那些数学悖论

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1+1=3

$4 - 10 + 6\frac{1}{4} = 9 - 15 + 6\frac{1}{4}$ (两边同时 $ + 6\frac{1}{4}$ ) ${2}^{2} - 2 \times 2 \times \frac{5}{2} + {(\frac{5}{2})^{2}} = {3}^{3} - 2 \times 3 \times \frac{5}{2} + {(\frac{5}{2})^{2}} $ (等换等) ${(2 - \frac{5}{2})}^{2} = {(3 - \frac{5}{2})}^{2}$ (完全平方公式) $2 - \frac{5}{2} = 3 - \frac{5}{2}$ (两边同时开平方) $2 = 3$ (两边同时 $+ \frac{5}{2}$ ) $1 + 1 = 3$ (等换等) 来，找错吧 # $ ∞ = -\frac{1}{12}

老悖论了\ 先问大家一个问题：1 + 2 + 3 + 4 + 5 + \dots = ?\ 看上去是∞是吧，那看好了：\ 令S = 1 + 2 + 3 + 4 + 5 + \dots，\

$B = 1 - 2 + 3 - 4 + 5 - \dots$。 则$A = 1 - (1 - 1 + 1 - 1 + 1 - 1 + \dots) = 1 - A$\ 解此方程得：$A = \frac{1}{2}

则B = 1 - (1 + 1) + (2 + 1) - (3 + 1) + (4 + 1) -\dots = 1 - 1 - 1 + 2 + 1 - 3 - 1 + 4 + 1 - \dots = (1 - 1 + 1 - 1 + 1 -\dots) - (1 - 2 + 3 - 4 + \dots) = A - B\ 解此方程得：B = \frac{1}{4}

则S = 1 + 2 + 3 + 4 + 5 + \dots

S - B = (1 - 1) + (2 + 2) + (3 - 3) + (4 + 4) + (5 - 5) + \dots = 4 + 8 + 12 + 16 + 20 + \dots = 4S

解此方程得： S = -\frac{1}{12} \ So...why?\ 为什么一个看似是∞的数其实 = -\frac{1}{12} ？\ 来，找错吧 \times 2

1 + 1 = -1

\begin{aligned}1 + 1 &= 1^2 + 1^2 \\ &= 1^2 + 2 \times 1 \times 1 + 1^2 - 2 \times 1 \times 1 \\ &= (1 + 1)^2 - 2 \times 1 \times 1 \\ &= (1 + 1)^2 - 2\end{aligned}

令 a = 1 + 1，则上式可改写成 a = a^2 - 2。\ 解得 a_1 = 2 , a_2 = -1。\ 带入 a = 1 + 1 得：\

来，找错吧$ \times 3

1 = -1

\sqrt{(-1)})^2 \\ &= -1 \end{aligned}

来，找错吧 \times 4