標題: 如何證明 root2 is an irrational no. [打印本頁] 作者: 法皇拿破崙 時間: 3/9/2005 04:01 PM 標題: 如何證明 root2 is an irrational no. 請大家指教[E-QuM][E-QuM][E-QuM]作者: oscar 時間: 3/9/2005 08:58 PM 我看過一本書說明了Irrationality of square root 2的Proof。 -------------------------------------------------------------------------------------------------------- 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. -------------------------------------------------------------------------------------------------------- Copyright by Oxford University Press Copy from the book "A Very Short Introduction - Mathematics"