Proof that π is irrational

In the 1760s, Johann Heinrich Lambert was the first to prove that the number is irrational, meaning it cannot be expressed as a fraction $a/b$, where $a$ and $b$ are both integers. In the 19th century, Charles Hermite found a proof that requires no prerequisite knowledge beyond basic calculus. Three simplifications of Hermite's proof are due to Mary Cartwright, Ivan Niven, and Nicolas Bourbaki. Another proof, which is a simplification of Lambert's proof, is due to Miklós Laczkovich. Many of these are proofs by contradiction.

In 1882, Ferdinand von Lindemann proved that is not just irrational, but transcendental as well.

Lambert's proof

In 1761, Lambert proved that is irrational by first showing that this continued fraction expansion holds:

:$\tan(x) = \cfrac.$

Then Lambert proved that if ''x'' is non-zero and rational, then this expression must be irrational. Since tan(/4) = 1, it follows that /4 is irrational, and thus is also irrational. A simplification of Lambert's proof is given below.

Hermite's proof

Written in 1873, this proof uses the characterization of as the smallest positive number whose half is a zero of the cosine function and it actually proves that ² is irrational. As in many proofs of irrationality, it is a proof by contradiction.

Consider the sequences of functions ''A''_''n'' and ''U''_''n'' from $\R$ into $\R$ for $n \in \N_0$ defined by:

: \begin
A_0(x) &= \sin(x), && A_(x) =\int_0^xyA_n(y)\,dy \\ ptU_0(x) &= \fracx, && U_(x) =-\fracx
\end

Using induction we can prove that

:\begin
A_n(x) &=\frac-\frac+\frac\mp\cdots \\ ptU_n(x) &=\frac1-\frac+\frac\mp\cdots
\end

and therefore we have:

:$U_n(x)=\frac.\,$

So

:
\begin
\frac & =U_(x)=-\fracx=-\frac1x\frac \left(\frac\right) \\ pt& = -\frac \left( \frac \right ) = \frac
\end

which is equivalent to

:$A_(x)=(2n+1)A_n(x)-x^2A_(x).\,$

Using the definition of the sequence and employing induction we can show that

:$A_n(x) = P_n(x^2) \sin(x) + x Q_n(x^2) \cos(x),\,$

where ''P''_''n'' and ''Q''_''n'' are polynomial functions with integer coefficients and the degree of ''P''_''n'' is smaller than or equal to ⌊''n''/2⌋. In particular, ''A''_''n''(/2) = ''P''_''n''(²/4).

Hermite also gave a closed expression for the function ''A''_''n'', namely

:$A_n(x)=\frac\int_0^1(1-z^2)^n\cos(xz)\,\mathrmz.\,$

He did not justify this assertion, but it can be proved easily. First of all, this assertion is equivalent to

:$\frac\int_0^1(1-z^2)^n\cos(x z)\,\mathrmz=\frac=U_n(x).$

Proceeding by induction, take ''n'' = 0.

:$\int_0^1\cos(xz)\,\mathrmz=\fracx=U_0(x)$

and, for the inductive step, consider any $n \in \N$. If

:$\frac\int_0^1(1-z^2)^n\cos(xz)\,\mathrmz=U_n(x),$

then, using integration by parts and Leibniz's rule, one gets

:\begin
\frac &\int_0^1(1-z^2)^\cos(xz)\,\mathrmz \\
&=\frac\left (\overbrace^ + \int_0^12(n+1)(1-z^2)^nz \fracx\,\mathrmz\right )\\ pt&= \frac1x\cdot\frac1\int_0^1(1-z^2)^nz\sin(xz)\,\mathrmz\\ pt&= -\frac1x\cdot\frac\left(\frac1\int_0^1(1-z^2)^n\cos(xz)\,\mathrmz\right) \\ pt&= -\fracx \\ pt&= U_(x).
\end

If ²/4 = ''p''/''q'', with ''p'' and ''q'' in $\N$, then, since the coefficients of ''P''_''n'' are integers and its degree is smaller than or equal to ⌊''n''/2⌋, ''q''^{⌊''n''/2⌋}''P''_''n''(²/4) is some integer ''N''. In other words,

:$N=q^A_n\left(\frac\pi2\right) =q^\frac\int_0^1(1-z^2)^n \cos \left(\frac\pi2z \right)\,\mathrmz.$

But this number is clearly greater than 0. On the other hand, the limit of this quantity as ''n'' goes to infinity is zero, and so, if ''n'' is large enough, ''N'' < 1. Thereby, a contradiction is reached.

Hermite did not present his proof as an end in itself but as an afterthought within his search for a proof of the transcendence of . He discussed the recurrence relations to motivate and to obtain a convenient integral representation. Once this integral representation is obtained, there are various ways to present a succinct and self-contained proof starting from the integral (as in Cartwright's, Bourbaki's or Niven's presentations), which Hermite could easily see (as he did in his proof of the transcendence of ''e'').

Moreover, Hermite's proof is closer to Lambert's proof than it seems. In fact, ''A''_''n''(''x'') is the "residue" (or "remainder") of Lambert's continued fraction for tan(''x'').

Cartwright's proof

Harold Jeffreys wrote that this proof was set as an example in an exam at Cambridge University in 1945 by Mary Cartwright, but that she had not traced its origin. It still remains on the 4th problem sheet today for the Analysis IA course at Cambridge University.

Consider the integrals

:$I_n(x)=\int_^1(1 - z^2)^n\cos(xz)\,dz,$

where ''n'' is a non-negative integer.

Two integrations by parts give the recurrence relation

:$x^2I_n(x)=2n(2n-1)I_(x)-4n(n-1)I_(x). \qquad (n \geq 2)$

If

:$J_n(x)=x^I_n(x),$

then this becomes

:$J_n(x)=2n(2n-1)J_(x)-4n(n-1)x^2J_(x).$

Furthermore, ''J''₀(''x'') = 2sin(''x'') and ''J''₁(''x'') = −4''x'' cos(''x'') + 4sin(''x''). Hence for all ''n'' ∈ Z₊,

:$J_n(x)=x^I_n(x)=n!\bigl(P_n(x)\sin(x)+Q_n(x)\cos(x)\bigr),$

where ''P''_''n''(''x'') and ''Q''_''n''(''x'') are polynomials of degree ≤ ''n'', and with integer coefficients (depending on ''n'').

Take ''x'' = /2, and suppose if possible that /2 = ''a''/''b'', where ''a'' and ''b'' are natural numbers (i.e., assume that is rational). Then

: $\fracI_n\left(\frac\pi2\right) = P_n\left(\frac\pi2\right)b^.$

The right side is an integer. But 0 < ''I''_''n''(/2) < 2 since the interval minus;1, 1has length 2 and the function that is being integrated takes only values between 0 and 1. On the other hand,

: $\frac \to 0 \quad \textn \to \infty.$

Hence, for sufficiently large ''n''

: $0 < \frac < 1,$

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

This proof is similar to Hermite's proof. Indeed,

:\begin
J_n(x)&=x^\int_^1 (1 - z^2)^n \cos(xz)\,dz\\ pt &=2x^\int_0^1 (1 - z^2)^n \cos(xz)\,dz\\ pt &=2^n!A_n(x).
\end

However, it is clearly simpler. This is achieved by omitting the inductive definition of the functions ''A''_''n'' and taking as a starting point their expression as an integral.

Niven's proof

This proof uses the characterization of as the smallest positive zero of the sine function.

Suppose that is rational, i.e. for some integers ''a'' and , which may be taken without loss of generality to be positive. Given any positive integer ''n'', we define the polynomial function:

:$f(x) = \frac$

and, for each ''x'' ∈ ℝ let

:$F(x) = f(x)-f''(x)+f^(x)+\cdots+(-1)^n f^(x).$

Claim 1: is an integer.

Proof:
Expanding ''f'' as a sum of monomials, the coefficient of ''x^k'' is a number of the form where ''c_k'' is an integer, which is 0 if . Therefore, is 0 when and it is equal to if ; in each case, is an integer and therefore ''F''(0) is an integer.

On the other hand, = ''f''(''x'') and so = for each non-negative integer ''k''. In particular, = Therefore, is also an integer and so ''F''() is an integer (in fact, it is easy to see that ''F''() = ''F''(0), but that is not relevant to the proof). Since ''F''(0) and ''F''() are integers, so is their sum.

Claim 2:

:$\int_0^\pi f(x)\sin(x)\,dx=F(0)+F(\pi)$

Proof: Since is the zero polynomial, we have

:$F'' + F = f.$

The derivatives of the sine and cosine function are given by sin' = cos and cos' = −sin. Hence the product rule implies

:$(F'\cdot\sin - F\cdot\cos)' = f\cdot\sin$

By the fundamental theorem of calculus

: $\left. \int_0^\pi f(x)\sin(x)\,dx= \bigl(F'(x)\sin x - F(x)\cos x\bigr) \_0^\pi.$

Since and (here we use the above-mentioned characterization of as a zero of the sine function), Claim 2 follows.

Conclusion: Since and for (because is the ''smallest'' positive zero of the sine function), Claims 1 and 2 show that is a ''positive'' integer. Since and for , we have, by the original definition of ''f'',

:$\int_0^\pi f(x)\sin(x)\,dx\le\pi\frac$

which is smaller than 1 for large ''n'', hence for these ''n'', by Claim 2. This is impossible for the positive integer . This shows that the original assumption that π is rational leads to a contradiction, which concludes the proof.

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

:$\int_0^\pi f(x)\sin(x)\,dx = \sum_^n (-1)^j \left (f^(\pi)+f^(0)\right )+(-1)^\int_0^\pi f^(x)\sin(x)\,dx,$

which is obtained by integrations by parts. Claim 2 essentially establishes this formula, where the use of ''F'' hides the iterated integration by parts. The last integral vanishes because is the zero polynomial. Claim 1 shows that the remaining sum is an integer.

Niven's proof is closer to Cartwright's (and therefore Hermite's) proof than it appears at first sight. In fact,

:$\begin J_n(x)&=x^\int_^1(1-z^2)^n\cos(xz)\,dz\\ &=\int_^1\left (x^2-(xz)^2\right )^nx\cos(xz)\,dz. \end$

Therefore, the substitution ''xz'' = ''y'' turns this integral into

:$\int_^x(x^2-y^2)^n\cos(y)\,dy.$

In particular,

:\begin
J_n\left(\frac\pi2\right)&=\int_^\left(\frac4-y^2\right)^n\cos(y)\,dy\\ pt&=\int_0^\pi\left(\frac4-\left(y-\frac\pi2\right)^2\right)^n\cos\left(y-\frac\pi2\right)\,dy\\ pt&=\int_0^\pi y^n(\pi-y)^n\sin(y)\,dy\\ pt&=\frac\int_0^\pi f(x)\sin(x)\,dx.
\end

Another connection between the proofs lies in the fact that Hermite already mentions that if ''f'' is a polynomial function and

:$F=f-f^+f^\mp\cdots,$

then

:$\int f(x)\sin(x)\,dx=F'(x)\sin(x)-F(x)\cos(x)+C,$

from which it follows that

:$\int_0^\pi f(x)\sin(x)\,dx=F(\pi)+F(0).$

Bourbaki's proof

Bourbaki's proof is outlined as an exercise in his calculus treatise. For each natural number ''b'' and each non-negative integer ''n'', define

:$A_n(b)=b^n\int_0^\pi\frac\sin(x)\,dx.$

Since ''A''_''n''(''b'') is the integral of a function defined on ,that takes the value 0 on 0 and on and which is greater than 0 otherwise, ''A''_''n''(''b'') > 0. Besides, for each natural number ''b'', ''A''_''n''(''b'') < 1 if ''n'' is large enough, because

: $x(\pi-x) \le \left(\frac\pi2\right)^2$

and therefore

:$A_n(b)\le\pi b^n \frac \left(\frac\pi2\right)^ = \pi \frac.$

On the other hand, repeated integration by parts allows us to deduce that, if ''a'' and ''b'' are natural numbers such that  = ''a''/''b'' and ''f'' is the polynomial function from ,into R defined by

: $f(x)=\frac,$

then:

:\begin
A_n(b) &= \int_0^\pi f(x)\sin(x)\,dx \\ pt &= \Big f(x)\cos(x)\Big^-\Big f'(x) \sin(x) \Big^ + \cdots \pm \Big f^(x) \cos(x) \Big^ \pm \int_0^\pi f^(x)\cos(x)\,dx.
\end

This last integral is 0, since ''f'' ^{(2''n'' + 1)} is the null function (because ''f'' is a polynomial function of degree 2''n''). Since each function ''f'' ^(''k'') (with ) takes integer values on 0 and on and since the same thing happens with the sine and the cosine functions, this proves that ''A''_''n''(''b'') is an integer. Since it is also greater than 0, it must be a natural number. But it was also proved that ''A''_''n''(''b'') < 1 if ''n'' is large enough, thereby reaching a contradiction.

This proof is quite close to Niven's proof, the main difference between them being the way of proving that the numbers ''A''_''n''(''b'') are integers.

Laczkovich's proof

Miklós Laczkovich's proof is a simplification of Lambert's original proof. He considers the functions

:$f_k(x) = 1 - \frack+\frac-\frac + \cdots \quad (k\notin\).$

These functions are clearly defined for all ''x'' ∈ R. Besides

:$f_(x) = \cos(2x),$
:$f_(x) = \frac.$

Claim 1: The following recurrence relation holds:

:$\forall x\in\R: \qquad \fracf_(x)=f_(x)-f_k(x).$

Proof: This can be proved by comparing the coefficients of the powers of ''x''.

Claim 2: For each ''x'' ∈ R, $\lim_f_k(x)=1.$

Proof: In fact, the sequence ''x''^2''n''/''n''! is bounded (since it converges to 0) and if ''C'' is an upper bound and if ''k'' > 1, then

:$\left, f_k(x)-1\\leqslant\sum_^\infty\frac C=C\frac=\frac C.$

Claim 3: If ''x'' ≠ 0 and if ''x''² is rational, then

:$\forall k\in\Q\smallsetminus\: \qquad f_k(x)\neq0 \quad \text \quad \frac\notin\Q.$

Proof: Otherwise, there would be a number ''y'' ≠ 0 and integers ''a'' and ''b'' such that ''f''_''k''(''x'') = ''ay'' and ''f''_''k'' + 1(''x'') = ''by''. In order to see why, take ''y'' = ''f''_''k'' + 1(''x''), ''a'' = 0 and ''b'' = 1 if ''f''_''k''(''x'') = 0; otherwise, choose integers ''a'' and ''b'' such that ''f''_''k'' + 1(''x'')/''f_k''(''x'') = ''b''/''a'' and define ''y'' = ''f''_''k''(''x'')/''a'' = ''f''_''k'' + 1(''x'')/''b''. In each case, ''y'' cannot be 0, because otherwise it would follow from claim 1 that each ''f''_{''k'' + ''n''}(''x'') (''n'' ∈ N) would be 0, which would contradict claim 2. Now, take a natural number ''c'' such that all three numbers ''bc''/''k'', ''ck''/''x''² and ''c''/''x''² are integers and consider the sequence

:$g_n=\beginf_k(x) & n=0\\ \dfracf_(x) & n \neq 0 \end$

Then

:$g_0=f_k(x)=ay\in\Z y \quad \text \quad g_1=\frac ckf_(x)=\fracky\in\Z y.$

On the other hand, it follows from claim 1 that

:\begin
g_&=\frac\cdot\fracf_(x)\\ pt& =\fracf_(x)-\fracf_(x)\\ pt&=\fracg_-\fracg_n\\ pt&=\left(\frac+\frac cn\right)g_-\fracg_n,
\end

which is a linear combination of ''g''_''n'' + 1 and ''g''_''n'' with integer coefficients. Therefore, each ''g''_''n'' is an integer multiple of ''y''. Besides, it follows from claim 2 that each ''g_n'' is greater than 0 (and therefore that ''g''_''n'' ≥ , ''y'', ) if ''n'' is large enough and that the sequence of all ''g''_''n'' converges to 0. But a sequence of numbers greater than or equal to , ''y'', cannot converge to 0.

Since ''f''_1/2(/4) = cos(/2) = 0, it follows from claim 3 that ²/16 is irrational and therefore that is irrational.

On the other hand, since

:$\tan x=\frac=x\frac,$

another consequence of Claim 3 is that, if ''x'' ∈ Q \ , then tan ''x'' is irrational.

Laczkovich's proof is really about the hypergeometric function. In fact, ''f''_''k''(''x'') = ₀''F''₁(''k''; −''x''²) and Gauss found a continued fraction expansion of the hypergeometric function using its functional equation. This allowed Laczkovich to find a new and simpler proof of the fact that the tangent function has the continued fraction expansion that Lambert had discovered.

Laczkovich's result can also be expressed in Bessel functions of the first kind ''J''_''ν''(''x''). In fact, Γ(''k'')''J''_{''k'' − 1}(2''x'') = ''x''^{''k'' − 1}''f''_''k''(''x''). So Laczkovich's result is equivalent to: If ''x'' ≠ 0 and if ''x''² is rational, then

:$\forall k\in\Q\smallsetminus\: \qquad \frac\notin\Q.$

See also

* Proof that e is irrational
* Proof that is transcendental

References

{{DEFAULTSORT:Proof That Pi Is Irrational
Pi
Article proofs
Irrational numbers