- Proof that π is irrational
Although the

mathematical constant known as π (pi) has been studied since ancient times, and so has the concept ofirrational number , it was not until the 18th century that π was proved to be irrational.In the 20th century, proofs were found that require no prerequisite knowledge beyond

integral calculus . One of those, due toIvan Niven , is widely known. A somewhat earlier similar proof is byMary Cartwright . She set it as an examination problem but did not publish it. Cartwright's proof is reproduced in "Jeffreys", in an appendix.**Niven's proof**The proof uses the characterization of π as the smallest positive zero of the

sine function. As in many proofs of irrationality, the argument proceeds byreductio ad absurdum .**Preparation:**Suppose that π is rational, i.e. π = "a" / "b" for some integers "a" and "b" ≠ 0, which may be takenwithout loss of generality to be positive. Given any positive integer "n", we define the polynomial function:$f(x)\; =\; \{x^n(a\; -\; bx)^n\; over\; n!\},quad\; xinmathbb\{R\}!$

and denote by

:$F(x)\; =\; f(x)\; +\; cdots\; +\; (-1)^j\; f^\{(2j)\}(x)\; +\; cdots\; +\; (-1)^n\; f^\{(2n)\}(x).\; !$

the alternating sum of "f" and its first "n" even derivatives.

**Claim 1:**"F"(0) = "F"(π)**Proof:**Since:$f(x)=b^n\{x^n(pi\; -\; x)^n\; over\; n!\}=f(pi-x),quad\; xinmathbb\{R\}!$

the

chain rule andmathematical induction imply: $f^\{(j)\}(x)\; =\; (-1)^j\; f^\{(j)\}(pi\; -\; x),quad\; xinmathbb\{R\}!$

for all the derivatives, in particular

:$f^\{(2j)\}(0)=f^\{(2j)\}(pi)!$

for "j" = 1, 2, ...,"n" and Claim 1 follows from the definition of "F".

**Claim 2:**"F"(0) is an integer.**Proof:**Using thebinomial formula to expand ("a" – "bx")^{"n"}and the index transformation "j" = "k" + "n", we get the representation:$f(x)=\{1over\; n!\}sum\_\{j=n\}^\{2n\}\{n\; choose\; j-n\}a^\{2n-j\}(-b)^\{j-n\}x^\{j\}.!$

Since the coefficients of "x"

^{0}, "x"^{1}, ..., "x"^{"n" − 1}are zero and the degree of the polynomial "f" is at most 2"n", we have "f"^{ ("j")}(0) = 0 for "j" < "n" and "j" > 2"n". Furthermore,: $f^\{(j)\}(0)=\{j!over\; n!\}\{n\; choose\; j-n\}a^\{2n-j\}(-b)^\{j-n\}quadmbox\{for\; \}\; nle\; jle\; 2n!$

Since "j" ≥ "n", the fraction of the two

factorial s is an integer. The same holds for thebinomial coefficient , as can be seen from its combinatorical interpretation orPascal's triangle . Hence "f" and every derivative of "f" at 0 is an integer and so is "F"(0).**Claim 3:**:$frac12\; int\_0^pi\; f(x)sin(x),dx=F(0)!$**Proof:**Since "f"^{ (2"n" + 2)}is thezero polynomial , we have:$F"\; +\; F\; =\; f.,$

The derivatives of the

sine andcosine function are given by (sin "x")' = cos "x" and (cos "x")' = −sin "x", hence theproduct rule implies:$(F\text{'}cdotsin\; -\; Fcdotcos)\text{'}\; =\; fcdotsin!$

By the

fundamental theorem of calculus :$frac12\; int\_0^pi\; f(x)sin(x),dx=\; frac12\; igl(F\text{'}(x)sin\; x\; -\; F(x)cos\; xigr)Big|\_\{x=0\}^\{x=pi\}.!$

Since sin 0 = sin π = 0 and cos 0 = –cos π = 1 (here we use the abovementioned characterization of π as a zero of the sine function), Claim 3 follows from Claim 1.

**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 2 and 3 show that "F"(0) is a "positive" integer. Since:$x(pi\; -x)\; =\; Bigl(fracpi2Bigr)^2-Bigl(x-fracpi2Bigr)^2leBigl(fracpi2Bigr)^2,quad\; xinmathbb\{R\}!$

and 0 ≤ sin "x" ≤ 1 for 0 ≤ "x" ≤ π, we have

:$frac12\; int\_0^pi\; f(x)sin(x),dxle\; frac\{b^n\}\{n!\}Bigl(fracpi2Bigr)^\{2n+1\}!$

which is smaller than 1 for large "n", hence "F"(0) < 1 by Claim 3 for these "n". This is impossible for the positive integer "F"(0).

**Analysis of Niven's proof**The above proof is a polished version, which is kept as simple as possible concerning the prerequisites, of an analysis of the formula

:$egin\{align\}frac12int\_0^pi\; f(x)sin(x),dx=frac12sum\_\{j=0\}^n\; (-1)^j\; igl(f^\{(2j)\}(pi)+f^\{(2j)\}(0)igr)\backslash qquad+frac\{(-1)^\{n+12int\_0^pi\; f^\{(2n+2)\}(x)sin(x),dx,end\{align\}$

which is obtained by 2"n" + 2 partial integrations. Claim 3 essentially establishes this formula, where the use of "F" hides the iterated partial integrations. The last integral vanishes because "f"

^{ (2"n" + 2)}is the zero polynomial. Claims 1 and 2 show that the remaining sum is an integer.**Cartwright's proof**Jeffreys, page 268, says:

The following was set as an example in the Mathematics Preliminary Examination at Cambridge in 1945 by Dame

Mary Cartwright , but she has not traced its origin.Consider the integrals

: $I\_n\; =\; int\_\{-1\}^1\; (1\; -\; x^2)^n\; cos(alpha\; x),dx.$

Two integrations by parts give the

recurrence relation : $alpha^2\; I\_n\; =\; 2n(2n\; -\; 1)\; I\_\{n-1\}\; -\; 4n(n-1)I\_\{n-2\},quad\; n\; ge\; 2.$

If

: $J\_n\; =\; alpha^\{2n+1\}I\_n,,$

then this becomes

: $J\_n\; =\; 2n(2n\; -\; 1)J\_\{n-1\}\; -\; 4n(n\; -\; 1)alpha^2\; J\_\{n-2\}.$

Also

: $J\_0\; =\; 2sinalpha,quad\; J\_1\; =\; -4alphacosalpha\; +\; 4sinalpha.,$

Hence for all "n",

: $J\_n\; =\; alpha^\{2n+1\}I\_n\; =\; n!(P\_nsinalpha\; +\; Q\_ncosalpha),$

where "P"

_{"n"}, "Q"_{"n"}arepolynomial s in "α" of degree ≤ 2"n", and with integral coefficients depending on "n".Take "α" = (1/2)"π", and suppose if possible that

: $frac\{1\}\{2\}pi\; =\; frac\{b\}\{a\}$

where "a" and "b" are integers. Then

: $frac\{b^\{2n+1\{n!\}\; I\_n\; =\; P\_n\; a^\{2n+1\}.$

The right side is an integer. But 0 < "I"

_{"n"}< 2 since: $0\; <\; (1\; -\; x^2)^ncosleft(frac\{1\}\{2\}pi\; x\; ight)\; <\; 1\; ext\{\; for\; \}-1\; <\; x\; <\; 1$

and

: $frac\{b^\{2n+1\{n!\}\; o\; 0\; ext\{\; as\; \}n\; o\; infty.$

Hence for sufficiently large "n"

: $0\; <\; frac\{b^\{2n+1\}I\_n\}\{n!\}\; <\; 1,$

that is, we could find an integer between 0 and 1. That is the contradiction that follows from the assumption that "π" is rational.

**References***

Ivan Niven , "A simple proof that π is irrational", Bull. Amer. Math. Soc., Vol. 53, No. 6, p. 509, (1947) [*http://www.ams.org/bull/1947-53-06/S0002-9904-1947-08821-2/S0002-9904-1947-08821-2.pdf Online*]

*Harold Jeffreys , "Scientific Inference", 3rd edition, Cambridge University Press, 1973, page 268.

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