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如何證明 root2 is an irrational no.

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1#
發表於 3/9/2005 04:01 PM | 只看該作者 回帖獎勵 |倒序瀏覽 |閱讀模式
請大家指教[E-QuM][E-QuM][E-QuM]
2#
發表於 3/9/2005 08:58 PM | 只看該作者
我看過一本書說明了Irrationality of square root 2的Proof。
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Steps for proving square root 2 is irrational:
1) Assume square root 2 is rational.
2) If square root 2 is rational, then we can find whole numbers p and q such that (root2)=p/q by definition of rational number.
3) Any fraction p/q is equal to some fraction r/s where r and s are not both even.
4) Therefore, if square root 2 is rational, then we can find whole numbers r and s, not both even, such that (root2)=r/s.
5) If (root2)=r/s, then 2=r^2 / s^2 by squaring both sides.
6) r is a whole number and r^2 is even. We would like to show that r must also be even. Let us assume that r is odd and seek a contradiction.
7) Let us call a whole number r good if it is either a multiple of two or one more than a multiple of two. If r is good, then r=2s or r=2s+1, where s is also a whole number. If r=2s then r+1=2s+1, and if r=2s+1, then r+1=2s+2=2(s+1). Either way, r+1 is also good.
8) 1 is good, since 0=0x2 is a multiple of 2 and 1=0+1.
9) Applying step 8 repeatly, we can deduce that 2 is good, then that 3 is good, then that 4 is good, and so on.
10) Therefore, every positive whole number is good, as we were trying to prove.
11) Back to step 6, since r is odd, there is a whole number t such that r=2t+1.
12) It follows that r^2=(2t+1)^2=4t^2+4t+1.
13) But 4t^2+4t+1=2(2t^2+2t)+1, which is odd, contradicting the fact that r^2 is even.
14) Therefore r is even.
15) If r is even, then r=2t for some whole number t by the definition of even number.
16) If 2s^2=r^2 and r=2t, then 2s^2=(2t)^2=4t^2, from which it follows that s^2=2t^2, then s^2 is even, which means that s is evem.
17) Under the assumption that square root 2 is rational, we have shown that (root2)=r/s, with r and s not both evem. We have then shown that r is even and that s is even. This is a clear contradiction. Since the assumption that square root 2 is rational has consequences that are clearly false, the assumption itself must be false. Therefore, square root 2 is irrational.
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Copyright by Oxford University Press
Copy from the book "A Very Short Introduction - Mathematics"
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