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

網友 Koopa 提供了兩個方法證明 \sqrt{10} 是無理數[這(竟)是 2008-AL-Pure Mathematics (I) Q.11(a)],非常感謝。我也提供兩個方法,相信大多數考生也會想到的。

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。)

相信還有其他證法,歡迎提供。讓我順道舊帖重貼。

張貼時間:2003-04-11 02:04:03
『數學平常談:證某類無理數的小法』@濟濟一堂學術討論區

易知 \sqrt{2} 是無理數,那麼 , … 諸如此類又是否無理數呢?如何證明?原來我們可以利用以下命題作一併考慮。

命題

若 α 是以下整系數多項式方程(polynomial equation with integral coefficients)的有理根(rational root),

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

則 α 必然是整數。

證明

若 α 是 f(x) = 0 的一個有理根;
若 α = 0,則命題成立。
若 α ≠ 0 ,我們可設 α = \frac{a}{b},(其中 a,b 除 ±1 外再無別的公 因數,即是說 \frac{a}{b} 已是約至最簡。)從而有

(\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})

因 RHS 是整數,故 LHS 亦是整數,但因 a,b 除 ±1 外再無別的公因數,迫使 b 必然要是 ±1 ,從而 α 必然是整數。(Q.E.D.)

有了上述命題,我們輕易證明 \sqrt{2} 是無理數。

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

可見,\sqrt{2} 是上式的一個根(root)。(留意到上式是leading coefficient 為 1 的整系數多項式方程。)

利用反證法,若 \sqrt{2} 是有理數,即 \sqrt{2} 是上式的有理根,由命題,即 \sqrt{2} 是整數,但這是無可能,故 \sqrt{2} 是無理數。

又例如

證明 是無理數。

x =
x^7 - 1997 = 0

是上式的一個根, 由命題,若 是有理根,它必是整數;但這是無可能的(\because 128 < 1997 < 2187 => 2 < < 3),故 必是無理數。

習題:試證 是無理數。

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 | 回覆

  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 | 回覆

  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 | 回覆


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