{"history":[],"article":{"id":"c025f219-c8d1-4c36-b146-23673f5ce2e5","title":"Proof that e is irrational","slug":"proof-that-e-is-irrational","content":":::summary\n:::\n\n# \nThe number e was introduced by [Jacob Bernoulli](https://bliptext.com/articles/jacob-bernoulli) in 1683.\nMore than half a century later, [Euler](https://bliptext.com/articles/leonhard-euler), who had been a student of Jacob's younger brother [Johann](https://bliptext.com/articles/johann-bernoulli), proved that e is [irrational](https://bliptext.com/articles/irrational-number); that is, that it cannot be expressed as the quotient of two integers.\n\n# Euler's proof\nEuler wrote the first proof of the fact that e is irrational in 1737 (but the text was only published seven years later).\nHe computed the representation of e as a [simple continued fraction](https://bliptext.com/articles/simple-continued-fraction), which is\nSince this continued fraction is infinite and every rational number has a terminating continued fraction, e is irrational.\nA short proof of the previous equality is known.\nSince the simple continued fraction of e is not [periodic](https://bliptext.com/articles/periodic-continued-fraction), this also proves that e is not a root of a quadratic polynomial with rational coefficients; in particular, e2 is irrational.\n\n# Fourier's proof\nThe most well-known proof is [Joseph Fourier](https://bliptext.com/articles/joseph-fourier)'s [proof by contradiction](https://bliptext.com/articles/proof-by-contradiction), which is based upon the equality\nInitially e is assumed to be a rational number of the form a⁄b.\nThe idea is to then analyze the scaled-up difference (here denoted x) between the series representation of e and its strictly smaller b-th partial sum, which approximates the limiting value e.\nBy choosing the scale factor to be the [factorial](https://bliptext.com/articles/factorial) of b, the fraction a⁄b and the b-th partial sum are turned into [integers](https://bliptext.com/articles/integer), hence x must be a positive integer.\nHowever, the fast convergence of the series representation implies that x is still strictly smaller than 1.\nFrom this contradiction we deduce that e is irrational.\nNow for the details.\nIf e is a [rational number](https://bliptext.com/articles/rational-number), there exist positive integers a and b such that e = a⁄b.\nDefine the number\nUse the assumption that e = a⁄b to obtain\nThe first term is an integer, and every fraction in the sum is actually an integer because n ≤ b for each term.\nTherefore, under the assumption that e is rational, x is an integer.\nWe now prove that 0 < x < 1.\nFirst, to prove that x is strictly positive, we insert the above series representation of e into the definition of x and obtain\nbecause all the terms are strictly positive.\nWe now prove that x < 1.\nFor all terms with n ≥ b + 1 we have the upper estimate\nThis inequality is strict for every n ≥ b + 2.\nChanging the index of summation to k = n – b and using the formula for the [infinite geometric series](https://bliptext.com/articles/geometric-series), we obtain\nAnd therefore x<1.\nSince there is no integer strictly between 0 and 1, we have reached a contradiction, and so e is irrational, [Q.E.D.](https://bliptext.com/articles/q-e-d)\n\n# Alternate proofs\nAnother proof can be obtained from the previous one by noting that\nand this inequality is equivalent to the assertion that bx < 1.\nThis is impossible, of course, since b and x are positive integers.\nStill another proof can be obtained from the fact that\nDefine s_n as follows:\nThen\nwhich implies\nfor any positive integer n.\nNote that is always an integer.\nAssume that e^{-1} is rational, so where p, q are co-prime, and q \\neq 0.\nIt is possible to appropriately choose n so that is an integer, i.e. Hence, for this choice, the difference between and would be an integer.\nBut from the above inequality, that is not possible.\nSo, e^{-1} is irrational.\nThis means that e is irrational.\n\n# Generalizations\nIn 1840, [Liouville](https://bliptext.com/articles/joseph-liouville) published a proof of the fact that e2 is irrational followed by a proof that e2 is not a root of a second-degree polynomial with rational coefficients.\nThis last fact implies that e4 is irrational.\nHis proofs are similar to Fourier's proof of the irrationality of e.\nIn 1891, [Hurwitz](https://bliptext.com/articles/adolf-hurwitz) explained how it is possible to prove along the same line of ideas that e is not a root of a third-degree polynomial with rational coefficients, which implies that e3 is irrational.\nMore generally, eq is irrational for any non-zero rational q.\n[Charles Hermite](https://bliptext.com/articles/charles-hermite) further proved that e is a [transcendental number](https://bliptext.com/articles/transcendental-number), in 1873, which means that is not a root of any polynomial with rational coefficients, as is\neα\nfor any non-zero [algebraic](https://bliptext.com/articles/algebraic-number) α.","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T19:17:10.970Z","updated_at":"2025-01-25T19:17:10.970Z","search_vector":"'e':3 'irrat':5 'proof':1","search_content":" \n \n\n# \nThe number e was introduced by [Jacob Bernoulli](https://bliptext.com/articles/jacob-bernoulli) in 1683.\nMore than half a century later, [Euler](https://bliptext.com/articles/leonhard-euler), who had been a student of Jacob's younger brother [Johann](https://bliptext.com/articles/johann-bernoulli), proved that e is [irrational](https://bliptext.com/articles/irrational-number); that is, that it cannot be expressed as the quotient of two integers.\n\n# Euler's proof\nEuler wrote the first proof of the fact that e is irrational in 1737 (but the text was only published seven years later).\nHe computed the representation of e as a [simple continued fraction](https://bliptext.com/articles/simple-continued-fraction), which is\nSince this continued fraction is infinite and every rational number has a terminating continued fraction, e is irrational.\nA short proof of the previous equality is known.\nSince the simple continued fraction of e is not [periodic](https://bliptext.com/articles/periodic-continued-fraction), this also proves that e is not a root of a quadratic polynomial with rational coefficients; in particular, e2 is irrational.\n\n# Fourier's proof\nThe most well-known proof is [Joseph Fourier](https://bliptext.com/articles/joseph-fourier)'s [proof by contradiction](https://bliptext.com/articles/proof-by-contradiction), which is based upon the equality\nInitially e is assumed to be a rational number of the form a⁄b.\nThe idea is to then analyze the scaled-up difference (here denoted x) between the series representation of e and its strictly smaller b-th partial sum, which approximates the limiting value e.\nBy choosing the scale factor to be the [factorial](https://bliptext.com/articles/factorial) of b, the fraction a⁄b and the b-th partial sum are turned into [integers](https://bliptext.com/articles/integer), hence x must be a positive integer.\nHowever, the fast convergence of the series representation implies that x is still strictly smaller than 1.\nFrom this contradiction we deduce that e is irrational.\nNow for the details.\nIf e is a [rational number](https://bliptext.com/articles/rational-number), there exist positive integers a and b such that e = a⁄b.\nDefine the number\nUse the assumption that e = a⁄b to obtain\nThe first term is an integer, and every fraction in the sum is actually an integer because n ≤ b for each term.\nTherefore, under the assumption that e is rational, x is an integer.\nWe now prove that 0 < x < 1.\nFirst, to prove that x is strictly positive, we insert the above series representation of e into the definition of x and obtain\nbecause all the terms are strictly positive.\nWe now prove that x < 1.\nFor all terms with n ≥ b + 1 we have the upper estimate\nThis inequality is strict for every n ≥ b + 2.\nChanging the index of summation to k = n – b and using the formula for the [infinite geometric series](https://bliptext.com/articles/geometric-series), we obtain\nAnd therefore x<1.\nSince there is no integer strictly between 0 and 1, we have reached a contradiction, and so e is irrational, [Q.E.D.](https://bliptext.com/articles/q-e-d)\n\n# Alternate proofs\nAnother proof can be obtained from the previous one by noting that\nand this inequality is equivalent to the assertion that bx < 1.\nThis is impossible, of course, since b and x are positive integers.\nStill another proof can be obtained from the fact that\nDefine s_n as follows:\nThen\nwhich implies\nfor any positive integer n.\nNote that is always an integer.\nAssume that e^{-1} is rational, so where p, q are co-prime, and q \\neq 0.\nIt is possible to appropriately choose n so that is an integer, i.e. Hence, for this choice, the difference between and would be an integer.\nBut from the above inequality, that is not possible.\nSo, e^{-1} is irrational.\nThis means that e is irrational.\n\n# Generalizations\nIn 1840, [Liouville](https://bliptext.com/articles/joseph-liouville) published a proof of the fact that e2 is irrational followed by a proof that e2 is not a root of a second-degree polynomial with rational coefficients.\nThis last fact implies that e4 is irrational.\nHis proofs are similar to Fourier's proof of the irrationality of e.\nIn 1891, [Hurwitz](https://bliptext.com/articles/adolf-hurwitz) explained how it is possible to prove along the same line of ideas that e is not a root of a third-degree polynomial with rational coefficients, which implies that e3 is irrational.\nMore generally, eq is irrational for any non-zero rational q.\n[Charles Hermite](https://bliptext.com/articles/charles-hermite) further proved that e is a [transcendental number](https://bliptext.com/articles/transcendental-number), in 1873, which means that is not a root of any polynomial with rational coefficients, as is\neα\nfor any non-zero [algebraic](https://bliptext.com/articles/algebraic-number) α.","revisionCount":0},"hasMore":false,"nextCursor":null}

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