Misplaced Pages

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76-611: This method is also known as the cyclic method and contains traces of mathematical induction . Chakra in Sanskrit means cycle. As per popular legend, Chakravala indicates a mythical range of mountains which orbits around the Earth like a wall and not limited by light and darkness. Brahmagupta in 628 CE studied indeterminate quadratic equations, including Pell's equation for minimum integers x and y . Brahmagupta could solve it for several N , but not all. Jayadeva and Bhaskara offered

152-624: A {\displaystyle a} is odd, x = a 2 ( a 2 − 3 ) , y = b 2 ( a 2 − 1 ) {\displaystyle x={\frac {a}{2}}(a^{2}-3),y={\frac {b}{2}}(a^{2}-1)} When k = − 4 {\displaystyle k=-4} , then ( a 2 ) 2 − N ( b 2 ) 2 = − 1 {\displaystyle ({\frac {a}{2}})^{2}-N({\frac {b}{2}})^{2}=-1} . Composing with itself yields (

228-441: A 2 − 1 , y = 2 a b {\displaystyle x=2a^{2}-1,y=2ab} Again using the equation, ( 2 a 2 − k ) 2 − N ( 2 a b ) 2 = k 2 {\displaystyle (2a^{2}-k)^{2}-N(2ab)^{2}=k^{2}} ⇒ {\displaystyle \Rightarrow } ( 2

304-500: A 2 − 2 2 , a b 2 , 1 ) {\displaystyle ({\frac {a^{2}-2}{2}},{\frac {ab}{2}},1)} . Composing the triples gives ( a 2 ( a 2 − 3 ) ) 2 − N ( b 2 ( a 2 − 1 ) ) 2 = 1 {\displaystyle ({\frac {a}{2}}(a^{2}-3))^{2}-N({\frac {b}{2}}(a^{2}-1))^{2}=1} When

380-404: A 2 − 4 4 ) 2 − N ( 2 a b 4 ) 2 = 1 {\displaystyle ({\frac {2a^{2}-4}{4}})^{2}-N({\frac {2ab}{4}})^{2}=1} . Leading to the triples ( a 2 , b 2 , 1 ) {\displaystyle ({\frac {a}{2}},{\frac {b}{2}},1)} and (

456-572: A 2 − 61 b 2 = 1 {\displaystyle a^{2}-61b^{2}=1} ), issued as a challenge by Fermat many centuries later, was given by Bhaskara as an example. We start with a solution a 2 − 61 b 2 = k {\displaystyle a^{2}-61b^{2}=k} for any k found by any means. In this case we can let b be 1, thus, since 8 2 − 61 ⋅ 1 2 = 3 {\displaystyle 8^{2}-61\cdot 1^{2}=3} , we have

532-475: A 2 − k ) 2 − N ( 2 a b ) 2 = k 2 {\displaystyle (2a^{2}-k)^{2}-N(2ab)^{2}=k^{2}} The new triple can be expressed as ( 2 a 2 − k , 2 a b , k 2 ) {\displaystyle (2a^{2}-k,2ab,k^{2})} . Substituting k = − 1 {\displaystyle k=-1} gives

608-398: A 2 − k k ) 2 − N ( 2 a b k ) 2 = 1 {\displaystyle \left({\frac {2a^{2}-k}{k}}\right)^{2}-N\left({\frac {2ab}{k}}\right)^{2}=1} Substituting k = 2 {\displaystyle k=2} , x = a 2 − 1 , y =

684-482: A 2 + 2 ) ( a 4 + 4 a 2 + 1 ) 2 ) 2 − N ( a b ( a 2 + 3 ) ( a 2 + 1 ) 2 ) 2 = 1 {\displaystyle ({\frac {(a^{2}+2)(a^{4}+4a^{2}+1)}{2}})^{2}-N({\frac {ab(a^{2}+3)(a^{2}+1)}{2}})^{2}=1} ⇒ {\displaystyle \Rightarrow } ( (

760-575: A 2 + 2 ) ( a 4 + 4 a 2 + 2 ) + N a 2 b 2 ( a 2 + 2 ) 4 ) 2 − N ( a b ( a 4 + 4 a 2 + 3 ) 2 ) 2 = 1 {\displaystyle ({\frac {(a^{2}+2)(a^{4}+4a^{2}+2)+Na^{2}b^{2}(a^{2}+2)}{4}})^{2}-N({\frac {ab(a^{4}+4a^{2}+3)}{2}})^{2}=1} ⇒ {\displaystyle \Rightarrow } ( (

836-462: A 2 + 2 ) [ ( a 2 + 1 ) ( a 2 + 3 ) − 2 ) ] 2 ) 2 − N ( a b ( a 2 + 3 ) ( a 2 + 1 ) 2 ) 2 = 1 {\displaystyle ({\frac {(a^{2}+2)[(a^{2}+1)(a^{2}+3)-2)]}{2}})^{2}-N({\frac {ab(a^{2}+3)(a^{2}+1)}{2}})^{2}=1} . This give us

SECTION 10

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912-397: A 2 + N b 2 4 ) 2 − N ( a b 2 ) 2 = 1 {\displaystyle ({\frac {a^{2}+Nb^{2}}{4}})^{2}-N({\frac {ab}{2}})^{2}=1} ⇒ {\displaystyle \Rightarrow } ( a 2 + 2 2 ) 2 − N (

988-581: A 2 = N b 2 + k {\displaystyle a^{2}=Nb^{2}+k} , when k is ±1, ±2, or ±4. Using Brahmagupta's identity to compose the triple ( a , b , k ) {\displaystyle (a,b,k)} with itself: ( a 2 + N b 2 ) 2 − N ( 2 a b ) 2 = k 2 {\displaystyle (a^{2}+Nb^{2})^{2}-N(2ab)^{2}=k^{2}} ⇒ {\displaystyle \Rightarrow } ( 2

1064-416: A 4 + 4 a 2 + 2 2 ) 2 − N ( a b ( a 2 + 2 ) 2 ) 2 = 1 {\displaystyle ({\frac {a^{4}+4a^{2}+2}{2}})^{2}-N({\frac {ab(a^{2}+2)}{2}})^{2}=1} Finally, from the earlier equations, compose the triples ( a 2 + 2 2 ,

1140-431: A , b , k ) = ( 8 , 1 , − 3 ) {\displaystyle (a,b,k)=(8,1,-3)} . We want a positive integer m such that k divides a  +  bm , i.e. 3 divides 8 + m, and | m  − 67| is minimal. The first condition implies that m is of the form 3 t + 1 (i.e. 1, 4, 7, 10,… etc.), and among such m , the minimal value is attained for m = 7. Replacing (

1216-473: A b {\displaystyle x=a^{2}-1,y=ab} Substituting k = − 2 {\displaystyle k=-2} , x = a 2 + 1 , y = a b {\displaystyle x=a^{2}+1,y=ab} Substituting k = 4 {\displaystyle k=4} into the equation ( 2 a 2 − 4 4 ) 2 − N ( 2

1292-615: A b 2 ) 2 = 1 {\displaystyle ({\frac {a^{2}+2}{2}})^{2}-N({\frac {ab}{2}})^{2}=1} . Again composing itself yields ( ( a 2 + 2 ) 2 + N a 2 b 2 ) 4 ) 2 − N ( a b ( a 2 + 2 ) 2 ) 2 = 1 {\displaystyle ({\frac {(a^{2}+2)^{2}+Na^{2}b^{2})}{4}})^{2}-N({\frac {ab(a^{2}+2)}{2}})^{2}=1} ⇒ {\displaystyle \Rightarrow } (

1368-401: A b 2 , 1 ) {\displaystyle ({\frac {a^{2}+2}{2}},{\frac {ab}{2}},1)} and ( a 4 + 4 a 2 + 2 2 , a b ( a 2 + 2 ) 2 , 1 ) {\displaystyle ({\frac {a^{4}+4a^{2}+2}{2}},{\frac {ab(a^{2}+2)}{2}},1)} , to get ( (

1444-398: A b 4 ) 2 = 1 {\displaystyle ({\frac {2a^{2}-4}{4}})^{2}-N({\frac {2ab}{4}})^{2}=1} creates the triple ( a 2 − 2 2 , a b 2 , 1 ) {\displaystyle ({\frac {a^{2}-2}{2}},{\frac {ab}{2}},1)} . Which is a solution if a {\displaystyle a}

1520-461: A  +  bm , and | m  − 67| is minimal. We then update a , b , and k to a m + N b | k | , a + b m | k | {\displaystyle {\frac {am+Nb}{|k|}},{\frac {a+bm}{|k|}}} and m 2 − N k {\displaystyle {\frac {m^{2}-N}{k}}} respectively. We have (

1596-781: A ,  b ,  k ) with ( a m + N b | k | , a + b m | k | , m 2 − N k ) {\displaystyle \left({\frac {am+Nb}{|k|}},{\frac {a+bm}{|k|}},{\frac {m^{2}-N}{k}}\right)} , we get the new values a = ( 8 ⋅ 7 + 67 ⋅ 1 ) / 3 = 41 , b = ( 8 + 1 ⋅ 7 ) / 3 = 5 , k = ( 7 2 − 67 ) / ( − 3 ) = 6 {\displaystyle a=(8\cdot 7+67\cdot 1)/3=41,b=(8+1\cdot 7)/3=5,k=(7^{2}-67)/(-3)=6} . That is, we have

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1672-503: A combination of such coins. Let S ( k ) denote the statement " k dollars can be formed by a combination of 4- and 5-dollar coins". The proof that S ( k ) is true for all k ≥ 12 can then be achieved by induction on k as follows: Base case: Showing that S ( k ) holds for k = 12 is simple: take three 4-dollar coins. Induction step: Given that S ( k ) holds for some value of k ≥ 12 ( induction hypothesis ), prove that S ( k + 1) holds, too. Assume S ( k )

1748-415: A finite chain of deductive reasoning involving the variable n {\displaystyle n} , which can take infinitely many values. The result is a rigorous proof of the statement, not an assertion of its probability. In 370 BC, Plato 's Parmenides may have contained traces of an early example of an implicit inductive proof, however, the earliest implicit proof by mathematical induction

1824-447: A ladder: Mathematical induction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the basis ) and that from each rung we can climb up to the next one (the step ). A proof by induction consists of two cases. The first, the base case , proves the statement for n = 0 {\displaystyle n=0} without assuming any knowledge of other cases. The second case,

1900-587: A more general version, | sin ⁡ n x | ≤ n | sin ⁡ x | {\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|} for any real numbers n , x {\displaystyle n,x} , could be proven without induction; but the case n = 1 2 , x = π {\textstyle n={\frac {1}{2}},\,x=\pi } shows it may be false for non-integer values of n {\displaystyle n} . This suggests we examine

1976-467: A new triple In the general method, the main idea is that any triple ( a , b , k ) {\displaystyle (a,b,k)} (that is, one which satisfies a 2 − N b 2 = k {\displaystyle a^{2}-Nb^{2}=k} ) can be composed with the trivial triple ( m , 1 , m 2 − N ) {\displaystyle (m,1,m^{2}-N)} to get

2052-433: A solution a 2 − 67 b 2 = k {\displaystyle a^{2}-67b^{2}=k} for any k found by any means; in this case we can let b be 1, thus producing 8 2 − 67 ⋅ 1 2 = − 3 {\displaystyle 8^{2}-67\cdot 1^{2}=-3} . At each step, we find an m  > 0 such that k divides

2128-427: A solution: x = 2 a 2 + 1 , y = 2 a b {\displaystyle x=2a^{2}+1,y=2ab} For k = 1 {\displaystyle k=1} , the original ( a , b ) {\displaystyle (a,b)} was already a solution. Substituting k = 1 {\displaystyle k=1} yields a second: x = 2

2204-610: Is −4, we can use Brahmagupta's idea: it can be scaled down to the rational solution ( 39 / 2 , 5 / 2 , − 1 ) {\displaystyle (39/2,5/2,-1)\,} , which composed with itself three times, with m = 7 , 11 , 9 {\displaystyle m={7,11,9}} respectively, when k becomes square and scaling can be applied, this gives ( 1523 / 2 , 195 / 2 , 1 ) {\displaystyle (1523/2,195/2,1)\,} . Finally, such procedure can be repeated until

2280-408: Is actually a special case of the previous form, because if the statement to be proved is P ( n ) then proving it with these two rules is equivalent with proving P ( n + b ) for all natural numbers n with an induction base case 0 . Assume an infinite supply of 4- and 5-dollar coins. Induction can be used to prove that any whole amount of dollars greater than or equal to 12 can be formed by

2356-415: Is an inference rule used in formal proofs , and is the foundation of most correctness proofs for computer programs. Despite its name, mathematical induction differs fundamentally from inductive reasoning as used in philosophy , in which the examination of many cases results in a probable conclusion. The mathematical method examines infinitely many cases to prove a general statement, but it does so by

Chakravala method - Misplaced Pages Continue

2432-867: Is an identity used as a lemma during the chakravala method . It states that: for integers m , x , y , N , {\displaystyle m,\,x,\,y,\,N,} and non-zero integer k {\displaystyle k} . The proof follows from simple algebraic manipulations as follows: multiply both sides of the equation by m 2 − N {\displaystyle m^{2}-N} , add N 2 x 2 + 2 N m x y + N y 2 {\displaystyle N^{2}x^{2}+2Nmxy+Ny^{2}} , factor, and divide by k 2 {\displaystyle k^{2}} . So long as neither k {\displaystyle k} nor m 2 − N {\displaystyle m^{2}-N} are zero,

2508-457: Is even: x = a 2 − 2 2 , y = a b 2 {\displaystyle x={\frac {a^{2}-2}{2}},y={\frac {ab}{2}}} If a is odd, start with the equations ( a 2 ) 2 − N ( b 2 ) 2 = 1 {\displaystyle ({\frac {a}{2}})^{2}-N({\frac {b}{2}})^{2}=1} and ( 2

2584-459: Is of the form 6 t  + 5 (i.e. 5, 11, 17,… etc.), and among such m , | m  − 67| is minimal for m  = 5. This leads to the new solution a  = (41⋅5 + 67⋅5)/6, etc.: For 7 to divide 90 + 11 m , we must have m = 2 + 7 t (i.e. 2, 9, 16,… etc.) and among such m , we pick m = 9. At this point, we could continue with the cyclic method (and it would end, after seven iterations), but since

2660-470: Is scaled down (or Bhaskara's lemma is directly used) to get: For 3 to divide 8 + m {\displaystyle 8+m} and | m 2 − 61 | {\displaystyle |m^{2}-61|} to be minimal, we choose m = 7 {\displaystyle m=7} , so that we have the triple ( 39 , 5 , − 4 ) {\displaystyle (39,5,-4)} . Now that k

2736-492: Is the earliest extant proof of the sum formula for integral cubes . In India, early implicit proofs by mathematical induction appear in Bhaskara 's " cyclic method ". None of these ancient mathematicians, however, explicitly stated the induction hypothesis. Another similar case (contrary to what Vacca has written, as Freudenthal carefully showed) was that of Francesco Maurolico in his Arithmeticorum libri duo (1575), who used

2812-409: Is true for some arbitrary k ≥ 12 . If there is a solution for k dollars that includes at least one 4-dollar coin, replace it by a 5-dollar coin to make k + 1 dollars. Otherwise, if only 5-dollar coins are used, k must be a multiple of 5 and so at least 15; but then we can replace three 5-dollar coins by four 4-dollar coins to make k + 1 dollars. In each case, S ( k + 1)

2888-414: Is true for some natural number n , it also holds for some strictly smaller natural number m . Because there are no infinite decreasing sequences of natural numbers, this situation would be impossible, thereby showing ( by contradiction ) that Q ( n ) cannot be true for any n . The validity of this method can be verified from the usual principle of mathematical induction. Using mathematical induction on

2964-421: Is true. Therefore, by the principle of induction, S ( k ) holds for all k ≥ 12 , and the proof is complete. In this example, although S ( k ) also holds for k ∈ { 4 , 5 , 8 , 9 , 10 } {\textstyle k\in \{4,5,8,9,10\}} , the above proof cannot be modified to replace the minimum amount of 12 dollar to any lower value m . For m = 11 ,

3040-1832: Is true. Using the angle addition formula and the triangle inequality , we deduce: | sin ⁡ ( k + 1 ) x | = | sin ⁡ k x cos ⁡ x + sin ⁡ x cos ⁡ k x | (angle addition) ≤ | sin ⁡ k x cos ⁡ x | + | sin ⁡ x cos ⁡ k x | (triangle inequality) = | sin ⁡ k x | | cos ⁡ x | + | sin ⁡ x | | cos ⁡ k x | ≤ | sin ⁡ k x | + | sin ⁡ x | ( | cos ⁡ t | ≤ 1 ) ≤ k | sin ⁡ x | + | sin ⁡ x | (induction hypothesis ) = ( k + 1 ) | sin ⁡ x | . {\displaystyle {\begin{aligned}\left|\sin(k+1)x\right|&=\left|\sin kx\cos x+\sin x\cos kx\right|&&{\text{(angle addition)}}\\&\leq \left|\sin kx\cos x\right|+\left|\sin x\,\cos kx\right|&&{\text{(triangle inequality)}}\\&=\left|\sin kx\right|\left|\cos x\right|+\left|\sin x\right|\left|\cos kx\right|\\&\leq \left|\sin kx\right|+\left|\sin x\right|&&(\left|\cos t\right|\leq 1)\\&\leq k\left|\sin x\right|+\left|\sin x\right|&&{\text{(induction hypothesis}})\\&=(k+1)\left|\sin x\right|.\end{aligned}}} The inequality between

3116-844: Is useful to find a solution to Pell's Equation , but it is not always the smallest integer pair. e.g. 36 2 − 52 ∗ 5 2 = − 4 {\displaystyle 36^{2}-52*5^{2}=-4} . The equation will give you x = 1093436498 , y = 151632270 {\displaystyle x=1093436498,y=151632270} , which when put into Pell's Equation yields 1195601955878350801 − 1195601955878350800 = 1 {\displaystyle 1195601955878350801-1195601955878350800=1} , which works, but so does x = 649 , y = 90 {\displaystyle x=649,y=90} for N = 52 {\displaystyle N=52} . The n  = 61 case (determining an integer solution satisfying

Chakravala method - Misplaced Pages Continue

3192-419: The implication P ( k ) ⟹ P ( k + 1 ) {\displaystyle P(k)\implies P(k+1)} for any natural number k {\displaystyle k} . Assume the induction hypothesis: for a given value n = k ≥ 0 {\displaystyle n=k\geq 0} , the single case P ( k ) {\displaystyle P(k)}

3268-667: The induction step , proves that if the statement holds for any given case n = k {\displaystyle n=k} , then it must also hold for the next case n = k + 1 {\displaystyle n=k+1} . These two steps establish that the statement holds for every natural number n {\displaystyle n} . The base case does not necessarily begin with n = 0 {\displaystyle n=0} , but often with n = 1 {\displaystyle n=1} , and possibly with any fixed natural number n = N {\displaystyle n=N} , establishing

3344-412: The proof of commutativity accompanying addition of natural numbers . More complicated arguments involving three or more counters are also possible. The method of infinite descent is a variation of mathematical induction which was used by Pierre de Fermat . It is used to show that some statement Q ( n ) is false for all natural numbers n . Its traditional form consists of showing that if Q ( n )

3420-456: The base case is actually false; for m = 10 , the second case in the induction step (replacing three 5- by four 4-dollar coins) will not work; let alone for even lower m . It is sometimes desirable to prove a statement involving two natural numbers, n and m , by iterating the induction process. That is, one proves a base case and an induction step for n , and in each of those proves a base case and an induction step for m . See, for example,

3496-646: The chosen value. This results in a new triple ( a , b , k ). The process is repeated until a triple with k = 1 {\displaystyle k=1} is found. This method always terminates with a solution (proved by Lagrange in 1768). Optionally, we can stop when k is ±1, ±2, or ±4, as Brahmagupta's approach gives a solution for those cases. In AD 628, Brahmagupta discovered a general way to find x {\displaystyle x} and y {\displaystyle y} of x 2 = N y 2 + 1 , {\displaystyle x^{2}=Ny^{2}+1,} when given

3572-471: The equation x 2 − N y 2 = k {\displaystyle x^{2}-Ny^{2}=k} , this allows the "composition" ( samāsa ) of two solution triples ( x 1 , y 1 , k 1 ) {\displaystyle (x_{1},y_{1},k_{1})} and ( x 2 , y 2 , k 2 ) {\displaystyle (x_{2},y_{2},k_{2})} into

3648-554: The exact nature of the property to be proven. All variants of induction are special cases of transfinite induction ; see below . If one wishes to prove a statement, not for all natural numbers, but only for all numbers n greater than or equal to a certain number b , then the proof by induction consists of the following: This can be used, for example, to show that 2 ≥ n + 5 for n ≥ 3 . In this way, one can prove that some statement P ( n ) holds for all n ≥ 1 , or even for all n ≥ −5 . This form of mathematical induction

3724-402: The extreme left hand and right hand sides, we deduce that: 0 + 1 + 2 + ⋯ + k + ( k + 1 ) = ( k + 1 ) ( ( k + 1 ) + 1 ) 2 . {\displaystyle 0+1+2+\cdots +k+(k+1)={\frac {(k+1)((k+1)+1)}{2}}.} That is, the statement P ( k + 1) also holds true, establishing

3800-454: The extreme left-hand and right-hand quantities shows that P ( k + 1 ) {\displaystyle P(k+1)} is true, which completes the induction step. Conclusion: The proposition P ( n ) {\displaystyle P(n)} holds for all natural numbers n . {\displaystyle n.}   Q.E.D. In practice, proofs by induction are often structured differently, depending on

3876-454: The first complete solution to the equation, using the chakravala method to find for x 2 = 61 y 2 + 1 , {\displaystyle \,x^{2}=61y^{2}+1,} the solution This case was notorious for its difficulty, and was first solved in Europe by Brouncker in 1657–58 in response to a challenge by Fermat , using continued fractions. A method for

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3952-1238: The following statement P ( n ) for all natural numbers n . P ( n ) :     0 + 1 + 2 + ⋯ + n = n ( n + 1 ) 2 . {\displaystyle P(n)\!:\ \ 0+1+2+\cdots +n={\frac {n(n+1)}{2}}.} This states a general formula for the sum of the natural numbers less than or equal to a given number; in fact an infinite sequence of statements: 0 = ( 0 ) ( 0 + 1 ) 2 {\displaystyle 0={\tfrac {(0)(0+1)}{2}}} , 0 + 1 = ( 1 ) ( 1 + 1 ) 2 {\displaystyle 0+1={\tfrac {(1)(1+1)}{2}}} , 0 + 1 + 2 = ( 2 ) ( 2 + 1 ) 2 {\displaystyle 0+1+2={\tfrac {(2)(2+1)}{2}}} , etc. Proposition. For every n ∈ N {\displaystyle n\in \mathbb {N} } , 0 + 1 + 2 + ⋯ + n = n ( n + 1 ) 2 . {\displaystyle 0+1+2+\cdots +n={\tfrac {n(n+1)}{2}}.} Proof. Let P ( n ) be

4028-459: The general problem was first completely described rigorously by Lagrange in 1766. Lagrange's method, however, requires the calculation of 21 successive convergents of the simple continued fraction for the square root of 61, while the chakravala method is much simpler. Selenius, in his assessment of the chakravala method, states Hermann Hankel calls the chakravala method From Brahmagupta's identity , we observe that for given N , For

4104-592: The induction hypothesis that for a particular k , the single case n = k holds, meaning P ( k ) is true: 0 + 1 + ⋯ + k = k ( k + 1 ) 2 . {\displaystyle 0+1+\cdots +k={\frac {k(k+1)}{2}}.} It follows that: ( 0 + 1 + 2 + ⋯ + k ) + ( k + 1 ) = k ( k + 1 ) 2 + ( k + 1 ) . {\displaystyle (0+1+2+\cdots +k)+(k+1)={\frac {k(k+1)}{2}}+(k+1).} Algebraically ,

4180-493: The induction step that ∀ k ( P ( k ) → P ( k + 1 ) ) {\displaystyle \forall k\,(P(k)\to P(k+1))} whereupon the induction principle "automates" n applications of this step in getting from P (0) to P ( n ) . This could be called "predecessor induction" because each step proves something about a number from something about that number's predecessor. Bhaskara%27s lemma Bhaskara's Lemma

4256-492: The induction step, that the statement holds for a particular n , is called the induction hypothesis or inductive hypothesis . To prove the induction step, one assumes the induction hypothesis for n and then uses this assumption to prove that the statement holds for n + 1 . Authors who prefer to define natural numbers to begin at 0 use that value in the base case; those who define natural numbers to begin at 1 use that value. Mathematical induction can be used to prove

4332-651: The induction step. Conclusion: Since both the base case and the induction step have been proved as true, by mathematical induction the statement P ( n ) holds for every natural number n . Q.E.D. Induction is often used to prove inequalities . As an example, we prove that | sin ⁡ n x | ≤ n | sin ⁡ x | {\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|} for any real number x {\displaystyle x} and natural number n {\displaystyle n} . At first glance, it may appear that

4408-459: The infinitely many cases P ( 0 ) , P ( 1 ) , P ( 2 ) , P ( 3 ) , … {\displaystyle P(0),P(1),P(2),P(3),\dots }   all hold. This is done by first proving a simple case, then also showing that if we assume the claim is true for a given case, then the next case is also true. Informal metaphors help to explain this technique, such as falling dominoes or climbing

4484-442: The new solution: At this point, one round of the cyclic algorithm is complete. We now repeat the process. We have ( a , b , k ) = ( 41 , 5 , 6 ) {\displaystyle (a,b,k)=(41,5,6)} . We want an m  > 0 such that k divides a  +  bm , i.e. 6 divides 41 + 5 m , and | m  − 67| is minimal. The first condition implies that m

4560-413: The new triple ( a m + N b , a + b m , k ( m 2 − N ) ) {\displaystyle (am+Nb,a+bm,k(m^{2}-N))} for any m . Assuming we started with a triple for which gcd ( a , b ) = 1 {\displaystyle \gcd(a,b)=1} , this can be scaled down by k (this is Bhaskara's lemma ): Since

4636-623: The right hand side simplifies as: k ( k + 1 ) 2 + ( k + 1 ) = k ( k + 1 ) + 2 ( k + 1 ) 2 = ( k + 1 ) ( k + 2 ) 2 = ( k + 1 ) ( ( k + 1 ) + 1 ) 2 . {\displaystyle {\begin{aligned}{\frac {k(k+1)}{2}}+(k+1)&={\frac {k(k+1)+2(k+1)}{2}}\\&={\frac {(k+1)(k+2)}{2}}\\&={\frac {(k+1)((k+1)+1)}{2}}.\end{aligned}}} Equating

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4712-419: The right-hand side is among ±1, ±2, ±4, we can also use Brahmagupta's observation directly. Composing the triple (221, 27, −2) with itself, we get Mathematical induction Mathematical induction is a method for proving that a statement P ( n ) {\displaystyle P(n)} is true for every natural number n {\displaystyle n} , that is, that

4788-433: The signs inside the squares do not matter, the following substitutions are possible: When a positive integer m is chosen so that ( a  +  bm )/ k is an integer, so are the other two numbers in the triple. Among such m , the method chooses one that minimizes the absolute value of m  −  N and hence that of ( m  −  N )/ k . Then the substitution relations are applied for m equal to

4864-440: The solution is found (requiring 9 additional self-compositions and 4 additional square-scalings): ( 1766319049 , 226153980 , 1 ) {\displaystyle (1766319049,\,226153980,\,1)} . This is the minimal integer solution. Suppose we are to solve x 2 − 67 y 2 = 1 {\displaystyle x^{2}-67y^{2}=1} for x and y . We start with

4940-497: The solutions x = ( a 2 + 2 ) [ ( a 2 + 1 ) ( a 2 + 3 ) − 2 ) ] 2 y = a b ( a 2 + 3 ) ( a 2 + 1 ) 2 {\displaystyle x={\frac {(a^{2}+2)[(a^{2}+1)(a^{2}+3)-2)]}{2}}y={\frac {ab(a^{2}+3)(a^{2}+1)}{2}}} (Note, k = − 4 {\displaystyle k=-4}

5016-619: The statement | sin ⁡ n x | ≤ n | sin ⁡ x | {\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|} . We induce on n {\displaystyle n} . Base case: The calculation | sin ⁡ 0 x | = 0 ≤ 0 = 0 | sin ⁡ x | {\displaystyle \left|\sin 0x\right|=0\leq 0=0\left|\sin x\right|} verifies P ( 0 ) {\displaystyle P(0)} . Induction step: We show

5092-609: The statement 0 + 1 + 2 + ⋯ + n = n ( n + 1 ) 2 . {\displaystyle 0+1+2+\cdots +n={\tfrac {n(n+1)}{2}}.} We give a proof by induction on n . Base case: Show that the statement holds for the smallest natural number n = 0 . P (0) is clearly true: 0 = 0 ( 0 + 1 ) 2 . {\displaystyle 0={\tfrac {0(0+1)}{2}}\,.} Induction step: Show that for every k ≥ 0 , if P ( k ) holds, then P ( k + 1) also holds. Assume

5168-467: The statement P ( n ) defined as " Q ( m ) is false for all natural numbers m less than or equal to n ", it follows that P ( n ) holds for all n , which means that Q ( n ) is false for every natural number n . If one wishes to prove that a property P holds for all natural numbers less than or equal to n , proving P satisfies the following conditions suffices: The most common form of proof by mathematical induction requires proving in

5244-690: The statement specifically for natural values of n {\displaystyle n} , and induction is the readiest tool. Proposition. For any x ∈ R {\displaystyle x\in \mathbb {R} } and n ∈ N {\displaystyle n\in \mathbb {N} } , | sin ⁡ n x | ≤ n | sin ⁡ x | {\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|} . Proof. Fix an arbitrary real number x {\displaystyle x} , and let P ( n ) {\displaystyle P(n)} be

5320-423: The technique to prove that the sum of the first n odd integers is n . The earliest rigorous use of induction was by Gersonides (1288–1344). The first explicit formulation of the principle of induction was given by Pascal in his Traité du triangle arithmétique (1665). Another Frenchman, Fermat , made ample use of a related principle: indirect proof by infinite descent . The induction hypothesis

5396-475: The triple ( a , b , k ) = ( 8 , 1 , 3 ) {\displaystyle (a,b,k)=(8,1,3)} . Composing it with ( m , 1 , m 2 − 61 ) {\displaystyle (m,1,m^{2}-61)} gives the triple ( 8 m + 61 , 8 + m , 3 ( m 2 − 61 ) ) {\displaystyle (8m+61,8+m,3(m^{2}-61))} , which

5472-447: The truth of the statement for all natural numbers n ≥ N {\displaystyle n\geq N} . The method can be extended to prove statements about more general well-founded structures, such as trees ; this generalization, known as structural induction , is used in mathematical logic and computer science . Mathematical induction in this extended sense is closely related to recursion . Mathematical induction

5548-419: The two basic components of a modern argument by induction, namely the truth of the statement for n = 1 (1 = 1 ) and the deriving of the truth for n = k from that of n = k - 1. Of course, this second component is not explicit since, in some sense, al-Karaji's argument is in reverse; this is, he starts from n = 10 and goes down to 1 rather than proceeding upward. Nevertheless, his argument in al-Fakhri

5624-581: Was also employed by the Swiss Jakob Bernoulli , and from then on it became well known. The modern formal treatment of the principle came only in the 19th century, with George Boole , Augustus De Morgan , Charles Sanders Peirce , Giuseppe Peano , and Richard Dedekind . The simplest and most common form of mathematical induction infers that a statement involving a natural number n (that is, an integer n ≥ 0 or 1) holds for all values of n . The proof consists of two steps: The hypothesis in

5700-462: Was that of an inductive argument for dealing with certain arithmetic sequences. Thus al-Karaji used such an argument to prove the result on the sums of integral cubes already known to Aryabhata [...] Al-Karaji did not, however, state a general result for arbitrary n . He stated his theorem for the particular integer 10 [...] His proof, nevertheless, was clearly designed to be extendable to any other integer. [...] Al-Karaji's argument includes in essence

5776-511: Was written by al-Karaji around 1000 AD, who applied it to arithmetic sequences to prove the binomial theorem and properties of Pascal's triangle . Whilst the original work was lost, it was later referenced by Al-Samawal al-Maghribi in his treatise al-Bahir fi'l-jabr (The Brilliant in Algebra) in around 1150 AD. Katz says in his history of mathematics Another important idea introduced by al-Karaji and continued by al-Samaw'al and others

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