數學知識交流--無理數(1)

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問：

請prove root2係無理數,用數學方法計

答：

這問題很多人問過，我不敢全自己答案，參一下 website .

反証法就一定了：

若 √2 是有理數，則能表示為 p/q ( p, q 均為整數, 而且 p, q 為最簡方式, 即 p, q 是互質數 )

√2 = p/q
2 = p^2/q^2
2q^2 = p^2

由上式看到 p 必為雙數，設 p = 2r

2 (q^2) = (2r)^2 = 4r^2
q^2 = 2r^2

由上式看到 q 必為雙數。

可是，如 p, q 也是雙數，便不會是互質數。
所以，証明最初的假設有誤，亦即開方 2 不是有理數。

( 可參

(1) http://zh.wikipedia.org/zh-hk/%E7%84%A1%E7%90%86%E6%95%B8
(2) http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php
(3) http://en.wikipedia.org/wiki/Square_root_of_2
(4) http://zh.wikipedia.org/zh-hk/2%E7%9A%84%E7%AE%97%E8%A1%93%E5%B9%B3%E6%96%B9%E6%A0%B9 )

2011-09-02 22:06:14 補充：
期待 LII 的無理數(2)~~

感謝3位大大的回答~
現在我明白了prove root2 ,但仲有一條問題~
http://hk.knowledge.yahoo.com/question/question?qid=7011090200912&mode=w&from=question&recommend=0&.crumb=9bZTfOJOl6z

2011-09-02 22:12:38 補充：
haha,已經出左,不過如果照呢個方法直做好似唔得,要轉轉方法?

Let√2=a/b, which there is no common factor between a and b ---[#]

√2=a/b
2=a²/b²
a²=2b² <--[a² has a factor: 2]
Since only even number's square come into even number,
so a is also a even number, which has the factor: 2

So, let a=2c, which c is a natural number.
a²=2b²
(2c)²=2b²
4c²=2b²
b²=2c² <--[b² has a factor: 2]
Since only even number's square come into even number,
so b is also a even number, which has the factor: 2

So, a and b have the common factor:2--- [@]

Since [#] and [@] must not come out at the same time,
so √2≠a/b, which is a irrational number.

Let √2 is rational, i.e. √2 = m/n and m,n are co-prime and m,n have the same signs
√2 = m/n => 2 = m^2 /n^2
But since m is co-prime from n, i.e. m^2 ,n^2 is co-prime,
therefore, contradiction arises since m^2 /n^2 = 2 => m^2 = 2n^2
Therefore, √2 is irrational