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> 今天我们要聊的是——为什么N0 = 1(N≠0N \NEQ 0N=0)
相信有些人一直不懂为什么N0 = 1(N≠0N \NEQ 0N=0),我就来为大家解释一下:
因为 NA ÷ NB =NA-B ,那为什么 NA ÷ NB =NA-B 呢?
* 因为 nan^ana 就是
n×n×n×⋯×n×n⏟a个n\underbrace{n × n × n ×\cdots × n × n}_{a个n}a个nn×n×n×⋯×n×n
nbn^bnb 就是
n×n×n×⋯×n×n⏟b个n\underbrace{n × n × n ×\cdots × n × n}_{b个n}b个nn×n×n×⋯×n×n ,
* 那么nan^ana ÷ nbn^bnb 就等于
(n×n×n×⋯×n×n⏟a个n)(\underbrace{n × n × n ×\cdots × n × n}_{a个n})(a个nn×n×n×⋯×n×n ) ÷ (n×n×n×⋯×n×n⏟b个n)(\underbrace{n × n × n ×\cdots × n × n}_{b个n})(b个nn×n×n×⋯×n×n )
利用"前面是减除,拆括号,变符号"就等于
n×n×n×⋯×n×n⏟a个n\underbrace{n × n × n ×\cdots × n × n}_{a个n}a个nn×n×n×⋯×n×n ÷ n÷n÷n÷⋯÷n÷n⏟b个n\underbrace{n ÷ n ÷ n ÷\cdots ÷ n ÷ n}_{b个n}b个nn÷n÷n÷⋯÷n÷n ,
化简,因为 n ÷ n = 1,相互抵消,所以化简后就成了这样:n×n×n×⋯×n×n⏟a−b个n\underbrace{n × n × n ×\cdots × n × n}_{a - b个n}a−b个nn×n×n×⋯×n×n
n×n×n×⋯×n×n⏟a−b个n\underbrace{n × n × n ×\cdots × n × n}_{a - b个n}a−b个nn×n×n×⋯×n×n 就等于 na−bn^{a-b}na−b
* 所以,全部合起来就是
nan^ana ÷ nbn^bnb = (n×n×n×⋯×n×n⏟a个n)(\underbrace{n × n × n ×\cdots × n × n}_{a个n})(a个nn×n×n×⋯×n×n ) ÷ (n×n×n×⋯×n×n⏟b个n)(\underbrace{n × n × n ×\cdots × n × n}_{b个n})(b个nn×n×n×⋯×n×n ) = n×n×n×⋯×n×n⏟a个n\underbrace{n × n × n ×\cdots × n × n}_{a个n}a个nn×n×n×⋯×n×n ÷ n÷n÷n÷⋯÷n÷n⏟b个n\underbrace{n ÷ n ÷
n ÷\cdots ÷ n ÷ n}_{b个n}b个nn÷n÷n÷⋯÷n÷n = n×n×n×⋯×n×n⏟a−b个n\underbrace{n × n × n ×\cdots × n × n}_{a - b个n}a−b个nn×n×n×⋯×n×n = na−bn^{a-b}na−b
解释完nan^ana ÷ nbn^bnb = na−bn^{a-b}na−b后,我们回到正题,有一种特殊情况,就是a = b,那么 nan^ana ÷ nbn^bnb = nan^ana ÷ nan^ana = na−an^{a-a}na−a,我们分开解:
1.nan^ana ÷ nan^ana = 1(因为两个相同的数相除等于1)
2.na−an^{a-a}na−a = n0n^0n0(因为两个相同的数相减等于0)
将值带进去,就成了 1(nan^ana ÷ nan^ana) = n0n^0n0(na−an^{a-a}na−a)
所以n0n^0n0 = 1。
但是,为什么零除外呢?请看下面。
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为什么零除外呢?
* 很简单因为0a ÷ 0b,要知道 0a0^a0a 是 0×0×0×⋯×0×0⏟a个0\underbrace{0 × 0 × 0 ×\cdots × 0 × 0}_{a个0}a个00×0×0×⋯×0×0 = 0的,0b0^b0b也一样等于0(0×0×0×⋯×0×0⏟b个0\underbrace{0 × 0 × 0 ×\cdots × 0 × 0}_{b个0}b个00×0×0×⋯×0×0 = 0)
然而0 ÷ 0 没有被定义,所以0除外。
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