Another proof, which is a simplification of Lambert's proof, is due to Miklós Laczkovich.
Bukti lain, yang merupakan penyederhanaan dari bukti Lambert, diberikan oleh Miklós Laczkovich.
Many of these are proofs by contradiction.
Banyak dari bukti-bukti ini merupakan pembuktian melalui kontradiksi.
In 1882, Ferdinand von Lindemann proved that π is not just irrational, but transcendental as well.
Pada tahun 1882, Ferdinand von Lindemann membuktikan bahwa π tidak hanya irasional, namun juga transenden.
In 1761, Johann Heinrich Lambert proved that π is irrational by first showing that this continued fraction expansion holds: tan ⁡ ( x )x 1 − x 2 3 − x 2 5 − x 2 7 − ⋱ .
Pada 1761, Johann Heinrich Lambert membuktikan bahwa π merupakan bilangan irasional dengan menunjukkan bahwa berlakunya penjabaran pecahan berlanjut berikut: tan ( x )
{1-{\cfrac {x^{2}}{3-{\cfrac {x^{2}}{5-{\cfrac {x^{2}}{7-{}\ddots }}}}}}}}.} Then Lambert proved that if x is non-zero and rational, then this expression must be irrational.
x 1 − x 2 3 − x 2 5 − x 2 7 − ⋱ {1-{\cfrac {x^{2}}{3-{\cfrac {x^{2}}{5-{\cfrac {x^{2}}{7-\ddots }}}}}}}}} kemudian Lambert membuktikan bahwa jika x merupakan bilangan rasional selain nol, maka ekspresi di atas haruslah irasional.
Since tan ⁡ π 41 {4}}=1} , it follows that π 4 {4}}} is irrational, and thus π is also irrational.
1 {4}}\right)=1} , maka π 4 {4}}} merupakan bilangan irasional, sehingga π juga irasional.
sin ⁡ ( x ) , A n + 1 ( x )∫ 0 x y A n ( y )
d x∫ 0 π x n ( a − b x ) n n ! sin ( x )
d y U 0 ( x )sin ⁡ ( x )
d x∫ 0 π x n ( a − b x ) n n ! sin ( x )
x , U n + 1 ( x )− U n ′ ( x ) x A_{0}(x)&=\sin(x),&&A_{n+1}(x)=\int _{0}^{x}yA_{n}(y)\,dy\\[4pt]U_{0}(x)&={\frac {\sin(x)}{x}},&&U_{n+1}(x)=-{\frac {U_{n}'(x)}{x}}\end{aligned}}} Using induction we can prove that A n ( x )x 2 n + 1 ( 2 n + 1 ) ! !
+ … + ( − 1 ) n d 2 n f d x 2 n ( x ) f(x)&:={\dfrac {x^{n}\left(a-bx\right)^{n}}{n!}}\\F(x)&:=f(x)-{\dfrac {{\text{d}}^{2}f}{{\text{d}}x^{2}}}(x)+{\dfrac {{\text{d}}^{4}f}{{\text{d}}x^{4}}}(x)-{\dfrac {{\text{d}}^{6}f}{{\text{d}}x^{6}}}(x)+\ldots +\left(-1\right)^{n}{\dfrac {{\text{d}}^{2n}f}{{\text{d}}x^{2n}}}(x)\end{aligned}}}
x 2 ( 2 n + 1 ) )( 2 n + 1 )
d 2 n + 2 f d x 2 n + 2 ( x )
x 2 n + 3 {\frac {A_{n+1}(x)}{x^{2n+3}}}&=U_{n+1}(x)=-{\frac {U_{n}'(x)}{x}}=-{\frac {1}{x}}{\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {A_{n}(x)}{x^{2n+1}}}\right)\\[6pt]&=-{\frac {1}{x}}\left({\frac {A_{n}'(x)\cdot x^{2n+1}-(2n+1)x^{2n}A_{n}(x)}{x^{2(2n+1)}}}\right)\\[6pt]&={\frac {(2n+1)A_{n}(x)-xA_{n}'(x)}{x^{2n+3}}}\end{aligned}}} which is equivalent to A n + 1 ( x )( 2 n + 1 )
+ … + ( − 1 ) n d 2 n f d x 2 n ( x ) f(x)&:={\dfrac {x^{n}\left(a-bx\right)^{n}}{n!}}\\F(x)&:=f(x)-{\dfrac {{\text{d}}^{2}f}{{\text{d}}x^{2}}}(x)+{\dfrac {{\text{d}}^{4}f}{{\text{d}}x^{4}}}(x)-{\dfrac {{\text{d}}^{6}f}{{\text{d}}x^{6}}}(x)+\ldots +\left(-1\right)^{n}{\dfrac {{\text{d}}^{2n}f}{{\text{d}}x^{2n}}}(x)\end{aligned}}}
&{\frac {1}{2^{n+1}(n+1)!}}\int _{0}^{1}\left(1-z^{2}\right)^{n+1}\cos(xz)\,\mathrm {d} z\\&\qquad{\frac {1}{2^{n+1}(n+1)!}}{\Biggl (}\,\overbrace {\left.(1-z^{2})^{n+1}{\frac {\sin(xz)}{x}}\right|_{z=0}^{z=1}} ^{=\,0}\ +\,\int _{0}^{1}2(n+1)\left(1-z^{2}\right)^{n}z{\frac {\sin(xz)}{x}}\,\mathrm {d} z{\Biggr )}\\[8pt]&\qquad{\frac {1}{x}}\cdot {\frac {1}{2^{n}n!}}\int _{0}^{1}\left(1-
+ … + ( − 1 ) n d 2 n f d x 2 n ( x ) f(x)&:={\dfrac {x^{n}\left(a-bx\right)^{n}}{n!}}\\F(x)&:=f(x)-{\dfrac {{\text{d}}^{2}f}{{\text{d}}x^{2}}}(x)+{\dfrac {{\text{d}}^{4}f}{{\text{d}}x^{4}}}(x)-{\dfrac {{\text{d}}^{6}f}{{\text{d}}x^{6}}}(x)+\ldots +\left(-1\right)^{n}{\dfrac {{\text{d}}^{2n}f}{{\text{d}}x^{2n}}}(x)\end{aligned}}}
Take x1 2 π , {2}}\pi ,} and suppose if possible that 1 2 πa / b {2}}\pia/b} where a and b are natural numbers (i.e., assume that π is rational).
Andaikan π merupakan bilangan rasional, atau dengan kata lain, πa b {b}}} , untuk suatu bilangan bulat a dan b , yang kedua-duanya dapat diasumsikan tanpa mengurangi keumuman sebagai bilangan positif.
That is the contradiction that follows from the assumption that π is rational.
Akibatnya, asumsi di awal (bahwasanya π merupakan bilangan rasional) tidaklah benar, sehingga terbukti bahwa π merupakan bilangan irasional.
J_{n}(x)&=x^{2n+1}\int _{-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz\\[5pt]&=2x^{2n+1}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,dz\\[5pt]&=2^{n+1}n!A_{n}(x).\end{aligned}}} However, it is clearly simpler.
+ … + ( − 1 ) n d 2 n f d x 2 n ( x ) f(x)&:={\dfrac {x^{n}\left(a-bx\right)^{n}}{n!}}\\F(x)&:=f(x)-{\dfrac {{\text{d}}^{2}f}{{\text{d}}x^{2}}}(x)+{\dfrac {{\text{d}}^{4}f}{{\text{d}}x^{4}}}(x)-{\dfrac {{\text{d}}^{6}f}{{\text{d}}x^{6}}}(x)+\ldots +\left(-1\right)^{n}{\dfrac {{\text{d}}^{2n}f}{{\text{d}}x^{2n}}}(x)\end{aligned}}}
This proof uses the characterization of π as the smallest positive zero of the sine function.
Pembuktian ini menggunakan karakterisasi dari π sebagai pembuat nol dari fungsi sinus.
Suppose that π is rational, i.e. πa / b for some integers a and b which may be taken without loss of generality to both be positive.
Andaikan π merupakan bilangan rasional, atau dengan kata lain, πa b {b}}} , untuk suatu bilangan bulat a dan b , yang kedua-duanya dapat diasumsikan tanpa mengurangi keumuman sebagai bilangan positif.
c k } if n ≤ k ≤ 2 n ; in each case, f ( k ) ( 0 ) (0)} is an integer and therefore F ( 0 ) is an integer.
x k ^{2n}{\dfrac {c_{k}}{n!}}x^{k}} dengan c k } merupakan suatu bilangan bulat dan c k00} jika k < n .
On the other hand, f ( π − x )f ( x ) and so ( − 1 ) k f ( k ) ( π − x )f ( k ) ( x ) f^{(k)}(\pi -x)=f^{(k)}(x)} for each non-negative integer k .
d xF ( 0 ) + F ( π ) ^{\pi }f\!\left(x\right)\cdot \sin \!\left(x\right)\,{\text{d}}x=F\!\left(0\right)+F\!\left(\pi \right)} Perhatikan bahwa
Therefore, f ( k ) ( π ) (\pi )} is also an integer and so F ( π ) is an integer (in fact, it is easy to see that F ( π )
Berdasarkan definisi dari fungsi F , maka terbukti bahwa F ( π ) merupakan bilangan bulat.
Claim 2: ∫ 0 π f ( x ) sin ⁡ ( x )
Pertama, akan dibuktikan bahwa ∫ 0 π f ( x ) ⋅ sin ( x )
d x( F ′ ( x ) sin ⁡ x − F ( x ) cos ⁡ x ) | 0 π .
d x∫ 0 π x n ( a − b x ) n n ! sin ( x )
^{\pi }f(x)\sin(x)\,dx={\bigl (}F'(x)\sin x-F(x)\cos x{\bigr )}\right|_{0}^{\pi }.}
d xF ( π ) + F ( 0 ) ^{\pi }f\!\left(x\right)\,\sin \!\left(x\right)\,{\text{d}}x=F\!\left(\pi \right)+F\!\left(0\right)}
Conclusion: Since f ( x ) > 0 and sin ⁡ x > 0 for 0 < x < π (because π is the smallest positive zero of the sine function), Claims 1 and 2 show that F ( 0 ) + F ( π ) is a positive integer.
Oleh karena 0 < x ( a − b x ) < a π sin ( x ) ≤ 1 maka berdasarkan definisi dari fungsi f ( x ) , didapatkan F ( π ) + F ( 0 )∫ 0 π f ( x ) sin ( x )
d x , ^{\pi }f(x)\sin(x)\,dx=\sum _{j=0}^{n}(-1)^{j}\left(f^{(2j)}(\pi )+f^{(2j)}(0)\right)+(-1)^{n+1}\int _{0}^{\pi }f^{(2n+2)}(x)\sin(x)\,dx,} which is obtained by 2 n + 2 integrations by parts.
+ … + ( − 1 ) n d 2 n f d x 2 n ( x ) f(x)&:={\dfrac {x^{n}\left(a-bx\right)^{n}}{n!}}\\F(x)&:=f(x)-{\dfrac {{\text{d}}^{2}f}{{\text{d}}x^{2}}}(x)+{\dfrac {{\text{d}}^{4}f}{{\text{d}}x^{4}}}(x)-{\dfrac {{\text{d}}^{6}f}{{\text{d}}x^{6}}}(x)+\ldots +\left(-1\right)^{n}{\dfrac {{\text{d}}^{2n}f}{{\text{d}}x^{2n}}}(x)\end{aligned}}}
J_{n}(x)&=x^{2n+1}\int _{-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz\\&=\int _{-1}^{1}\left(x^{2}-(xz)^{2}\right)^{n}x\cos(xz)\,dz.\end{aligned}}} Therefore, the substitution x zy turns this integral into ∫ − x x ( x 2 − y 2 ) n cos ⁡ ( y )
+ … + ( − 1 ) n d 2 n f d x 2 n ( x ) f(x)&:={\dfrac {x^{n}\left(a-bx\right)^{n}}{n!}}\\F(x)&:=f(x)-{\dfrac {{\text{d}}^{2}f}{{\text{d}}x^{2}}}(x)+{\dfrac {{\text{d}}^{4}f}{{\text{d}}x^{4}}}(x)-{\dfrac {{\text{d}}^{6}f}{{\text{d}}x^{6}}}(x)+\ldots +\left(-1\right)^{n}{\dfrac {{\text{d}}^{2n}f}{{\text{d}}x^{2n}}}(x)\end{aligned}}}
d y∫ 0 π y n ( π − y ) n sin ⁡ ( y )
d x∫ 0 π x n ( a − b x ) n n ! sin ( x )
b n ∫ 0 π f ( x ) sin ⁡ ( x )
d x∫ 0 π x n ( a − b x ) n n ! sin ( x )
d xF ( π ) + F ( 0 ) .
d x ≤ ∫ 0 π ( a π ) n n !
b n ∫ 0 π x n ( π − x ) n n ! sin ⁡ ( x )
d x∫ 0 π x n ( a − b x ) n n ! sin ( x )
(x)=1-{\frac {x^{2}}{k}}+{\frac {x^{4}}{2!k(k+1)}}-{\frac {x^{6}}{3!k(k+1)(k+2)}}+\cdots \quad (k\notin \{0,-1,-2,\ldots \}).}
x k ^{2n}{\dfrac {c_{k}}{n!}}x^{k}} dengan c k } merupakan suatu bilangan bulat dan c k00} jika k < n .
f k + n ( x ) n ≠ 0{\begin{cases}f_{k}(x)&n=0\\{\dfrac {c^{n}}{k(k+1)\cdots (k+n-1)}}f_{k+n}(x)&n\neq 0\end{cases}}} Then g 0f k ( x )
⋅ c k n ≤ k ≤ 2 n }^{k}f}{{\text{d}}x^{k}}}\left(0\right)={\begin{cases}0&k<n\\{\dfrac {k!}{n!}}\cdot c_{k}&n\leq k\leq 2n\end{cases}}} Pada kedua kasus di atas, maka dapat disimpulkan bahwa
cos ⁡ 1 2 π0 , ({\tfrac {1}{4}}\pi )=\cos {\tfrac {1}{2}}\pi0,} it follows from claim 3 that 1 16 π 2 {16}}\pi ^{2}} is irrational and therefore that π is irrational.
1 {4}}\right)=1} , maka π 4 {4}}} merupakan bilangan irasional, sehingga π juga irasional.
Proof that e is irrational Proof that π is transcendental
Bukti bahwa e irasional Bukti bahwa π transenden