Fix typo, thanks ZRY

Signed-off-by: szdytom <szdytom@qq.com>
This commit is contained in:
方而静 2025-07-11 11:20:54 +08:00
parent de6c54d0fa
commit 07ca699a59
Signed by: szTom
GPG Key ID: 072D999D60C6473C

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@ -28,7 +28,7 @@
$ (alpha beta) lambda &= ((a + b ii)(c + d ii))(e + f ii) \ $ (alpha beta) lambda &= ((a + b ii)(c + d ii))(e + f ii) \
&= (a c - b d + (a d + b c) ii)(e + f ii) \ &= (a c - b d + (a d + b c) ii)(e + f ii) \
&= (a c e - b d f - (a d + b c)f + (a d + b c)e) + ((a d + b c)e + (a c - b d)f) ii \ &= (a c e - b d e - (a d + b c)f) + ((a d + b c)e + (a c - b d)f) ii \
&= alpha (beta lambda) $ &= alpha (beta lambda) $
] ]
@ -69,7 +69,7 @@
根据定义,令 $alpha = a + b ii$(其中 $a,b in RR$),则取 $beta = (a / (a^2 + b^2)) - (b / (a^2 + b^2)) ii$,则有 根据定义,令 $alpha = a + b ii$(其中 $a,b in RR$),则取 $beta = (a / (a^2 + b^2)) - (b / (a^2 + b^2)) ii$,则有
$ alpha beta &= (a + b ii)(a/(a^2 + b^2) - b/(a^2 + b^2) ii) \ $ alpha beta &= (a + b ii)(a/(a^2 + b^2) - b/(a^2 + b^2) ii) \
&= (a^2 + b^2)(a^2 + b^2) \ &= (a^2 + b^2) / (a^2 + b^2) \
&= 1 $ &= 1 $
#tab 因此,这样的 $beta$ 存在。为了说明其唯一性,我们假设存在另一个 $beta'$,也满足 $alpha beta' = 1$,则有 #tab 因此,这样的 $beta$ 存在。为了说明其唯一性,我们假设存在另一个 $beta'$,也满足 $alpha beta' = 1$,则有