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転送 [[]]== 計算の順序 ==
テトレーションは
結合法則
を満たさないため、計算の順序を変えると値が変わってしまいます。
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{\displaystyle {\begin{aligned}c\uparrow ^{2}\left(b\uparrow ^{2}a\right)=&c\uparrow ^{2}\left(\underbrace {b\uparrow b\uparrow b\cdots \uparrow b} _{a}\right)=c\uparrow ^{2}\underbrace {b^{b^{b^{\vdots ^{b^{b}}}}}} _{{\text{高さ}}a}=\underbrace {c^{c^{c^{\cdot ^{\cdot ^{\cdot ^{c^{c^{c}}}}}}}}} _{{\text{高さ}}\underbrace {b^{b^{b^{\vdots ^{b^{b}}}}}} _{{\text{高さ}}a}}\\\left(c\uparrow ^{2}b\right)\uparrow ^{2}a{~\,}=&\underbrace {\left(c\uparrow ^{2}b\right)\uparrow \left(c\uparrow ^{2}b\right)\cdots \uparrow \left(c\uparrow ^{2}b\right)\uparrow \left(c\uparrow ^{2}b\right)} _{a}=\left(c\uparrow ^{2}b\right)\uparrow \underbrace {\left(c\uparrow ^{2}b\right)\cdots \uparrow \left(c\uparrow ^{2}b\right)\uparrow \left(c\uparrow ^{2}b\right)} _{\left(a-1\right)}\\=&\left(c\uparrow ^{2}b\right)\uparrow \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-1\right)\\=&\left(\underbrace {c\uparrow c\uparrow c\cdots \uparrow c\uparrow c} _{b}\right)\uparrow \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-1\right)\\=&\left(c\uparrow \underbrace {c\uparrow c\cdots \uparrow c\uparrow c} _{\left(b-1\right)}\right)\uparrow \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-1\right)\\=&\left(c\uparrow c\uparrow ^{2}\left(b-1\right)\right)\uparrow \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-1\right)\\=&c\uparrow \left(c\uparrow ^{2}\left(b-1\right)\times \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-1\right)\right)\\=&c\uparrow \left(c\uparrow ^{2}\left(b-1\right)\times \left(\underbrace {c\uparrow c\uparrow c\cdots \uparrow c\uparrow c} _{b}\right)\uparrow \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-2\right)\right)\\=&c\uparrow \left(c\uparrow ^{2}\left(b-1\right)\times \left(c\uparrow \underbrace {c\uparrow c\cdots \uparrow c\uparrow c} _{\left(b-1\right)}\right)\uparrow \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-2\right)\right)\\=&c\uparrow \left(c\uparrow ^{2}\left(b-1\right)\times \left(c\uparrow c\uparrow ^{2}\left(b-1\right)\right)\uparrow \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-2\right)\right)\\=&c\uparrow \left(c\uparrow ^{2}\left(b-1\right)\times c\uparrow \left(c\uparrow ^{2}\left(b-1\right)\times \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-2\right)\right)\right)\\=&c\uparrow \left(c\uparrow c\uparrow ^{2}\left(b-2\right)\times c\uparrow \left(c\uparrow ^{2}\left(b-1\right)\times \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-2\right)\right)\right)\\=&c\uparrow \left(c\uparrow \left(c\uparrow ^{2}\left(b-2\right)+\left(c\uparrow ^{2}\left(b-1\right)\times \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-2\right)\right)\right)\right)\\=&c\uparrow \left(c\uparrow \left(c\uparrow ^{2}\left(b-2\right)+\left(c\uparrow ^{2}\left(b-1\right)\times \left(c\uparrow ^{2}b\right)\uparrow \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-3\right)\right)\right)\right)\\=&c\uparrow \left(c\uparrow \left(c\uparrow ^{2}\left(b-2\right)+\left(c\uparrow ^{2}\left(b-1\right)\times \left(c\uparrow c\uparrow ^{2}\left(b-1\right)\right)\uparrow \left(c\uparrow ^{2}b\right)\uparrow ^{2}\left(a-3\right)\right)\right)\right)\end{aligned}}}
--beautiful icosagon 2020年5月5日 (火) 17:01 (UTC)
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