How Can We Prove That Pi Is Irrational Using Geometrical and Series Methods?

[Aditya89]

Hey guys, how do you prove that 'pi' is irrational? I think that it is related to infinite series? Is there any geometrical method?

 

[Tx]

The series for Pi is Gregory's Series to proove the irrationality is quite difficult, just google it.

 

[HallsofIvy]

Here's my favorite proof:

Lemma 1: Let c be a positive real number. If there exist a function, f, continuous on [0,c] and positive on (0,c) and such that f and its iterated anti-derivatives can be taken to be integer valued at both 0 and c, the c is irrational.
("can be taken to be"- we can always choose the constant of integration such that we have any given value at either 0 or c. Here we require that it the function be integer valued at BOTH 0 and c. I will post the proof separately. I remember seeing it in "Mathematics Magazine" many years ago but don't remember the author.)

Theorem $\pi$ is irrational:
f(x)= sin(x) is continuous for all x and positive on (0, $\pi$). All anti-derivatives can be taken to be $\pm sin(x)$ or $\pm cos(x)$, all of which are integer valued (0 or $\pm 1$) at $x= \pi$. Therefore, by lemma 1, $\pi$ is irrational.

One can also use lemma 1 to prove that e is irrational.

Lemma 2: If a is a positive real number, not equal to 1, such that ln(a) is rational, then a itself is irrational.
Proof: First note that ln(1/a)= -ln(a) is rational if and only if ln(a) is and 1/a is rational if and only if a is rational so it is sufficient to prove this for a> 1. (If a<1, apply the lemma to 1/a.)
If a>1 then ln(a)> 0. Suppose that ln(a) is rational and, contradicting the hypothesis, that a is rational: a= m/n reduced to lowest terms. Apply lemma 1 with c= ln(a)= ln(m/n) and f(x)= ne^x. Then f(x) is positive and continuous for all x and so for the required intervals. We can take ALL anti-derivatives to be f(x)=ne^x by taking the constant of integration to be 0. f(0)= n, an integer, and f(c)= f(ln(m/n)= ne^ln(m/n)= m, an integer. Therefore, by lemma 1, a is rational, a contradiction.

Now: Theorem: e is irrational.
e is a positive real number, not equal to 1 (since 1= e⁰ and e^x is 1 to 1). ln(e)= 1, a rational number. Therefore, by lemma 2, e is irrational.

 

[HallsofIvy]

Here is the proof of "lemma 1" above. As I said there, I remember reading it in "Mathematics Magazine" many years ago but don't remember the author. It is certainly not original with me! I find it intriguing for two reasons- first, irrationality is a numeric, not function, property, yet this depends upon calculus methods. Second, it is the "worst" kind of proof by contradiction! Contradicting the conclusion (that c is irrational) leads to two conclusions (I call them "statement A" and "statement B" below) neither of which seems to have much to do with irrationality but which contradict one another.
Lemma i: let c be a positive real number. If there exist a function f, continuous on [0,c] and positive on (0,c), such that f and all of its anti-derivatives can be taken to be integer valued at 0 and c (by appropriate choice of the constant of integration), then c is irrational.
Proof by contradiction:
First: define the set P of all polynomials, p(x), such that p and all of its derivatives are integer valued at 0 and c.
Notice "derivatives" rather than "anti-derivatives". That allows us to prove:
Lemma i: if f(x) is the function above and p(x) is any polynomial in P, then $\int_0^c f(x)p(x)dx$ is an integer.
To prove this, use repeated integeration by parts, repeatedly integrating the f "part" and differentiating the p "part". Since p is a polynomial and differentiating reduces the degree, that will eventually terminate giving the integral as a sum of anti-derivatives of f times derivatives of p, all of which are integer valued at 0 and c.
A similar proof gives
Lemma ii: the set P is closed under multiplication.
Suppose p and q are both in P. The pq is a polynomial and pq(0)= p(0)q(0) and pq(c)= p(c)q(c) are products of integers. All derivatives of pq can be done by repeated application of the product rule: every derivative is a sum of products of various derivatives of p times derivatives of q- all integer valued at 0 and c.

Now, suppose c is rational: c= m/n reduced to lowest terms. Let p₀(x)= m- 2nx. Clearly p is a polynomial. p(0)= m, an integer and p(c)= p(m/n)= m- 2n(m/n)= -m, an integer. p'(x)= -2n, an integer for all x, and all subsequent derivatives are 0. Therefore, p₀ is in P.

For i any positive integer, let $p_i(x)= \frac{(mx- nx^2)^i}{i!}$. We will prove, by induction, that p_i is in P for all i.,
If i= 1, p₁(x)= mx- nx^2= x(m- nx). p₁(0)= 0 because of that 'x' factor and p₁(c)= p₁(m/n)= 0 because of the 'm-nx' factor. p'₁= m- 2nx= p₀(x). Since that is in P we have immediately that all derivatives of p₁ are integer valued at p and so p₁ is in P.

Assume that p_i is in P for some i. $p_{i+1}= \frac{(mx-nx^2)^{i+1}}{(i+1)!}= \frac{x^{i+1}(m-nx)^{i+1}}{(i+1)!}$ is 0 at both x=0 and x= c= m/n because of the factors. $p'(x)= \frac{i(mx-nx^2)^i(m-2nx)}{(i+1)!}= \frac{(mx-nx^2)^i}{i!}(m- 2nx)= p_i(x)p_0(x)$. Since both p_i and p₀ are in P and P is closed under multiplication, so is p'_i+1- all further derivatives are integer valued at 0 and c and so p_i+1 is in P.
Since f(x) is continuous on [0,c], it takes on a maximum value there: let M= max f(x) on [0,c]. Further, since f is positive on (0, c), M> 0. Since, for all i, p_i is differentiable on [0, c] it not only takes on a maximum value but that maximum value occurs either at an endpoint (0 or c) or in the interior where p'_i(x)= 0. We have already seen that p_i is 0 at 0 and c, for all i, (and only at x= 0 or c) and that p'_i(x)= p_i-1(x)p₀(x). p'_i(x)= 0 only where p₀(x)= m- 2nx= 0 or x= m/2n= c/2. p_i(m/2n)= \frac{\left(\frac{m^2}{4n}\right)^i}{i!}. Since that is positive, that is the maximum value of p<sub>i on [0, c].

Now we can prove:
$$\int_0^c f(x)p_i(x)dx\le \int_0^cM\frac{\left(\frac{m^2}{4n}\right)^i}{i!}dx= Mc\frac{\left(\frac{m^2}{4n}\right)^i}{i!}$$
That is a constant times a constant to the i power, divided by i!. As i goes to infinity, the factorial dominates and the limit is 0- we can make that as small as we please:
Therefore, Statement A:
For some i, $\int_0^c f(x)p_i(x)dx< \frac{1}{2}$.
But since f(x) and all pi(x) are positive on (0, c) (every pi is 0 only at 0 and c and is positive at c/2),
$\int_0^cf(x)p_i(x)dx$ is a positive integer and so (statement B) must be larger than or equal to 1 for all i. That contradicts statement A and so the theorem is true.</sub>

 

[Tx]

Thanks for posting that, I've never seen a proof for 'lemma 1'.

 

[Aditya89]

Hey thanks For that!