hcju/proofwikiqa huggingface dataset:
Let $S_n$ denote the set of all positive integers less than and coprime to $n$, excluding $1$. Let $\map P n$ denote the propositional function: :All positive integers less than and coprime to $n$, excluding $1$, are prime. We establish that $\map P n = \mathrm T$ for all the positive integers given: {{begin-eqn}} {{eqn | l = S_1       | r = \varnothing       | c = trivially }} {{eqn | l = S_2       | r = \varnothing       | c = trivially }} {{eqn | l = S_3       | r = \set 2       | c = which is prime }} {{eqn | l = S_4       | r = \set 3       | c = which is prime }} {{eqn | l = S_6       | r = \set 5       | c = which is prime }} {{eqn | l = S_8       | r = \set {3, 5, 7}       | c = all prime }} {{eqn | l = S_{12}       | r = \set {5, 7, 11}       | c = all prime }} {{eqn | l = S_{18}       | r = \set {5, 7, 11, 13, 17}       | c = all prime }} {{eqn | l = S_{24}       | r = \set {5, 7, 11, 13, 17, 19, 23}       | c = all prime }} {{eqn | l = S_{30}       | r = \set {7, 11, 13, 17, 19, 23, 29}       | c = all prime }} {{end-eqn}} From Schatunowsky's Theorem: :$30$ is the greatest positive integer $n$ such that $\map P n$ is true We note that for all primes $p$ greater than $3$, $p - 1$ is composite, and so $\map P p = \mathrm F$. The remaining composite numbers less than $30$ are investigated: {{begin-eqn}} {{eqn | l = S_9       | r = \set {2, 4, 5, 7, 8}       | c = of which $2, 4, 8$ are composite }} {{eqn | l = S_{10}       | r = \set {3, 7, 9}       | c = of which $9$ is composite, }} {{eqn | l = S_{14}       | r = \set {3, 5, 9, 11, 13}       | c = of which $9$ is composite }} {{eqn | l = S_{15}       | r = \set {2, 4, 7, 8, 11, 13, 14}       | c = of which $4, 8, 14$ are composite }} {{eqn | l = S_{16}       | r = \set {3, 5, 7, 9, 11, 13, 15}       | c = of which $9, 15$ are composite }} {{eqn | l = S_{20}       | r = \set {3, 7, 9, 11, 13, 17, 19}       | c = of which $9$ is composite }} {{eqn | l = S_{21}       | r = \set {2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20}       | c = of which $4, 8, 10, 16, 20$ are composite }} {{eqn | l = S_{22}       | r = \set {3, 5, 7, 9, 13, 15, 17, 19, 21}       | c = of which $9, 15, 21$ are composite }} {{eqn | l = S_{25}       | r = \set {2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24}       | c = of which $4, 6, 8, 9, 12, 14, 16, 18, 21, 22, 24$ are composite }} {{eqn | l = S_{26}       | r = \set {3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25}       | c = of which $9, 15, 21, 25$ are composite }} {{eqn | l = S_{27}       | r = \set {2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26}       | c = of which $4, 8, 10, 14, 16, 20, 22, 25, 26$ are composite }} {{eqn | l = S_{28}       | r = \set {3, 5, 9, 11, 13, 15, 17, 19, 23, 25, 27}       | c = of which $9, 15, 25, 27$ are composite }} {{end-eqn}} That exhausts the list. Hence the result. {{qed}}
From Area of Right Triangle, the area $\mathcal A$ is: :$\mathcal A = \dfrac {a b} 2$ where $a$ and $b$ are the legs. $(1): \quad$ The area of the $\tuple {6, 8, 10}$ triangle is $\dfrac {6 \times 8} 2 = 24$. Its perimeter equals $6 + 8 + 10 = 24$. $(2): \quad$ The area of the $\tuple {5, 12, 13}$ triangle is $\dfrac {5 \times 12} 2 = 30$. Its perimeter equals $5 + 12 + 13 = 30$. It remains to prove that these are the only ones. Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle $T$. Thus $a, b, c$ form a Pythagorean triple. By definition of Pythagorean triple, $a, b, c$ are in the form: :$2 m n, m^2 - n^2, m^2 + n^2$ We have that $m^2 + n^2$ is always the hypotenuse. Thus the area of $T$ is given by: :$\mathcal A = m n \paren {m^2 - n^2}$ The perimeter of $T$ is given by: :$\mathcal P = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$ We need to find all $m$ and $n$ such that $\mathcal P = \mathcal A$.  Thus: {{begin-eqn}} {{eqn | l = 2 m^2 + 2 m n       | r = m n \paren {m^2 - n^2}       | c =  }} {{eqn | ll= \leadsto       | l = 2 m \paren {m + n}       | r = n \paren {m + n} \paren {m - n}       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = n \paren {m - n}       | r = 2       | c =  }} {{end-eqn}} As $m$ and $n$ are both (strictly) positive integers, it follows immediately that either: :$n = 1$ :$m - n = 2$ and so: :$m = 3, n = 1$ leading to the triangle: : $a = 6, b = 8, c = 10$ or: :$n = 2$ :$m - n = 1$ and so: :$m = 3, n = 2$ leading to the triangle: : $a = 12, b = 5, c = 13$ and the result follows. {{qed}}
We have: :$\map \Frob 1 = 1$ By Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, $\Frob$ is injective. {{qed}}
By Characteristic of Galois Field is Prime, $\Char \GF$ is a prime number, say $p$. By Frobenius Endomorphism on Field is Injective, $\Frob$ is injective. By Injection from Finite Set to Itself is Surjection, $\Frob$ is bijective. By Bijective Ring Homomorphism is Isomorphism, $\Frob$ is an automorphism. {{qed}} Category:Galois Fields Category:Perfect Fields pf6ghzyfw6taydho9xlsl66zwf7tcxw
Note that Group Homomorphism is Invariant under Congruence Modulo Kernel. === Uniqueness === By Universal Property of Quotient Set, there exists a unique such mapping $\overline f$. A fortiori, there exists at most one such group homomorphism. {{qed|lemma}} === Existence === {{finish}}
We demonstrate that this is indeed an arithmetic sequence: {{begin-eqn}} {{eqn | l = 66 - 30       | r = 36 }} {{eqn | l = 102 - 66       | r = 36 }} {{eqn | l = 138 - 102       | r = 36 }} {{end-eqn}} demonstrating the common difference of $36$. Then we note: {{begin-eqn}} {{eqn | l = 30       | r = 2 \times 3 \times 5 }} {{eqn | l = 66       | r = 2 \times 3 \times 11 }} {{eqn | l = 102       | r = 2 \times 3 \times 17 }} {{eqn | l = 138       | r = 2 \times 3 \times 23 }} {{end-eqn}} It is understood that '''smallest arithmetic sequence''' means: :integer sequence of length $4$ whose terms are in arithmetic sequence such that their sum is the smallest of all such sequences. To demonstrate that this sequence is indeed the smallest, according to this definition, we list all positive integers less than $138$, each with $3$ distinct prime factors. They are: {{begin-eqn}} {{eqn | l = 30       | r = 2 \times 3 \times 5 }} {{eqn | l = 42       | r = 2 \times 3 \times 7 }} {{eqn | l = 66       | r = 2 \times 3 \times 11 }} {{eqn | l = 70       | r = 2 \times 5 \times 7 }} {{eqn | l = 78       | r = 2 \times 3 \times 13 }} {{eqn | l = 102       | r = 2 \times 3 \times 17 }} {{eqn | l = 105       | r = 3 \times 5 \times 7 }} {{eqn | l = 110       | r = 2 \times 5 \times 11 }} {{eqn | l = 114       | r = 2 \times 3 \times 17 }} {{eqn | l = 130       | r = 2 \times 5 \times 13 }} {{end-eqn}} One cannot pick any other $4$ numbers from the list to form an arithmetic sequence. Hence any further such arithmetic sequence will contain larger terms whose sum will be consequently larger. {{qed}}
Performing the calculation using long division: <pre>    0.03225806451612903...   ----------------------- 31)1.00000000000000000000      93     155      --     ---       70      50       62      31       --      --        80     190        62     186        --     ---        180      40        155      31        ---      --         250      90         248      62         ---      --           200    280           186    279           ---    ---            140     100            124      93            ---     ---             160    ...             155             --- </pre> Therefore $\dfrac 1 {31} = 0 \cdotp \dot 03225 \, 80645 \, 1612 \dot 9$, with an odd period of $15$. The prime numbers less than $31$ are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$. We have: :$\dfrac 1 2 = 0 \cdotp 5$: not recurring. :$\dfrac 1 3 = 0 \cdotp \dot 3$: recurring with period $1$. :$\dfrac 1 5 = 0 \cdotp 2$: not recurring. :$\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$: recurring with period $6$. :$\dfrac 1 {11} = 0 \cdotp \dot 0 \dot 9$: recurring with period $2$. :$\dfrac 1 {13} = 0 \cdotp \dot 07692 \dot 3$: recurring with period $6$. :$\dfrac 1 {17} = 0 \cdotp \dot 058823529411764 \dot 7$: recurring with period $16$. :$\dfrac 1 {19} = 0 \cdotp \dot 05263157894736842 \dot 1$: recurring with period $18$. :$\dfrac 1 {23} = 0 \cdotp \dot 043478260869565217391 \dot 3$: recurring with period $22$. :$\dfrac 1 {29} = 0 \cdotp \dot 034482758620689655172413793 \dot 1$: recurring with period $28$. All of the above either terminates, has period $1$ or has even periods. {{qed}}
This can be determined by testing all the positive integers in succession for happiness. Checking $31$ and $32$: {{begin-eqn}} {{eqn | l = 31       | o = \to       | r = 3^2 + 1^2 = 9 + 1 = 10       | c =  }} {{eqn | o = \to       | r = 1^2 = 1       | c = and so $31$ is happy. }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 32       | o = \to       | r = 3^2 + 2^2 = 9 + 4 = 13       | c =  }} {{eqn | o = \to       | r = 1^2 + 3^2 = 1 + 9 = 10 }} {{eqn | o = \to       | r = 1^2 = 1       | c = and so $32$ is happy. }} {{end-eqn}} {{qed}}
Let $y \in f\left[{X}\right]$. By definition of image of set: :$\exists z \in X: y = f \left({z}\right)$ By definition of lower bound: :$x \preceq z$ Thus by definition of increasing mapping: :$f \left({x}\right) \preceq' y$ {{qed}}
By definition of order isomorphism: :$f$ is an order embedding. === Sufficient Condition === Assume that :$x$ is lower bound for $X$. By Order Embedding is Increasing Mapping: :$f$ is an increasing mapping. Thus by Increasing Mapping Preserves Lower Bounds: :$f \left({x}\right)$ is lower bound for $f \left[{X}\right]$. {{qed|lemma}} === Necessary Condition === Assume that :$f \left({x}\right)$ is lower bound for $f \left[{X}\right]$. Let $y \in X$. By definition of image of set: :$f \left({y}\right) \in f \left[{X}\right]$ By definition of lower bound: :$f \left({y}\right) \preceq' f \left({x}\right)$ Thus by definition of order embedding: :$y \preceq x$ {{qed}}
By definition of order embedding: :$\forall x, y \in S_1: x \preceq_1 y \implies f\left({x}\right) \preceq_2 f\left({y}\right)$ Hence $f$ is an increasing mapping. {{qed}} Category:Order Embeddings Category:Increasing Mappings 2dys8l0sul24o8x0jrg12yha84gkk4f
By definition of order isomorphism: :$f$ is an order embedding. === Sufficient Condition === Assume that :$x$ is upper bound for $X$. By Order Embedding is Increasing Mapping: :$f$ is an increasing mapping. Thus by Increasing Mapping Preserves Upper Bounds: :$\map f x$ is an upper bound for $f \sqbrk X$. {{qed|lemma}} === Necessary Condition === Assume that: :$\map f x$ is upper bound for $f \sqbrk X$. Let $y \in X$. By definition of image of set: :$\map f y \in f \sqbrk X$ By definition of upper bound: :$\map f x \preceq' \map f y$ Thus by definition of order embedding: :$x \preceq y$ {{qed}}
Let $y \in f\left[{X}\right]$. By definition of image of set: :$\exists z \in X: y = f\left({z}\right)$ By definition of upper bound: :$z \preceq x$ Thus by definition of increasing mapping: :$y \preceq' f\left({x}\right)$ {{qed}}
=== $f$ preserves infima === Let $X$ be a subset of $S$ such that :$X$ admits an infimum in $L$. By definition of infimum: :$\inf X$ is lower bound for $X$. Thus by Order Isomorphism Preserves Lower Bounds: :$f\left({\inf X}\right)$ is lower bound for $f\left[{X}\right]$. We will prove that :$\forall x \in S': x$ is lower bound for $f\left[{X}\right] \implies x \preceq' f\left({\inf X}\right)$ Let $x \in S'$ such that :$x$ is lower bound for $f\left[{X}\right]$. By definitions of order isomorphism and bijection: :$f$ is a surjection. By definition of surjection: :$S' = f\left[{S}\right]$ By definition of image of set: :$\exists y \in S: x = f\left({y}\right)$ By Order Isomorphism Preserves Lower Bounds: :$y$ is lower bound for $X$. By definition of infimum: :$y \preceq \inf X$ Thus by definition of order isomorphism: :$x \preceq' f\left({\inf X}\right)$ {{qed|lemma}} Thus by definition of infimum: :$f\left[{X}\right]$ admits an infimum in $L'$ and $\inf\left({f\left[{X}\right]}\right) = f\left( {\inf X}\right)$ {{qed|lemma}} === $f$ preserves suprema === This fallows by mutatis mutandis. {{qed}}
{{begin-eqn}} {{eqn | l = 2^{64} - 1       | r = \left({2^{32} }\right)^2 - 1       | c =  }} {{eqn | r = \left({2^{32} + 1}\right) \left({2^{32} - 1}\right)       | c = Difference of Two Squares }} {{eqn | r = \left({2^{32} + 1}\right) \left({\left({2^{16} }\right)^2 - 1}\right)       | c =  }} {{eqn | r = \left({2^{32} + 1}\right) \left({2^{16} + 1}\right) \left({2^{16} - 1}\right)       | c = Difference of Two Squares }} {{eqn | r = \left({2^{32} + 1}\right) \left({2^{16} + 1}\right) \left({\left({2^8}\right)^2 - 1}\right)       | c =  }} {{eqn | r = \left({2^{32} + 1}\right) \left({2^{16} + 1}\right) \left({2^8 + 1}\right) \left({2^8 - 1}\right)       | c = Difference of Two Squares }} {{eqn | r = \left({2^{32} + 1}\right) \left({2^{16} + 1}\right) \left({2^8 + 1}\right) \left({\left({2^4}\right)^2 - 1}\right)       | c =  }} {{eqn | r = \left({2^{32} + 1}\right) \left({2^{16} + 1}\right) \left({2^8 + 1}\right) \left({2^4 + 1}\right) \left({2^4 - 1}\right)       | c = Difference of Two Squares }} {{eqn | r = \left({2^{32} + 1}\right) \left({2^{16} + 1}\right) \left({2^8 + 1}\right) \left({2^4 + 1}\right) \left({\left({2^2}\right)^2 - 1}\right)       | c =  }} {{eqn | r = \left({2^{32} + 1}\right) \left({2^{16} + 1}\right) \left({2^8 + 1}\right) \left({2^4 + 1}\right) \left({2^2 + 1}\right) \left({2^2 - 1}\right)       | c = Difference of Two Squares }} {{eqn | r = \left({2^{32} + 1}\right) \left({2^{16} + 1}\right) \left({2^8 + 1}\right) \left({2^4 + 1}\right) \left({2^2 + 1}\right) \left({2 + 1}\right) \left({2 - 1}\right)       | c = Difference of Two Squares }} {{eqn | r = \left({2^{\left({2^5}\right)} + 1}\right) \left({2^{\left({2^4}\right)} + 1}\right) \left({2^{\left({2^3}\right)} + 1}\right) \left({2^{\left({2^2}\right)} + 1}\right) \left({2^{\left({2^1}\right)} + 1}\right) \left({2^{\left({2^0}\right)} + 1}\right)       | c =  }} {{end-eqn}} From Sequence of Fermat Primes: {{begin-eqn}} {{eqn | l = 2^{\left({2^0}\right)} + 1       | r = 3 }} {{eqn | l = 2^{\left({2^1}\right)} + 1       | r = 5 }} {{eqn | l = 2^{\left({2^2}\right)} + 1       | r = 17 }} {{eqn | l = 2^{\left({2^3}\right)} + 1       | r = 257 }} {{eqn | l = 2^{\left({2^4}\right)} + 1       | r = 65\ 537 }} {{end-eqn}} all of which are Fermat primes. Then: {{begin-eqn}} {{eqn | l = 2^{\left({2^5}\right)} + 1       | r = 4 \, 294 \, 967 \, 297       | c = Sequence of Fermat Numbers }} {{eqn | r = 641 \times 6 \, 700 \, 417       | c = Prime Decomposition of $5$th Fermat Number }} {{end-eqn}} Hence the result. {{qed}} Category:18,446,744,073,709,551,615 03u2upjlnlwktpp3kbvtzslvwb0st28
From Divisor of Fermat Number, if $2^{\paren {2^n} } + 1$ has a divisor, it will be in the form: :$k \, 2^{n + 2} + 1$ In the case of $n = 5$, a divisor of $2^{\paren {2^n} } + 1$ is then of the form: :$k \, 2^7 + 1 = k \times 128 + 1$ Further, such a number will (for small $k$ at least) be prime, otherwise it will itself have a divisor which is not of that form. Thus we investigate: {{begin-eqn}} {{eqn | l = 1 \times 128 + 1       | r = 129       | rr= = 3 \times 43       | c = which is not prime }} {{eqn | l = 2 \times 128 + 1       | r = 257       | c = which is prime }} {{eqn | l = 3 \times 128 + 1       | r = 385       | rr= = 5 \times 7 \times 11       | c = which is not prime }} {{eqn | l = 4 \times 128 + 1       | r = 513       | rr= = 3^3 \times 19       | c = which is not prime }} {{eqn | l = 5 \times 128 + 1       | r = 641       | c = which is prime }} {{eqn | l = 6 \times 128 + 1       | r = 769       | c = which is prime }} {{end-eqn}} We have that: {{begin-eqn}} {{eqn | l = 2^{32} - 1       | r = \paren {2^{16} + 1} \paren {2^{16} - 1}       | c = Difference of Two Squares }} {{eqn | r = \paren {2^{16} + 1} \paren {2^8 - 1} \paren {2^8 + 1}       | c =  }} {{end-eqn}} But :$2^8 + 1 = 257$ and so: :$2^{\paren {2^5} } - 1 \equiv 0 \pmod {257}$ which means: :$2^{\paren {2^5} } + 1 \equiv 2 \pmod {257}$ so there is no need to do a trial division of $2^{\paren {2^5} } + 1$ by $257$. Now dividing $4 \, 294 \, 967 \, 297$ by $641$ we find: <pre>           6 700 417     --------------- 641 ) 4 294 967 297       3 846       -----         448 9         448 7         -----             267 2             256 4             -----              10 89               6 41              -----               4 487               4 487               ===== </pre> and the divisor has been found. {{qed}}
Note the remarkable coincidence that $2^4 + 5^4 = 2^7 \cdot 5 + 1 = 641$. First we eliminate $y$ from $x^4 + y^4 = x^7 y + 1 = 0$: {{begin-eqn}} {{eqn | l = x^4 + y^4       | r = x^7 y + 1 = 0       | c =  }} {{eqn | ll= \leadsto       | l = -x^4       | r = y^4       | c =  }} {{eqn | ll= \leadsto       | l = x^{28} \paren {-x^4} + 1       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x^{32}       | r = -1       | c =  }} {{end-eqn}} Now we use the above result for $x = 2$ and $y = 4$ in modulo $641$: {{begin-eqn}} {{eqn | l = 2^4 + 5^4       | o = \equiv       | r = 0       | rr= \pmod {641}       | c =  }} {{eqn | l = 2^7 \cdot 5       | o = \equiv       | r = -1       | rr= \pmod {641}       | c =  }} {{eqn | ll= \leadsto       | l = 2^{28} \paren {-2^4}       | o = \equiv       | r = -1       | rr= \pmod {641}       | c =  }} {{eqn | ll= \leadsto       | l = 2^{32}       | o = \equiv       | r = -1       | rr= \pmod {641}       | c =  }} {{eqn | ll= \leadsto       | l = 2^{32} + 1       | o = \equiv       | r = 0       | rr= \pmod {641}       | c =  }} {{end-eqn}} Thus $2^{\paren {2^5} } + 1 = 6 \, 700 \, 417 \times 641$ and hence is not prime. {{qed}}
By definitions of compact subset, compact closure, and subset: :$\forall x \in S: x^{\mathrm{compact} } \subseteq K\left({L}\right)$ By definition of power set: :$\forall x \in S: x^{\mathrm{compact} } \in \mathcal P\left({K\left({L}\right)}\right)$ Define a mapping $f:S \to \mathcal P\left({K\left({L}\right)}\right)$ such that :$\forall x \in S: f\left({x}\right) = x^{\mathrm{compact} }$ By Compact Closure is Directed: :$\forall x \in S: x^{\mathrm{compact} }$ is directed. By definition of ordered subset: :$\forall x \in S: x^{\mathrm{compact} }$ is directed in $C$. We will prove that :$\forall x \in S: x^{\mathrm{compact} }$ is lower in $C$. Let $x \in S$. Let $y \in x^{\mathrm{compact} }, z \in K\left({L}\right)$ such that :$z \preceq y$ By definition of compact closure: :$y \preceq x$ By definition of transitivity: :$z \preceq x$ By definition of compact subset: :$z$ is compact. Thus by definition of compact closure: :$z \in x^{\mathrm{compact} }$ {{qed|lemma}} By Bottom in Compact Closure: :$\forall x \in S: \bot_L \in x^{\mathrm{compact} }$ where $\bot_L$ denotes the bottom in $L$. By definition: :$\forall x \in S: x^{\mathrm{compact} }$ is non-empty. By definition of ideal in ordered set: :$\forall x \in S: x^{\mathrm{compact} } \in \mathit{Ids}\left({C}\right)$ where $\mathit{Ids}\left({C}\right)$ denotes the set of all ideals in $C$. Then :$f:S \to \mathit{Ids}\left({C}\right)$ Define $I = \left({\mathit{Ids}\left({C}\right), \precsim'}\right)$ as an inclusion ordered set. By definitions of $\mathit{Ids}$ and power set: :$\mathit{Ids}\left({C}\right) \subseteq \mathcal P\left({K\left({L}\right)}\right)$ By Mapping Assigning to Element Its Compact Closure is Order Isomorphism: :$f$ is order isomorphism between $L$ and $I$. By Bottom in Compact Subset and Compact Subset is Join Subsemilattice: :$C$ is bounded below join semilattice. By Ideals are Continuous Lattice Subframe of Power Set: :$I$ is a continuous lattice subframe of $P$. By Power Set is Complete Lattice: :$P$ is a complete lattice. By Order Isomorphism Preserves Infima and Suprema: :$f:S \to \mathit{Ids}\left({C}\right)$ preserves infima and :$f:S \to \mathit{Ids}\left({C}\right)$ preserves suprema. Thus by Extension of Infima Preserving Mapping to Complete Lattice Preserves Infima: :$f:S \to \mathcal P\left({K\left({L}\right)}\right)$ preserves infima. By definition: :$f:S \to \mathit{Ids}\left({C}\right)$ preserves directed suprema. Thus by Extension of Directed Suprema Preserving Mapping to Complete Lattice Preserves Directed Suprema: :$f:S \to \mathcal P\left({K\left({L}\right)}\right)$ preserves directed suprema. By definition of order isomorphism: :$f:S \to \mathit{Ids}\left({C}\right)$ is a bijection. By definition of bijection: :$f:S \to \mathit{Ids}\left({C}\right)$ is an injection. Thus by definition of injection: :$f:S \to \mathcal P\left({K\left({L}\right)}\right)$ is an injection. Thus :$\forall x \in S: f\left({x}\right) = x^{\mathrm{compact} }$ {{qed}}
We have: {{begin-eqn}} {{eqn | l = 33       | r = 3 \times 11 }} {{eqn | l = 34       | r = 2 \times 17 }} {{eqn | l = 35       | r = 5 \times 7 }} {{end-eqn}} It can be seen from the sequence of semiprimes that there exist no smaller such triples. {{qed}}
It will be proved that the largest integer which cannot be expressed as the sum of distinct triangular numbers is $33$. The remaining integers in the sequence can be identified by inspection. We prove this using a variant of Second Principle of Mathematical Induction. Let $\map P n$ be the proposition: :$n$ can be expressed as the sum of distinct triangular numbers. === Basis for the induction === We verify the result up to $T_{13} = 91$. Then we conclude all $n$ with $34 \le n \le 91$ can be expressed as the sum of distinct triangular numbers. So $\map P n$ is true for all $34 \le n \le 91$. This is the basis for the induction. === Induction Hypothesis === Suppose for some $k > 91$, $\map P j$ is true for all $34 \le j < k$. That is, every integer between $34$ and $k - 1$ can be expressed as the sum of distinct triangular numbers. This is the induction hypothesis. === Induction Step === This is the induction step: We find the largest integer $i$ such that $T_i < k \le T_{i + 1}$. We show that $k - T_{i - 3}$ can be expressed as the sum of distinct triangular numbers, and the sum does not involve $T_{i - 3}$. Then $k$ can be expressed as the sum of distinct triangular numbers. Since $k > 91 = T_{13}$, we must have $i \ge 13$. Hence $k > k - T_{i - 3} > T_i - T_{i - 3} = i + \paren {i - 1} + \paren {i - 2} \ge 13 + 12 + 11 = 36$. Thus $k - T_{i - 3}$ can be expressed as the sum of distinct triangular numbers by Induction Hypothesis. A sufficient condition such that the sum does not involve $T_{i - 3}$ is $T_{i - 3} > k - T_{i - 3}$. We have: {{begin-eqn}} {{eqn | l = T_{i - 3} - \paren {k - T_{i - 3} }       | o = \ge       | r = 2 T_{i - 3} - T_{i + 1}       | c = From $k \le T_{i + 1}$ }} {{eqn | r = 2 \paren {\frac {\paren {i - 3} \paren {i - 2} } 2} - \frac {\paren {i + 1} \paren {i + 2} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac 1 2 \paren {2 i^2 - 10 i + 12 - i^2 - 3i - 2} }} {{eqn | r = \frac 1 2 \paren {i^2 - 13 i + 10} }} {{eqn | r = \frac 1 2 \paren {i \paren {i - 13} + 10} }} {{eqn | o = \ge       | r = \frac 1 2 \paren {10}       | c = From $i \ge 13$ }} {{eqn | r = 5 }} {{eqn | o = >       | r = 0 }} {{end-eqn}} Therefore we have $T_{i - 3} > k - T_{i - 3}$. Hence $\map P k$ is true. By the Second Principle of Mathematical Induction, $\map P n$ is true for all $n \ge 34$. Thus every integer greater than $33$ can be expressed as the sum of distinct triangular numbers. {{qed}}