17
$\begingroup$

Is there a way to show, that the only solution of $$\sin(x)=y$$ is $x=y=0$ with $x,y\in \mathbb{Q}$.

I am seaching a way to prove it with the things you learn in linear algebra and analysis 1+2 (with the knowledge of a second semester student).

$\endgroup$
1
  • 6
    $\begingroup$ This is at least as hard as proving that $\pi$ is irrational, so I doubt you'll get an answer at the level you want. $\endgroup$ Commented Feb 16, 2013 at 7:11

2 Answers 2

10
+50
$\begingroup$

Sorry for the previous spam. I shall prove for $\cos$, cosine of any rational numbers except for 0 cannot get rational numbers.

By using polynomial argument, we shall only have to prove for integers.

Suppose that $m\in\mathbb{N}$, $\cos(m)\in\mathbb{Q}$.

For any fixed prime number $p>m$, consider polynomial $x\in(0,m)$ $$f(x) = \frac{(x-m)^{2p}(m^2-(x-m)^2)^{p-1}}{(p-1)!}$$

And $$F(x) = \sum_{n = 0}^{2p-1} (-1)^{n+1}f^{2n}(x)$$ Which satisfies $$(F'(x)\sin(x)-F(x)\cos(x))' = F''(x)\sin(x) +F(x)\sin(x) = f(x)\sin(x)$$ since the other terms cancelled. $$\int_0^mf(x)\sin(x)dx = F'(m)\sin(m)-F(m)\cos(m)+F(0)$$ Consider $f$ is a polynomial of $(x-m)^2$, thus $F'(m) = 0$, and we can see that $$f(m-x) = x^{2p}(m^2-x^2)^{p-1}/(p-1)!$$ By computing, $p|f^{(l)}(m)$ for every $l$. That means $F(m)$ is a multiple of $p$ by definition of $F$, say $pM$.

If $\cos(m) = s/t$, then $$t\int_0^m f(x)\sin(x)dx = -spM+tN$$ is an integer, While $$f\le \frac{m^{4p-2}}{(p-1)!} $$ thus $$t\int_0^mf(x)\sin(x)dx\le t\frac{m^{4p-2}}{(p-1)!}\cdot m <1 $$ when $p$ is large enough. Contradiction of it is an integer.

As a result of this, $\sin$ should also satisfy this.

$\endgroup$
7
  • $\begingroup$ @DominicMichaelis To prove this you might need to use function $f(x) = \frac{x^p(m-x)^{2p}(2m-x)^{p-1}}{(p-1)!}$ and $F(x) = f(x)-f^{(2)}(x) + f^{(4)}(x) -\dots +f^{(4p-2)}(x)$. And give a contradiction for $\int_0^m f(x)\cos(x)$. $\endgroup$
    – Yimin
    Commented Feb 16, 2013 at 6:37
  • 2
    $\begingroup$ Actually, it is not a polynomial either, think $\sin 2\theta$. The cosine function is better behaved! $\endgroup$ Commented Feb 16, 2013 at 6:37
  • $\begingroup$ @AndréNicolas, oh,yes, it is for $\cos$. I will modify the proof for $\cos$ $\endgroup$
    – Yimin
    Commented Feb 16, 2013 at 6:38
  • 3
    $\begingroup$ @Yimin: This very closely resembles Niven’s proof of the irrationality of $ \pi $. $\endgroup$ Commented Feb 16, 2013 at 8:29
  • $\begingroup$ @HaskellCurry Yes, I think so, I have read his proof before, and I spent some time to re-prove this and wrote down. $\endgroup$
    – Yimin
    Commented Feb 16, 2013 at 8:32
8
$\begingroup$

One can hit it with a really heavy-duty theorem, the Lindemann-Weierstrass theorem. It is a consequence of this that if $x$ is non-zero algebraic, then $\sin x$ is transcendental.

Ivan Niven, in his book Rational and Irrational numbers, has a nice proof, "elementary" but not entirely simple, of the fact that the sine of a non-zero rational is irrational. That proof does not make any real use of linear algebra.

$\endgroup$
1
  • $\begingroup$ I know that theorem, but this problem was hidden in a linear algebra 2 exam due to a typo, so i wanted to know if it had been possible to solve it without "cheating" $\endgroup$ Commented Feb 16, 2013 at 6:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .