{"history":[],"article":{"id":"fd49b37c-d09e-4592-ac46-ea093822c06f","title":"Proof that 22/7 exceeds π","slug":"proof-that-22-7-exceeds","content":":::summary\n:::\n\n# \n[Proofs](https://bliptext.com/articles/mathematical-proof) of the mathematical result that the [rational number](https://bliptext.com/articles/rational-number) 22⁄7 is greater than [π](https://bliptext.com/articles/pi) (pi) date back to antiquity.\nOne of these proofs, more recently developed but requiring only elementary techniques from calculus, has attracted attention in modern mathematics due to its [mathematical elegance](https://bliptext.com/articles/mathematical-elegance) and its connections to the theory of [Diophantine approximations](https://bliptext.com/articles/diophantine-approximations).\nStephen Lucas calls this proof \"one of the more beautiful results related to approximating π\".\nJulian Havil ends a discussion of [continued fraction](https://bliptext.com/articles/continued-fraction) approximations of π with the result, describing it as \"impossible to resist mentioning\" in that context.\nThe purpose of the proof is not primarily to convince its readers that 22⁄7 (or 3 1⁄7) is indeed bigger than π; systematic methods of computing the value of π exist.\nIf one knows that π is approximately 3.14159, then it trivially follows that π < 22⁄7, which is approximately 3.142857.\nBut it takes much less work to show that π < 22⁄7 by the method used in this proof than to show that π is approximately 3.14159.\n# [This is not a perfect 22/7 circle, because 22/7 is not a perfect representation of pi. | upload.wikimedia.org/wikipedia/commons/9/92/22///Divided///by///7///Circle.png]\n\n# Background\n22⁄7 is a widely used [Diophantine approximation](https://bliptext.com/articles/diophantine-approximation) of π.\nIt is a convergent in the simple [continued fraction](https://bliptext.com/articles/continued-fraction) expansion of π.\nIt is greater than π, as can be readily seen in the [decimal expansions](https://bliptext.com/articles/decimal-expansion) of these values:\nThe approximation has been known since antiquity.\n[Archimedes](https://bliptext.com/articles/archimedes) wrote the first known proof that 22⁄7 is an overestimate in the 3rd century BCE, although he may not have been the first to use that approximation.\nHis proof proceeds by showing that 22⁄7 is greater than the ratio of the [perimeter](https://bliptext.com/articles/perimeter) of a [regular polygon](https://bliptext.com/articles/regular-polygon) with 96 sides to the diameter of a circle it circumscribes.\n\n# Proof\nThe proof can be expressed very succinctly:\nTherefore, 22⁄7 > π.\nThe evaluation of this integral was the first problem in the 1968 [Putnam Competition](https://bliptext.com/articles/william-lowell-putnam-mathematical-competition).\nIt is easier than most Putnam Competition problems, but the competition often features seemingly obscure problems that turn out to refer to something very familiar.\nThis integral has also been used in the entrance examinations for the [Indian Institutes of Technology](https://bliptext.com/articles/indian-institutes-of-technology).\n\n# Details of evaluation of the integral\nThat the [integral](https://bliptext.com/articles/integral) is positive follows from the fact that the [integrand](https://bliptext.com/articles/integrand) is [non-negative](https://bliptext.com/articles/non-negative); the denominator is positive and the numerator is a product of nonnegative numbers.\nOne can also easily check that the integrand is strictly positive for at least one point in the range of integration, say at 1⁄2.\nSince the integrand is continuous at that point and nonnegative elsewhere, the integral from 0 to 1 must be strictly positive.\nIt remains to show that the integral in fact evaluates to the desired quantity:\n(See [polynomial long division](https://bliptext.com/articles/polynomial-long-division).)\n\n# Quick upper and lower bounds\nIn, it is pointed out that if 1 is substituted for\nx\nin the denominator, one gets a lower bound on the integral, and if 0 is substituted for\nx\nin the denominator, one gets an upper bound:\nThus we have\nhence 3.1412 < π < 3.1421 in decimal expansion.\nThe bounds deviate by less than 0.015% from π.\nSee also.\n\n# Proof that 355/113 exceeds π\nAs discussed in, the well-known Diophantine approximation and far better upper estimate [355⁄113](https://bliptext.com/articles/355-113) for π follows from the relation\nwhere the first six digits after the decimal point agree with those of π.\nSubstituting 1 for\nx\nin the denominator, we get the lower bound\nsubstituting 0 for\nx\nin the denominator, we get twice this value as an upper bound, hence\nIn decimal expansion, this means 3.141 592 57 < π < 3.141 592 74, where the bold digits of the lower and upper bound are those of π.\n\n# Extensions\nThe above ideas can be generalized to get better approximations of π; see also and (in both references, however, no calculations are given).\nFor explicit calculations, consider, for every integer\nn ≥ 1\n,\nwhere the middle integral evaluates to\ninvolving π.\nThe last sum also appears in [Leibniz' formula for π](https://bliptext.com/articles/leibniz-formula-for-pi).\nThe correction term and [error bound](https://bliptext.com/articles/error-bound) is given by\nwhere the approximation (the tilde means that the quotient of both sides tends to one for large\nn\n) of the [central binomial coefficient](https://bliptext.com/articles/central-binomial-coefficient) follows from [Stirling's formula](https://bliptext.com/articles/stirling-s-formula) and shows the fast convergence of the integrals to π.\nCalculation of these integrals: For all integers\nk ≥ 0\nand\nℓ ≥ 2\nwe have\nApplying this formula recursively\n2n\ntimes yields\nFurthermore,\nwhere the first equality holds, because the terms for\n1 ≤ j ≤ 3n – 1\ncancel, and the second equality arises from the index shift\nj → j + 1\nin the first sum.\nApplication of these two results gives\nFor integers\nk, ℓ ≥ 0\n, using [integration by parts](https://bliptext.com/articles/integration-by-parts)\nℓ\ntimes, we obtain\nSetting\n, we obtain\nIntegrating equation (1) from 0 to 1 using equation (2) and\n, we get the claimed equation involving π.\nThe results for\nare given above.\nFor\nwe get\nand\nhence 3.141 592 31 < π < 3.141 592 89, where the bold digits of the lower and upper bound are those of π.\nSimilarly for\n,\nwith correction term and error bound\nhence 3.141 592 653 40 < π < 3.141 592 653 87.\nThe next step for\nis\nwith\nwhich gives 3.141 592 653 589 55 < π < 3.141 592 653 589 96.\n\n# Footnotes\n\n## Citations","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T19:24:12.520Z","updated_at":"2025-01-25T19:24:12.520Z","search_vector":"'22/7':3 'exceed':4 'proof':1 'π':5","search_content":" \n \n\n# \n[Proofs](https://bliptext.com/articles/mathematical-proof) of the mathematical result that the [rational number](https://bliptext.com/articles/rational-number) 22⁄7 is greater than [π](https://bliptext.com/articles/pi) (pi) date back to antiquity.\nOne of these proofs, more recently developed but requiring only elementary techniques from calculus, has attracted attention in modern mathematics due to its [mathematical elegance](https://bliptext.com/articles/mathematical-elegance) and its connections to the theory of [Diophantine approximations](https://bliptext.com/articles/diophantine-approximations).\nStephen Lucas calls this proof \"one of the more beautiful results related to approximating π\".\nJulian Havil ends a discussion of [continued fraction](https://bliptext.com/articles/continued-fraction) approximations of π with the result, describing it as \"impossible to resist mentioning\" in that context.\nThe purpose of the proof is not primarily to convince its readers that 22⁄7 (or 3 1⁄7) is indeed bigger than π; systematic methods of computing the value of π exist.\nIf one knows that π is approximately 3.14159, then it trivially follows that π π.\nThe evaluation of this integral was the first problem in the 1968 [Putnam Competition](https://bliptext.com/articles/william-lowell-putnam-mathematical-competition).\nIt is easier than most Putnam Competition problems, but the competition often features seemingly obscure problems that turn out to refer to something very familiar.\nThis integral has also been used in the entrance examinations for the [Indian Institutes of Technology](https://bliptext.com/articles/indian-institutes-of-technology).\n\n# Details of evaluation of the integral\nThat the [integral](https://bliptext.com/articles/integral) is positive follows from the fact that the [integrand](https://bliptext.com/articles/integrand) is [non-negative](https://bliptext.com/articles/non-negative); the denominator is positive and the numerator is a product of nonnegative numbers.\nOne can also easily check that the integrand is strictly positive for at least one point in the range of integration, say at 1⁄2.\nSince the integrand is continuous at that point and nonnegative elsewhere, the integral from 0 to 1 must be strictly positive.\nIt remains to show that the integral in fact evaluates to the desired quantity:\n(See [polynomial long division](https://bliptext.com/articles/polynomial-long-division).)\n\n# Quick upper and lower bounds\nIn, it is pointed out that if 1 is substituted for\nx\nin the denominator, one gets a lower bound on the integral, and if 0 is substituted for\nx\nin the denominator, one gets an upper bound:\nThus we have\nhence 3.1412 < π < 3.1421 in decimal expansion.\nThe bounds deviate by less than 0.015% from π.\nSee also.\n\n# Proof that 355/113 exceeds π\nAs discussed in, the well-known Diophantine approximation and far better upper estimate [355⁄113](https://bliptext.com/articles/355-113) for π follows from the relation\nwhere the first six digits after the decimal point agree with those of π.\nSubstituting 1 for\nx\nin the denominator, we get the lower bound\nsubstituting 0 for\nx\nin the denominator, we get twice this value as an upper bound, hence\nIn decimal expansion, this means 3.141 592 57 < π < 3.141 592 74, where the bold digits of the lower and upper bound are those of π.\n\n# Extensions\nThe above ideas can be generalized to get better approximations of π; see also and (in both references, however, no calculations are given).\nFor explicit calculations, consider, for every integer\nn ≥ 1\n,\nwhere the middle integral evaluates to\ninvolving π.\nThe last sum also appears in [Leibniz' formula for π](https://bliptext.com/articles/leibniz-formula-for-pi).\nThe correction term and [error bound](https://bliptext.com/articles/error-bound) is given by\nwhere the approximation (the tilde means that the quotient of both sides tends to one for large\nn\n) of the [central binomial coefficient](https://bliptext.com/articles/central-binomial-coefficient) follows from [Stirling's formula](https://bliptext.com/articles/stirling-s-formula) and shows the fast convergence of the integrals to π.\nCalculation of these integrals: For all integers\nk ≥ 0\nand\nℓ ≥ 2\nwe have\nApplying this formula recursively\n2n\ntimes yields\nFurthermore,\nwhere the first equality holds, because the terms for\n1 ≤ j ≤ 3n – 1\ncancel, and the second equality arises from the index shift\nj → j + 1\nin the first sum.\nApplication of these two results gives\nFor integers\nk, ℓ ≥ 0\n, using [integration by parts](https://bliptext.com/articles/integration-by-parts)\nℓ\ntimes, we obtain\nSetting\n, we obtain\nIntegrating equation (1) from 0 to 1 using equation (2) and\n, we get the claimed equation involving π.\nThe results for\nare given above.\nFor\nwe get\nand\nhence 3.141 592 31 < π < 3.141 592 89, where the bold digits of the lower and upper bound are those of π.\nSimilarly for\n,\nwith correction term and error bound\nhence 3.141 592 653 40 < π < 3.141 592 653 87.\nThe next step for\nis\nwith\nwhich gives 3.141 592 653 589 55 < π < 3.141 592 653 589 96.\n\n# Footnotes\n\n## Citations","revisionCount":0},"hasMore":false,"nextCursor":null}

History