hcju/proofwikips huggingface dataset:
Brahmagupta-Fibonacci Identity/Extension/General
== Extension to Brahmagupta-Fibonacci Identity ==  Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n, m$ be integers. Then: :$\displaystyle \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$. That is: the set of all integers of the form $a^2 + m b^2$ is closed under multiplication.  == Proof == The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$. $\map P 1$ is the trivial case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^1 \paren { {a_j}^2 + m {b_j}^2}       | r = {a_1}^2 + m {b_1}^2       | c =  }} {{eqn | r = c^2 + m d^2       | c = setting $c = a_1$ and $d = b_1$ }} {{end-eqn}} Thus $\map P 1$ is seen to hold. === Basis for the Induction === $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^2 \paren { {a_j}^2 + m {b_j}^2}       | r = \paren { {a_1}^2 + m {b_1}^2} \paren { {a_2}^2 + m {b_2}^2}       | c =  }} {{eqn | r = \paren {a_1 a_2 + m b_1 b_2}^2 + m \paren {a_1 b_2 - b_1 a_2}^2       | c = General Brahmagupta-Fibonacci Identity }} {{eqn | r = c^2 + m d^2       | c = setting $c = a_1 a_2 + m b_1 b_2$ and $d = a_1 b_2 - b_1 a_2$ }} {{end-eqn}} Thus $\map P 2$ 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: :$\displaystyle \prod_{j \mathop = 1}^k \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$. from which it is to be shown that: :$\displaystyle \prod_{j \mathop = 1}^{k + 1} \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$. === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^{k + 1} \paren { {a_j}^2 + m {b_j}^2}       | r = \prod_{j \mathop = 1}^k \paren { {a_j}^2 + m {b_j}^2} \paren { {a_{k + 1} }^2 + m {b_{k + 1} }^2}       | c =  }} {{eqn | r = \paren { {c'}^2 + m {d'}^2} \paren { {a_{k + 1} }^2 + m {b_{k + 1} }^2}       | c = Induction Hypothesis: for some $c', d' \in \Z$ }} {{eqn | r = c^2 + m d^2       | c = Basis for the Induction: for some $c, d \in \Z$ }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore, for all $n \in \Z_{> 0}$: :$\displaystyle \prod_{j \mathop = 1}^n \paren { {a_j}^2 + m {b_j}^2} = c^2 + m d^2$ for some $c, d \in \Z$.
Brahmagupta-Fibonacci Identity/Extension/Proof 2
== Extension to Brahmagupta-Fibonacci Identity == {{:Brahmagupta-Fibonacci Identity/Extension}} == Proof ==  The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \prod_{j \mathop = 1}^n \left({ {a_j}^2 + {b_j}^2}\right) = c^2 + d^2$ for some $c, d \in \Z$. $P \left({1}\right)$ is the trivial case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^1 \left({ {a_j}^2 + {b_j}^2}\right)       | r = {a_1}^2 + {b_1}^2       | c =  }} {{eqn | r = c^2 + d^2       | c = setting $c = a_1$ and $d = b_1$ }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. === Basis for the Induction === $P \left({2}\right)$ is the case: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^2 \left({ {a_j}^2 + {b_j}^2}\right)       | r = \left({ {a_1}^2 + {b_1}^2}\right) \left({ {a_2}^2 + {b_2}^2}\right)       | c =  }} {{eqn | r = \left({a_1 a_2 + b_1 b_2}\right)^2 + \left({a_1 b_2 - b_1 a_2}\right)^2       | c = Brahmagupta-Fibonacci Identity }} {{eqn | r = c^2 + d^2       | c = setting $c = a_1 a_2 + b_1 b_2$ and $d = a_1 b_2 - b_1 a_2$ }} {{end-eqn}} Thus $P \left({2}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \prod_{j \mathop = 1}^k \left({ {a_j}^2 + {b_j}^2}\right) = c^2 + d^2$ for some $c, d \in \Z$. from which it is to be shown that: :$\displaystyle \prod_{j \mathop = 1}^{k + 1} \left({ {a_j}^2 + {b_j}^2}\right) = c^2 + d^2$ for some $c, d \in \Z$. === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \prod_{j \mathop = 1}^{k + 1} \left({ {a_j}^2 + {b_j}^2}\right)       | r = \prod_{j \mathop = 1}^k \left({ {a_j}^2 + {b_j}^2}\right) \left({ {a_{k + 1} }^2 + {b_{k + 1} }^2}\right)       | c =  }} {{eqn | r = \left({ {c'}^2 + {d'}^2}\right) \left({ {a_{k + 1} }^2 + {b_{k + 1} }^2}\right)       | c = Induction Hypothesis: for some $c', d' \in \Z$ }} {{eqn | r = c^2 + m d^2       | c = Basis for the Induction: for some $c, d \in \Z$ }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore, for all $n \in \Z_{> 0}$: :$\displaystyle \prod_{j \mathop = 1}^n \left({ {a_j}^2 + {b_j}^2}\right) = c^2 + d^2$ for some $c, d \in \Z$.
Magic Square/Examples/Order 4
== Examples of Order $4$ Magic Squares ==  There are many order $4$ magic squares. ==== Dürer's Order $4$ Magic Square ==== This is one of the more famous ones, due to {{AuthorRef|Albrecht Dürer}}: {{:Magic Square/Examples/Order 4/Dürer}} ==== Moessner's Order $4$ Magic Square ==== This one, created by {{AuthorRef|Alfred Moessner}}, has extra interesting properties: {{:Magic Square/Examples/Order 4/Alfred Moessner}} == Also see == * Magic Constant of Order 4 Magic Square * Number of Magic Squares of Order 4
Magic Square/Examples/Order 4/Alfred Moessner
== Example of Order $4$ Magic Square == This example of an order $4$ magic square is due to {{AuthorRef|Alfred Moessner}}:  :$\begin{array}{|c|c|c|c|} \hline 12 & 13 & 1 & 8 \\ \hline 6 & 3 & 15 & 10 \\ \hline 7 & 2 & 14 & 11 \\ \hline 9 & 16 & 4 & 5 \\ \hline \end{array}$  == Also see == * Properties of Moessner's Order $4$ Magic Square == Historical Note == {{:Magic Square/Examples/Order 4/Alfred Moessner/Historical Note}} == Sources == * {{citation|author = Alfred Moessner|title = A Curious Magic Square|journal = Scripta Mathematica|volume = 13|date = 1947?|startpage = ???|endpage = ???}} * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Magic Square/Examples/Order 4/Dürer/Historical Note|next = Magic Square/Examples/Order 4/Alfred Moessner/Historical Note}}: $16$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Magic Square/Examples/Order 4/Dürer/Historical Note|next = Magic Square/Examples/Order 4/Alfred Moessner/Historical Note}}: $16$
Feit-Thompson Conjecture
== Conjecture == There exist no distinct prime numbers $p$ and $q$ such that: :$\dfrac {p^q - 1} {p - 1}$ divides $\dfrac {q^p - 1} {q - 1}$ === Stronger Feit-Thompson Conjecture === {{:Feit-Thompson Conjecture/Stronger}} {{Namedfor|Walter Feit|name2 = John Griggs Thompson|cat = Feit|cat2 = Thompson}}
Factorial/Examples/11
== Example of Factorial ==  :$11! = 39 \, 916 \, 800$  == Proof == From Examples of Factorials: :$10! = 3 \, 628 \, 800$ Then: <pre>  3 628 800 x       11 ----------  3 628 800 36 288 000 ---------- 39 916 800 </pre>
Factorial/Examples/12
== Example of Factorial ==  :$12! = 479 \, 001 \, 600$  == Proof == From $11$ factorial: :$11! = 39 \, 916 \, 800$ Then: <pre>  39 916 800 x        12 -----------  79 833 600 399 168 000 ----------- 479 001 600 </pre>
Factorial/Examples/13
== Example of Factorial ==  :$13! = 6 \, 227 \, 020 \, 800$  === Prime Factors of $13!$ === {{:Prime Factors of 13 Factorial}} == Proof == From $12$ factorial: :$12! = 479 \, 001 \, 600$ Then: <pre>   479 001 600 x          13 ------------- 1 437 004 800 4 790 016 000 ------------- 6 227 020 800 </pre>
Factorial/Examples/14
== Example of Factorial ==  :$14! = 87 \, 178 \, 291 \, 200$  == Proof == From $13$ factorial: :$13! = 6 \, 227 \, 020 \, 800$ Then: <pre>  6 227 020 800 x           14 --------------- 24 908 083 200 62 270 208 000 -------------- 87 178 291 200 </pre>
Factorial/Examples/15
== Example of Factorial ==  :$15! = 1 \, 307 \, 674 \, 368 \, 000$  == Proof == From $14$ Factorial: :$14! = 87 \, 178 \, 291 \, 200$ Then: <pre>    87 178 291 200 x              15 ------------------   435 891 456 000   871 782 912 000 ----------------- 1 307 674 368 000 </pre>
Factorial/Examples/16
== Example of Factorial ==  :$16! = 20 \, 922 \, 789 \, 888 \, 000$  == Proof == From $15$ Factorial: :$15! = 1 \, 307 \, 674 \, 368 \, 000$ Then: <pre>  1 307 674 368 000 x               16 ------------------  7 846 046 208 000 13 076 743 680 000 ------------------ 20 922 789 888 000 </pre>
Factorial/Examples/17
== Example of Factorial ==  :$17! = 355 \, 687 \, 428 \, 096 \, 000$  == Proof == From $16$ Factorial: :$16! = 20 \, 922 \, 789 \, 888 \, 000$ Then: <pre>  20 922 789 888 000 x                17 ------------------- 146 459 529 216 000 209 227 898 880 000 ------------------- 355 687 428 096 000 </pre>
Factorial/Examples/18
== Example of Factorial ==  :$18! = 6 \, 402 \, 373 \, 705 \, 728 \, 000$  == Proof == From $17$ Factorial: :$17! = 355 \, 687 \, 428 \, 096 \, 000$ Then: <pre>   355 687 428 096 000 x                  18 --------------------- 2 845 499 424 768 000 3 556 874 280 960 000 --------------------- 6 402 373 705 728 000 </pre>
Factorial/Examples/19
== Example of Factorial ==  :$19! = 121 \, 645 \, 100 \, 408 \, 832 \, 000$  == Proof == From $18$ Factorial: :$18! = 6 \, 402 \, 373 \, 705 \, 728 \, 000$ Then: <pre>   6 402 373 705 728 000 x                    19 -----------------------  57 621 363 351 552 000  64 023 737 057 280 000 ----------------------- 121 645 100 408 832 000 </pre>
Factorial/Examples/20
== Example of Factorial ==  :$20! = 2 \, 432 \, 902 \, 008 \, 176 \, 640 \, 000$  === Prime Factors of $20!$ === {{:Prime Factors of 20 Factorial}} == Proof == From $19$ Factorial: :$19! = 121 \, 645 \, 100 \, 408 \, 832 \, 000$ Then: <pre>   121 645 100 408 832 000 x                      20 ------------------------ 2 432 902 008 176 640 000 </pre> == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Cross-Sections of Leech Lattice|next = Factorial/Examples/450/Historical Note}}: $24$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Factorial|next = Factorial/Examples/1,000,000}}: $24$
Factorial/Examples/21
== Example of Factorial ==  :$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$  == Proof == From $20$ Factorial: :$20! = 2 \, 432 \, 902 \, 008 \, 176 \, 640 \, 000$ Then: <pre>   2 432 902 008 176 640 000 x                        21 ---------------------------   2 432 902 008 176 640 000  48 658 040 163 532 800 000 ---------------------------  51 090 942 171 709 440 000    </pre>
Factorial/Examples/22
== Example of Factorial ==  :$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$  == Proof == From $21$ Factorial: :$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$ Then: <pre>    51 090 942 171 709 440 000 x                          22 -----------------------------   102 181 884 343 418 880 000 1 021 818 843 434 188 800 000 ----------------------------- 1 124 000 727 777 607 680 000    </pre>
Factorial/Examples/23
== Example of Factorial ==  :$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$  == Proof == From $22$ Factorial: :$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$ Then: <pre>  1 124 000 727 777 607 680 000 x                           23 ------------------------------  3 372 002 183 332 823 040 000 22 480 014 555 552 153 600 000 ------------------------------ 25 852 016 738 884 976 640 000    </pre>
Factorial/Examples/24
== Example of Factorial ==  :$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$  == Proof == From $23$ Factorial: :$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$ Then: <pre>  25 852 016 738 884 976 640 000 x                            24 ------------------------------- 103 408 066 955 539 906 560 000 517 040 334 777 699 532 800 000 ------------------------------- 620 448 401 733 239 439 360 000    </pre>
Factorial/Examples/25
== Example of Factorial ==  :$25! = 15 \, 511 \, 210 \, 043 \, 330 \, 985 \, 984 \, 000 \, 000$  == Proof == From $24$ Factorial: :$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$ Then: <pre>    620 448 401 733 239 439 360 000 x                               25 ----------------------------------  3 102 242 008 666 197 196 800 000 12 408 968 034 664 788 787 200 000 ---------------------------------- 15 511 210 043 330 985 984 000 000    </pre>
Square of Small-Digit Palindromic Number is Palindromic/Examples/11
== Example of Square of Small-Digit Palindromic Number is Palindromic ==  ::$11^2 = 121$  == Proof == {{begin-eqn}} {{eqn | l = 11^2       | r = \left({10 + 1}\right)^2       | c =  }} {{eqn | r = 10^2 + 2 \times 10 \times 1 + 1^2       | c =  }} {{eqn | r = 100 + 20 + 1       | c =  }} {{eqn | r = 121       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Square of Small-Digit Palindromic Number is Palindromic/Examples/22|next = Square of Small-Digit Palindromic Number is Palindromic/Examples/111}}: $22$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 121|next = 121 is Square Number in All Bases greater than 2}}: $121$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Square of Small-Digit Palindromic Number is Palindromic/Examples/22|next = Square of Small-Digit Palindromic Number is Palindromic/Examples/111}}: $22$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 121|next = 121 is Square Number in All Bases greater than 2}}: $121$
Square of Small-Digit Palindromic Number is Palindromic/Examples/121
== Example of Square of Small-Digit Palindromic Number is Palindromic ==  ::$121^2 = 14641$  == Proof == <pre>   121 x 121 -----   121  2420 12100 ------- 14641 </pre> == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Square of Small-Digit Palindromic Number is Palindromic/Examples/1111|next = Square of Small-Digit Palindromic Number is Palindromic/Examples/212}}: $22$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Square of Small-Digit Palindromic Number is Palindromic/Examples/1111|next = Square of Small-Digit Palindromic Number is Palindromic/Examples/212}}: $22$
Sigma Function of Square-Free Integer/Examples/70
== Example of Sigma Function of Square-Free Integer ==  :$\map \sigma {70} = 144$  where $\sigma$ denotes the $\sigma$ function. == Proof 1 == {{:Sigma Function of Square-Free Integer/Examples/70/Proof 1}} == Proof 2 == {{:Sigma Function of Square-Free Integer/Examples/70/Proof 2}}
Birthday Paradox
== Paradox == Let there be $23$ or more people in a room. The probability that at least $2$ of them have the same birthday is greater than $50 \%$. == Proof == Let there be $n$ people in the room. Let $\map p n$ be the probability that no two people in the room have the same birthday. For simplicity, let us ignore leap years and assume there are $365$ days in the year. Let the birthday of person $1$ be established. The probability that person $2$ shares person $1$'s birthday is $\dfrac 1 {365}$. Thus, the probability that person $2$ does not share person $1$'s birthday is $\dfrac {364} {365}$. Similarly, the probability that person $3$ does not share the birthday of either person $1$ or person $2$ is $\dfrac {363} {365}$. And further, the probability that person $n$ does not share the birthday of any of the people indexed $1$ to $n - 1$ is $\dfrac {365 - \paren {n - 1} } {365}$. Hence the total probability that none of the $n$ people share a birthday is given by: :$\map p n = \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}$ {{begin-eqn}} {{eqn | l = \map p n       | r = \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}       | c =  }} {{eqn | r = \dfrac {365!} {365^n} \binom {365} n       | c =  }} {{end-eqn}} Setting $n = 23$ and evaluating the above gives: :$\map p {23} \approx 0.493$ Hence the probability that at least $2$ people share a birthday is $1 = 0.492 = 0.507 = 50.7 \%$  == Conclusion == This is a veridical paradox. Counter-intuitively, the probability of a shared birthday amongst such a small group of people is surprisingly high. == General Birthday Paradox == {{:Birthday Paradox/General}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 23 is Largest Integer not Sum of Distinct Perfect Powers/Mistake|next = Harmonic Series is Divergent}}: $23$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 23 is Largest Integer not Sum of Distinct Perfect Powers/Mistake|next = Birthday Paradox/General/3}}: $23$
Sigma Function of Integer/Examples/20
== Example of Sigma Function of Integer ==  :$\sigma \left({20}\right) = 42$  where $\sigma$ denotes the $\sigma$ function. == Proof == 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 $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$20 = 2^2 \times 5$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({20}\right)       | r = \frac {2^3 - 1} {2 - 1} \times \frac {5^2 - 1} {5 - 1}       | c =  }} {{eqn | r = \frac 7 1 \times \frac {6 \times 4} 4       | c =  }} {{eqn | r = 7 \times 6       | c =  }} {{eqn | r = 42       | c =  }} {{end-eqn}}
Sigma Function of Non-Square Semiprime/Examples/35
== Example of Sigma Function of Non-Square Semiprime ==  :$\map \sigma {35} = 48$  where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$35 = 5 \times 7$ and so by definition is a semiprime whose prime factors are distinct. Hence: {{begin-eqn}} {{eqn | l = \map \sigma {35}       | r = \paren {5 + 1} \paren {7 + 1}       | c = Sigma Function of Non-Square Semiprime }} {{eqn | r = 6 \times 8       | c =  }} {{eqn | r = 48       | c =  }} {{end-eqn}}
Binomial Theorem/Examples/Cube of Sum
== Example of Use of Binomial Theorem ==  :$\paren {x + y}^3 = x^3 + 3 x^2 y + 3 x y^2 + y^3$  === Corollary === {{:Cube of Sum/Corollary}} == Proof == Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ putting $n = 3$.  == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Square of Difference|next = Cube of Difference}}: $\S 2$: Special Products and Factors: $2.3$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Square of Sum/Algebraic Proof 2|next = Fourth Power of Sum}}: $\S 20$: Binomial Series: $20.6$ * {{BookReference|Elementary Number Theory|1980|David M. Burton|ed = revised|edpage = Revised Printing|prev = Square of Sum|next = Binomial Theorem/Examples/4th Power of Sum}}: Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem
Taylor Expansion for Polynomials/Order 1
Taylor Expansion for Polynomials/Order 1 0 54559 288840 2017-03-15T22:09:30Z Barto 3079 Created page with "== Theorem ==  Let $R$ be a commutative ring with unity.  Let $f(X)\in R[X]$ be a Definition:Polynomial (Abstract Al..." wikitext text/x-wiki == Theorem ==  Let $R$ be a commutative ring with unity. Let $f(X)\in R[X]$ be a polynomial. Let $a\in R$. Then there exists a polynomial $g(X)\in R[X]$ such that: :$f(X+a) = f(X) + a f'(X) +a^2 g(X)$ where $f'$ denotes the formal derivative of $f$.  == Proof == By linearity, it suffices to prove this for $f(X)=X^n$. This is now a direct consequence of the Binomial Theorem.
Smallest Set of Weights for One-Pan Balance
== Classic Problem == Consider a set of balance scales for determining the weight of a physical object. Let this set of scales be such that weights may be placed in one of the pans. What is the smallest set of weights needed to weigh any given integer weight up to a given amount? == Solution == A set of $m$ weights in the sequence $\sequence {2^n}$: :$1, 2, 4, 8, 16, \ldots$ allows one to weigh any given integer weight up to $2^m - 1$. == Proof == This is equivalent to the statement that every positive integer can be expressed uniquely in binary notation. This in turn is an application of the Basis Representation Theorem.  == Examples == {{:Smallest Set of Weights for One-Pan Balance/Examples}} == Also see == * Smallest Set of Weights for Two-Pan Balance == Historical Note == {{:Smallest Set of Weights for One-Pan Balance/Historical Note}} == Sources == * {{BookReference|Mathematical Recreations and Essays|1974|W.W. Rouse Ball|author2 = H.S.M. Coxeter|ed = 12th|edpage = Twelfth Edition}} * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 18,446,744,073,709,551,615|next = Smallest Set of Weights for One-Pan Balance/Historical Note}}: $31$ * {{BookReference|Curious and Interesting Puzzles|1992|David Wells|prev = Smallest Set of Weights for One-Pan Balance/Historical Note|next = Smallest Set of Weights for Two-Pan Balance}}: Bachet: $108$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 18,446,744,073,709,551,615|next = Smallest Set of Weights for One-Pan Balance/Historical Note}}: $31$
Smallest Set of Weights for Two-Pan Balance
== Classic Problem == Consider a set of balance scales for determining the weight of a physical object. Let this set of scales be such that weights may be placed in either of the two pans. What is the smallest set of weights needed to weigh any given integer weight up to a given amount? == Solution == A set of weights up to $3^m$ in the sequence $\sequence {3^n}$: :$1, 3, 9, 27, \ldots$ allows one to weigh any given integer weight up to $\dfrac {3^{m + 1} - 1} 2$. == Proof == Place the item to be weighed in the left hand pan of the balance. Let it weigh $n$. Let $n$ be expressed in balanced ternary representation. From Representation of Integers in Balanced Ternary, $n$ can be uniquely so represented. With $m$ digits, we can count up to $\dfrac {3^{m + 1} - 1} 2$. We use the balanced ternary representation to model how to place the weights. Let the digits of $n$ in such a representation be numbered $0, 1, \ldots, m$ from the least significant digit to the most significant digit. The $k$th digit represents the weight which weighs $3^k$. When the $k$th digit is $1$, place weight $3^k$ in the right hand pan. When the $k$th digit is $\underline 1$, place weight $3^k$ in the left hand pan. When the $k$th digit is $$, place weight $3^k$ in neither pan. This will make the scales balance.  == Examples == {{:Smallest Set of Weights for Two-Pan Balance/Examples}} == Also see == * Smallest Set of Weights for One-Pan Balance == Historical Note == {{:Smallest Set of Weights for Two-Pan Balance/Historical Note}} == Sources == * {{BookReference|Problèmes Plaisans et Delectables qui se font par les Nombres|1612|Claude-Gaspar Bachet}} * {{BookReference|Mathematical Recreations and Essays|1974|W.W. Rouse Ball|author2 = H.S.M. Coxeter|ed = 12th|edpage = Twelfth Edition}} * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Smallest Set of Weights for One-Pan Balance/Historical Note|next = Smallest Set of Weights for Two-Pan Balance/Historical Note}}: $31$ * {{BookReference|Curious and Interesting Puzzles|1992|David Wells|prev = Smallest Set of Weights for One-Pan Balance|next = Book:Claudio Gaspare Bacheto/Diophanti Alexandrini Arithmeticorum}}: Bachet: $108$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Smallest Set of Weights for One-Pan Balance/Historical Note|next = Smallest Set of Weights for Two-Pan Balance/Historical Note}}: $31$