# Quod Erat Demonstrandum

## 2008/04/12

### 數學平常談：證某類無理數的小法@濟濟一堂

Filed under: HKALE,Pure Mathematics — johnmayhk @ 4:26 下午

Let $\sqrt{10} = \frac{p}{q}$ where $p, q$ are non-zero integers and there is no common factor for $p$ and $q$ except $\pm 1$.

(method 1)

$\sqrt{10} = \frac{p}{q}$
$\Rightarrow p^2 = 10q^2$ – – – – – – (*)
$\Rightarrow p$ is even
$p = 2n$ for some even number $n$
Sub into (*)
$4n^2 = 10q^2$
$2n^2 = 5q^2$
$\Rightarrow q$ is even
Hence there is a common factor 2 for $p$ and $q$, contradiction.

(method 2)

By (*)
$p^2 = (10q)(q)$
$\because$ there is no common factor for $p$ and $q$ except $\pm 1$
This forces that $q = \pm 1$
Hence $p^2 = 10$, but $10$ is not a perfect square. （所以不止 10，任何非平方數的開方必然是無理數。）

『數學平常談：證某類無理數的小法』@濟濟一堂學術討論區

$f(x) = x^n + a_1x^{n-1} + \dots + a_{n-1}x + a_n = 0$

$(\frac{a}{b})^n + a_1(\frac{a}{b})^{n-1} + \dots + a_{n-1}(\frac{a}{b}) + a_n = 0$
$a^n + a_1a^{n-1}b + \dots + a_{n-1}ab^{n-1} + a_nb^n = 0$
$\frac{a^n}{b} = -(a_1a^{n-1} + \dots + a_{n-1}ab^{n-2} + a_nb^{n-1})$

$x = \sqrt{2}$
$x^2 - 2 = 0$

$x =$
$x^7 - 1997 = 0$

## 3 則迴響 »

1. Here is an easy way to prove 2^{1/3} + sqrt{3} is irrational. Using rational root theorem requires too many calculations.

Suppose 2^{1/3} + sqrt{3} = q is rational. Then we have 2^{1/3} is in Q(sqrt{3}).
Which implies Q(2^{1/3}) is contained in Q(sqrt{3}), a clear contradiction by degree considerations.

迴響 由 koopa — 2008/04/13 @ 2:16 上午 | 回應

2. Yes, $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(\sqrt{3})$ are of degree 3 and 2 respectively. Or it is quite easy to see $\sqrt[3]{2}$ is not the linear combination of 1 and $\sqrt{3}$. Just give another elementary way.

$\sqrt[3]{2} + \sqrt{3} = q$
$\Rightarrow \sqrt[3]{2} = q - \sqrt{3}$
$\Rightarrow 2 = (q - \sqrt{3})^3$
$\Rightarrow 2 = s + t\sqrt{3}$ for some $s, t \in \mathbb{Q}$ (with $t \ne 0$)
$\sqrt{3} = \frac{2 - s}{t} \in \mathbb{Q}$ impossible.

But, of course, koopa’s way is shorter.

迴響 由 johnmayhk — 2008/04/13 @ 3:36 下午 | 回應

3. An interesting proof of “e is irrational", from a past paper?

Let In = Int(1,0)[(e^t)(t^n)] dt,

Prove e is irrational by relating In and e/n

Try to do it. haha.

迴響 由 Cheung Hoi Kit — 2008/11/16 @ 9:01 下午 | 回應