Show that any right angled triangle with hypotenuse 1 may be approximated arbitrarily close by one with rational sides

Question

Taken from Mathematics and Its History Book by John Stillwell, page 8.

Please check if there are any flaws in my answer (have I actually understood the meaning of "arbitrarily close" and also, if you could suggest a more elementary answer, I would appreciate it.

Solution

Each and every possible right angled triangle with hypotenuse $1$, and sides $x$ and $y$ can be uniquely indexed with one of it's angle, $\theta$ defined between the side $x$ and the hypotenuse.

For $0 < \theta < \frac{\pi}{2}$, it covers all possible such triangles.

$$\{(x,y) | x^2 + y^2 = 1\} \sim \{\theta | (\sin \theta)^2 + (\cos \theta)^2 = 1\} $$

Thus we need to show that it is possible to find 2 rational sides $p$ and $q$ for any given level of precision (say n digits) such that the error between them and the real sides is less than $10^{-n}$. Note they say that $n$ can be arbitrarily chosen. i.e.

$$|p-\sin \theta| < 10^{-n} $$ and $$|q-\cos \theta| < 10^{-n} $$

There exists such a $p,q \in \mathbb{Q}$ for any $n \in \mathbb{N}$ and $0 < \theta < \frac{\pi}{2}$

We can show this using the Taylor series of $\sin$ and $\cos$, we can evaluate upto the required number of terms until we reach the required arbitrary threshold.

$${\displaystyle {\begin{aligned}\sin(x)&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots \\[8pt]&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}\\[8pt]\end{aligned}}}$$

$${\displaystyle {\begin{aligned}\cos x&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots \\&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}\end{aligned}}}$$

Since the finite sum of rational numbers are sure to be rational, the desired $p$ will be the truncated taylor series of $\sin \theta$ and $q$ will be the truncated taylor series of $\cos \theta$.

Answer 1

As pointed out in a comment, your proof is incomplete if not flawed, because the condition $p^2+q^2=1$ is not justified/satisfied.

An easier way to construct $(p,q)$ is to use Pythagorean triples of the form $(2mn,\, m^2-n^2,\, m^2+n^2)$. For any fixed positive integer $m$, let $p(n)=\frac{2mn}{m^2+n^2}$. It can be shown that $0\le p(n)-p(n-1)\le\frac2m$ for every $n\in\{1,2,\ldots,m\}$. Hence $p(n)$ can be made arbitrarily close to any $a\in(0,1)$. In turn, $q(n)=\frac{m^2-n^2}{m^2+n^2}=\sqrt{1-p(n)^2}$ is arbitrarily close to $b=\sqrt{1-a^2}$.

Answer 2

Start with $(1-t)^2+(2t)^2 =(1+t)^2 $, so that $(\frac{1-t}{1+t})^2+(\frac{2t}{1+t})^2 =1 $.

To approximate the right triangle with sides $(\sin z, \cos z, 1)$, we want $\sin z \approx \frac{1-t}{1+t} $ or $(1+t)\sin z \approx 1-t $ or $t(1+\sin z) \approx 1-\sin z $ or $t \approx \frac{1-\sin z}{1+\sin z} $.

Choose $t$ as a rational close to $\frac{1-\sin z}{1+\sin z} $ and the resulting right triangle will be rational and close to the original right triangle.

I'll let others work out how close if actually enough.

Answer 3

EDIT: This proof is incorrect. The fact that $y^{2} \in \mathbb{Q}$ does not guarantee that $y \in \mathbb{Q}$.

I am working through Stillwell's book too, and came up with this solution. I would also appreciate any comments on my proof.

Let $a, b \in \mathbb{R}$, so that $a^{2}+b^{2}=1$. If $a, b \in \mathbb{Q}$, then we are done, so the interesting case is when $a, b \in \mathbb{I}$.

By the density of $\mathbb{Q}$ in $\mathbb{R}$, we can find a rational between two real numbers. In particular, let $\epsilon > 0$ be the approximation threshold, making sure that $a-\epsilon>0$, and $a+\epsilon<1$.

Let $x \in \mathbb{Q}$ so that $a-\epsilon < x < a+\epsilon \ $ (by the density mentioned above). Because of the way we chose $\epsilon$, it holds that $x^{2} \in (0,1)$. Finally, let $y^{2} = 1 - x^{2}$. Since both $x$ and $y$ are rational, we proved the desired result.