S$. Define $P = \left({\mathcal P\left({S}\right), \precsim'}\right)$ where $\mathord\precsim' = \mathord\subseteq \cap \left({\mathcal P\left({S}\right) \times \mathcal P\left({S}\right)}\right)$ By Ideals are Continuous Lattice Subframe of Power Set: :$I$ is an continuous lattice subframe of $P$. By Lattice of Power Set is Algebraic: :$P$ is algebraic lattice. Thus by Continuous Lattice Subframe of Algebraic Lattice is Algebraic Lattice: :$I$ is algebraic lattice. {{qed}}
Checking the squares of all non-palindromic integers in turn from $10$ upwards, until a palindromic integer is reached: {{begin-eqn}} {{eqn | l = 10^2       | r = 100 }} {{eqn | l = 12^2       | r = 144 }} {{eqn | l = 13^2       | r = 169 }} {{eqn | l = 14^2       | r = 196 }} {{eqn | l = 15^2       | r = 225 }} {{eqn | l = 16^2       | r = 256 }} {{eqn | l = 17^2       | r = 289 }} {{eqn | l = 18^2       | r = 324 }} {{eqn | l = 19^2       | r = 361 }} {{eqn | l = 20^2       | r = 400 }} {{eqn | l = 21^2       | r = 441 }} {{eqn | l = 23^2       | r = 529 }} {{eqn | l = 24^2       | r = 576 }} {{eqn | l = 25^2       | r = 625 }} {{eqn | l = 26^2       | r = 676 }} {{end-eqn}} {{qed}}
{{questionable|A far cry from the Goldbach Conjecture. Puzzling statement.}}
By definition of algebraic ordered set: :$L$ is up-complete. By Up-Complete Lower Bounded Join Semilattice is Complete: :$L$ is a complete lattice. By definition: :$P$ is closure system of $L$. By Image of Operator Generated by Closure System is Set of Closure System :$\map {\operatorname {operator} } P \sqbrk S = T$ By Closure Operator Preserves Directed Suprema iff Image of Closure Operator Inherits Directed Suprema :$\map {\operatorname {operator} } P$ preserves directed suprema. Thus by Image of Directed Suprema Preserving Closure Operator is Algebraic Lattice: :$P$ is an algebraic lattice. {{qed}}
Performing the calculation using long division: <pre>    0.037...   -------- 27)1.00000      81      --      190      189      ---        100         81        ---         ... </pre> This is because $999 = 27 \times 37$. It can be determined by inspection of all smaller integers that this is indeed the smallest to have a period of $3$. {{qed}}