0.99...9... ≠ 1

非正函數學中，有一個證明是這樣的: 

]Prove.. 0.99...9... = 1

]If.. 0.99...9... = x

]10x = 9.99...9...

]10x - x = 9.99...9... - 0.99...9... = 9 = 9x

]x = 1 ......Proved

這個證明試圖說明循環小數0.99...9... 會與1 等值。

然而這個證明是錯誤的，在正函數學中，我們可以用極限的概念反駁上面的證明。

]Prove.. 0.99...9... ≠ 1

]If.. 0.99...9... = a

]a = lim(h→∞) Sum(k = 1 to h, 9 * 10^(-k))

]

]10a = lim(h→∞) Sum(k = 1 to h, 9 * 10^(-k)) * 10

]= lim(h→∞) Sum(k = 1 to h, 9 * 10^(-k + 1))

]= lim(h→∞) Sum(k = 0 to h - 1, 9 * 10^(-k))

]

]10a - a = 9a = lim(h→∞) Sum(k = 0 to h - 1, 9 * 10^(-k)) - Sum(k = 1 to h, 9 * 10^(-k))

]= lim(h→∞) Sum(k = 0 to 0, 9 * 10^(-k)) - Sum(k = h to h, 9 * 10^(-k))

]= lim(h→∞) 9 * 10^(-0) - 9 * 10^(-h)

]= lim(h→∞) 9 - 9 * 10^(-h)

]

]a = 9a / 9 = lim(h→∞) (9 - 9 * 10^(-h)) / 9 = lim(h→∞) 1 - 10 ^(-h)

]

]if (0.99...9... = 1) then.. 

]0.99...9... = a = 1

]a = lim(h→∞) a - 10^(-h)

]0 = lim(h→∞) -10^(-h)

]H = stone(x) (-10^x = 0) = stone(x) (10^x = 0 ||x ∈ Real||)

]Am(H) = 0 .......(0.99...9... = 1) is false

]Proved.. 0.99...9... ≠ 1