# Is it possible to manipulate Niven's proof of the irrationality of $\pi$ to prove the irrationality of $\sqrt{2}$?

Section 2 of Keith Conrad's note "Irrationality of $$\pi$$ and $$e$$" (PDF link via uconn.edu) recounts Ivan Niven's proof of irrationality of $$\pi$$. (See also Niven's original note "A simple proof that $$\pi$$ is irrational" (PDF link via ProjectEuclid.org).)

Is it possible to manipulate this proof to prove the irrationality of, say, $$\sqrt{2}$$?

• "However, there is an essential difference between proofs that √ 2 is irrational and proofs that π is irrational. One can prove √ 2 is irrational using only algebraic manipulations with a hypothetical rational expression for √ 2 to reach a contradiction. But all known proofs of the irrationality of π are based on techniques from calculus, which can be used to prove irrationality of other numbers, such as e and rational powers of e (aside from e 0 = 1)" Commented Feb 11, 2019 at 1:48
• I like this question. It seems you can adapt the proof but I need to look into it a bit more. Commented Feb 11, 2019 at 2:08
• @fleablood, I know proving the irrationality of numbers like $\sqrt{2}$ by Niven's technique is not a good idea. But I think If we can manipulate his proof for $\sqrt{2}$, we may try for some other 'good' numbers.
– ersh
Commented Feb 11, 2019 at 2:35
• @ersh: I added a link to a copy of Niven's original note, which seems (to me, anyway) easier to follow. Even so, it's bad form to require readers to traverse external links to understand a question. Given that Niven's argument is quite short, I recommend duplicating it within your question.
– Blue
Commented Mar 9, 2020 at 4:33

For to make the things more clear I rewrite Niven's proof a bit more simplified.

Let $$\pi=\frac{a}{b}$$, with $$a,b\in\textbf{N}$$. I define the function $$f(x)=\frac{x^n(a-bx)^n}{n!},$$ where $$n$$ is to be determined later. Proceeding it holds $$f\left(\frac{a}{b}-x\right)=f(x)\textrm{, hence }f(\pi-x)=f(x)$$ I define $$F(x)=f(x)-f''(x)+f^{(4)}(x)-\ldots+(-1)^n f^{(2n)}(x).$$ Then $$n!f(x)$$ is of the form $$A_1x^{2n}+A_2x^{2n-1}+\ldots+A_{n+1}x^n\textrm{, }A_k\in\textbf{Z}$$ Hence $$f(x),f'(x),f^{(2)}(x),\ldots,f^{(2n)}(x)$$, for $$x=0$$ are all integers. Hence $$f(\pi)$$ are also integers via. $$f(x-\pi)=f(x)\Rightarrow f(\pi)=f(0)$$. But
$$\frac{d}{dx}(F'(x)\sin x-F(x)\cos x)=F''(x)\sin x+F(x)\sin x=f(x)\sin x.$$ Hence $$\int^{\pi}_{0}f(x)\sin x dx=\left[F'(x)\sin x-F(x)\cos x\right]^{\pi}_{0}=F(\pi)+F(0)=\textrm{integer}$$ But for $$0, we have $$0 Hence $$\int^{\pi}_{0}f(x)\sin xdx\leq \frac{\pi^na^n}{n!}\pi$$ and when $$n$$ very large we have (clearly) $$0<\textrm{integer}<1$$ Contradiction. Hence $$\pi$$ is not rational. QED

If we assume that $$\xi$$ is any non rational number and we want to use Niven's proof for it, then we assume the integral $$A_n(x)=\int_{(n)}A_0(x)(dx)^{(n)}=\underbrace{\int\int\ldots\int}_{n-times}A_0(x)\underbrace{dxdx\ldots dx}_{n-times},$$ with some normalization conditions in the limit points of integration i.e. $$\int=\int^{x}_{c}$$ for some constant $$c$$ or even sequence $$c_1,c_2,\dots,c_n$$. Then seting as Niven $$\xi=\frac{a}{b}$$, $$a,b$$ posotive integers such $$gcd(a,b)=1$$ and $$a (for simplicity): $$f(x)=\frac{x^n(a-bx)^n}{n!}$$ and $$\int^{\xi}_{0}f(t)A_0(t)dt=\int^{\xi}_{0}f(t)A_1'(t)dt=f(\xi)A_1(\xi)-f(0)A_1(0)-\int^{\xi}_{0}f'(t)A_1(t)dt$$ also $$\int^{\xi}_{0}f'(t)A_1(t)dt=\int^{\xi}_{0}f'(t)A_2'(t)dt=f'(\xi)A_2(\xi)-f'(0)A_2(0)-\int^{\xi}_{0}f''(t)A_2(t)dt$$ $$\ldots$$ $$\int^{\xi}_{0}f^{(k-1)}(t)A_{k-1}(t)dt=\int^{\xi}_{0}f^{(k-1)}(t)A'_{k}(t)dt=f^{(k-1)}(\xi)A_{k}(\xi)-f^{(k-1)}(0)A_k(0)-$$ $$-\int^{\xi}_{0}f^{(k)}(t)A_k(t)dt.$$ Hence $$\int^{\xi}_{0}f(t)A_0(t)dt=\sum^{2n}_{k=n}f^{(k)}(0)\left(A_{k+1}(\xi)-A_{k+1}(0)\right)\tag 1$$ Hence if we assume that

I.

$$A_0(x)$$ is continuous and $$0, for some constant $$C$$.

II. $$\sum^{2n}_{k=n}f^{(k)}(0)\left[A_ {k+1}(\xi)-A_{k+1}(0)\right]\in\textbf{Z}\textrm{, }n=0,1,2,\ldots$$ Then holds (1) and $$0 hence $$0<\int^{\xi}_{0}f(t)A_0(t)dt\leq \frac{\pi^n a^n\xi C}{n!}$$ Hence again for $$n$$ large we have $$0<\textrm{integer}<1,$$ which is contradiction. Hence $$\xi$$ will then be irrational.

NOTE that

1.

$$f(x)=\frac{x^n(a-bx)^n}{n!},$$ is the same function as in Niven's proof.

1. If we replace $$A_0(x)=\sin x$$ and $$\xi=\pi$$, we get Niven's proof.

2. An application of the above method possibly can be found, if we assume $$A_0(x)=\sin(\log(x))$$, $$\xi=e^{\pi}$$.

• Great. However, I still wonder what should be our choice $A_0(x)$ for $\sqrt{2}$?
– ersh
Commented Mar 11, 2020 at 16:49
• You can take $\xi=\sqrt{2}$. If you can find any function $A_0(x)$ such that $I$ and $II$ hold for this constant ($\xi=\sqrt{2}$), then you are done. Commented Mar 11, 2020 at 20:21
• If for example $A_0(x)=x^2$, then $A_1(x)=\frac{x^3}{3}$, $A_2(x)=\frac{x^4}{12},\ldots$. But $A_k(\sqrt{2})$ is not for all $k=1,2,\ldots$ integer. Hence this $A_0(x)$ does not work. In the ''example'' of $\pi$ we have $A_0(x)=\sin x$, with $A_1(x)=-\cos x+c_1$, $A_2(x)=-\sin x+c_2,\ldots$ and in general $A_k(\pi)=integer$. Hence it works and thus $\pi$ is irrational. Commented Mar 11, 2020 at 20:42