Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question
6  
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. –  Chris Eagle Feb 16 '13 at 7:11
add comment

2 Answers

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.

share|improve this answer
    
@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 $\int_0^m f(x)\cos(x)$. –  Yimin Feb 16 '13 at 6:37
2  
Actually, it is not a polynomial either, think $\sin 2\theta$. The cosine function is better behaved! –  André Nicolas Feb 16 '13 at 6:37
    
@AndréNicolas, oh,yes, it is for $\cos$. I will modify the proof for $\cos$ –  Yimin Feb 16 '13 at 6:38
3  
@Yimin: This very closely resembles Niven’s proof of the irrationality of $ \pi $. –  Haskell Curry Feb 16 '13 at 8:29
    
@HaskellCurry Yes, I think so, I have read his proof before, and I spent some time to re-prove this and wrote down. –  Yimin Feb 16 '13 at 8:32
show 2 more comments

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.

share|improve this answer
    
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" –  Dominic Michaelis Feb 16 '13 at 6:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.