Quod Erat Demonstrandum

2016/08/29

講兩題

Filed under: Junior Form Mathematics,NSS — johnmayhk @ 5:02 下午
Tags: , ,


seize-the-moment-geek-green

對,seize the moment,抓緊此際講兩題。

(注:上圖式子是 k^{th} moment 的定義。特別地,當 k=2,它就是方差 variance。)

(一)

修物理的同學一定學過透鏡公式(lens formula)

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

它描述了 f 焦距(focal length),u 物距(object distance)及 v 像距(image distance)三者關係。

如何證明?以直線方程(equation of straight lines)輕易破之,從略。(同學試試)

以下講數,非物理。

「某倒數等於另外兩個倒數和」這種關係,不難見於幾何圖像,如

其中 AC 交 BD 於 F,E 在 AB 上,且 AD // EF // BC。

設 u = AD,f = EF,v = BC,則有

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

證明?不過是利用相似三角形:

\frac{AB}{u}=\frac{EB}{f}
\frac{AB}{v}=\frac{AE}{f}

合之,曰

AB(\frac{1}{u}+\frac{1}{v})=\frac{1}{f}(EB+AE)\Rightarrow \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

推廣一下:

其中 IE // GB // JF // HD,便有

\frac{1}{IE}+\frac{1}{HD}=\frac{1}{GB}+\frac{1}{JF}

證明?先把一些線段延伸一下:

其中 AL // CM // GB。由上面結果,

\frac{1}{a}=\frac{1}{p}+\frac{1}{s}
\frac{1}{b}=\frac{1}{r}+\frac{1}{s}
\frac{1}{c}=\frac{1}{p}+\frac{1}{q}
\frac{1}{d}=\frac{1}{r}+\frac{1}{q}

立即有:

\frac{1}{a}+\frac{1}{d}=\frac{1}{b}+\frac{1}{c}

Q.E.D.

(二)

m,n,p 為正整數,且 m+n 可被 p 整除;m^2+n^2 可被 p^2 整除;

證明

對於所有正整數 km^k+n^k 可被 p^k 整除。

(現在,以 a|b 表示 a 整除 b,例如 3|12。)

先證

p^2|mn

p|m+n\Rightarrow p^2|(m+n)^2

又,p^2|m^2+n^2,得

p^2|(m+n)^2-(m^2+n^2)\Rightarrow p^2|2mn

p 為奇數,立即得 p^2|mn

p 為偶數,由 p|m+nm,n 的奇偶(parity)相同,即同為奇數或同為偶數。

如果 m,n 同為奇,設

m=2r+1
n=2s+1
p=2t

其中 r,s,t 為整數。

p^2|m^2+n^2\Rightarrow (2t)^2|(2r+1)^2+(2s+1)^2\Rightarrow 4t^2|4(t^2+t+s^2+s)+2

這是不可能的,故

m,n 同為偶,必有正整數 u 使

三個整數 m_1=\frac{m}{2^u}, n_1=\frac{n}{2^u}, p_1=\frac{p}{2^u},其中一個為奇數。

p^2|2mnp_1^2|2m_1n_1

p_1 為奇數,立即得 p_1|m_1n_1\Rightarrow p^2|mn

p_1 為偶數,則 m_1,n_1 的奇偶不同,否則 p_1,m_1,n_1 同為偶數,有違「其中一個為奇數」這命題。但若 m_1,n_1 的奇偶不同,m_1+n_1 必為奇數,又因 p|m+n \Rightarrow p_1|m_1+n_1 不可能!

總結:p^2|mn

好了,設命題 P(k) 為 “m^k+n^k 可被 p^k 整除",已知

P(1),P(2) 為真。

假設

P(w),P(w+1) 為真,考慮

m^{w+2}+n^{w+2}

=(m+n)(m^{w+1}+n^{w+1})-mn(m^w+n^w)

p|m+n
p^{w+1}|m^{w+1}+n^{w+1}
p^2|mn
p^w|m^w+n^w

p^{w+2}|m^{w+2}+n^{w+2}

P(w+2) 也是真。證畢。

好啦,新學年,揀吓啲數學正能樣班T先,你喜歡哪件?

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1 則迴響 »

  1. (二)其實可以快D
    Assume p not equal to 1.
    Let q be a prime divisor of p.
    q^2|2mn implies q|mn which implies q|m or q|n.
    Either one with q|(m+n) implies the other.
    Now divides m,n,p by q and repeats until p=1.

    迴響 由 Sunny — 2016/09/20 @ 11:29 下午 | 回覆


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