Question Number 3160 by prakash jain last updated on 06/Dec/15
![Prove that e=Σ_(n=0) ^∞ (1/(n!)) is irrational.](https://www.tinkutara.com/question/Q3160.png)
$$\mathrm{Prove}\:\mathrm{that}\:{e}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\:\:\mathrm{is}\:\mathrm{irrational}. \\ $$
Answered by Yozzi last updated on 06/Dec/15
![Fourier′s Proof Suppose for contradiction that e is rational, so we have e=(a/b), (a,b∈Z, b≠0). Define x=b!(e−Σ_(n=0) ^b (1/(n!))) where x∈Z. To check that x∈Z, substitute e=(a/b). x=b!((a/b)−Σ_(n=0) ^b (1/(n!))) =(b−1)!a−Σ_(n=0) ^b ((b!)/(n!)) Note that (b−1)!a∈Z. As n varies in the series, b!≥n! ⇒((b!)/(n!))≥1 with ((b!)/(n!))∈Z so that (Σ_(n=0) ^b ((b!)/(n!)))∈Z. Hence, x∈Z. Since we also have e=Σ_(n=0) ^∞ (1/(n!)), ∴ x=b!(Σ_(n=0) ^∞ (1/(n!))−Σ_(n=0) ^b (1/(n!))) x=b!Σ_(n=b+1) ^∞ (1/(n!))=b!((1/((b+1)!))+(1/((b+2)!))+(1/((b+3)!))+(1/((b+4)!))+...) x>0 For every term in the series in the expression for x,where n≥b+1 ((b!)/(n!))=(1/((b+1)(b+2)(b+3)...(b+(n−b))))<(1/((b+1)^(n−b) )) ∴x=Σ_(n=b+1) ^∞ ((b!)/(n!))<Σ_(n=b+1) ^∞ (1/((b+1)^(n−b) )) let k=n−b. ∴x<Σ_(k=1) ^∞ (1/((b+1)^k ))=((1/(b+1))/(1−(1/(b+1))))=(1/b)<1 So x<1. But∄x∈Z where 0<x<1. So, by contradiction, e cannot be rational; e is irrational.](https://www.tinkutara.com/question/Q3173.png)
$${Fourier}'{s}\:{Proof} \\ $$$${Suppose}\:{for}\:{contradiction}\:{that} \\ $$$${e}\:{is}\:{rational},\:{so}\:{we}\:{have} \\ $$$${e}=\frac{{a}}{{b}},\:\left({a},{b}\in\mathbb{Z},\:{b}\neq\mathrm{0}\right). \\ $$$${Define}\:{x}={b}!\left({e}−\underset{{n}=\mathrm{0}} {\overset{{b}} {\sum}}\frac{\mathrm{1}}{{n}!}\right)\:{where}\:{x}\in\mathbb{Z}. \\ $$$${To}\:{check}\:{that}\:{x}\in\mathbb{Z},\:{substitute}\:{e}=\frac{{a}}{{b}}. \\ $$$${x}={b}!\left(\frac{{a}}{{b}}−\underset{{n}=\mathrm{0}} {\overset{{b}} {\sum}}\frac{\mathrm{1}}{{n}!}\right) \\ $$$$\:\:\:=\left({b}−\mathrm{1}\right)!{a}−\underset{{n}=\mathrm{0}} {\overset{{b}} {\sum}}\frac{{b}!}{{n}!} \\ $$$${Note}\:{that}\:\left({b}−\mathrm{1}\right)!{a}\in\mathbb{Z}. \\ $$$${As}\:{n}\:{varies}\:{in}\:{the}\:{series},\:{b}!\geqslant{n}! \\ $$$$\Rightarrow\frac{{b}!}{{n}!}\geqslant\mathrm{1}\:{with}\:\frac{{b}!}{{n}!}\in\mathbb{Z}\:{so}\:{that}\:\left(\underset{{n}=\mathrm{0}} {\overset{{b}} {\sum}}\frac{{b}!}{{n}!}\right)\in\mathbb{Z}. \\ $$$${Hence},\:{x}\in\mathbb{Z}.\: \\ $$$${Since}\:{we}\:{also}\:{have}\:{e}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}, \\ $$$$\therefore\:{x}={b}!\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}−\underset{{n}=\mathrm{0}} {\overset{{b}} {\sum}}\frac{\mathrm{1}}{{n}!}\right) \\ $$$$\:\:\:\:{x}={b}!\underset{{n}={b}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}={b}!\left(\frac{\mathrm{1}}{\left({b}+\mathrm{1}\right)!}+\frac{\mathrm{1}}{\left({b}+\mathrm{2}\right)!}+\frac{\mathrm{1}}{\left({b}+\mathrm{3}\right)!}+\frac{\mathrm{1}}{\left({b}+\mathrm{4}\right)!}+…\right) \\ $$$$\:\:\:\:{x}>\mathrm{0} \\ $$$${For}\:{every}\:{term}\:{in}\:{the}\:{series}\:{in}\:{the} \\ $$$${expression}\:{for}\:{x},{where}\:{n}\geqslant{b}+\mathrm{1} \\ $$$$\frac{{b}!}{{n}!}=\frac{\mathrm{1}}{\left({b}+\mathrm{1}\right)\left({b}+\mathrm{2}\right)\left({b}+\mathrm{3}\right)…\left({b}+\left({n}−{b}\right)\right)}<\frac{\mathrm{1}}{\left({b}+\mathrm{1}\right)^{{n}−{b}} } \\ $$$$\therefore{x}=\underset{{n}={b}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{{b}!}{{n}!}<\underset{{n}={b}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({b}+\mathrm{1}\right)^{{n}−{b}} } \\ $$$${let}\:{k}={n}−{b}. \\ $$$$\therefore{x}<\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({b}+\mathrm{1}\right)^{{k}} }=\frac{\mathrm{1}/\left({b}+\mathrm{1}\right)}{\mathrm{1}−\frac{\mathrm{1}}{{b}+\mathrm{1}}}=\frac{\mathrm{1}}{{b}}<\mathrm{1} \\ $$$${So}\:{x}<\mathrm{1}.\:{But}\nexists{x}\in\mathbb{Z}\:{where}\:\mathrm{0}<{x}<\mathrm{1}. \\ $$$${So},\:{by}\:{contradiction},\:{e}\:{cannot}\:{be} \\ $$$${rational};\:{e}\:{is}\:{irrational}. \\ $$$$ \\ $$$$ \\ $$