Re: [証明] 合成函數
※ 引述《GayerDior (蠟筆小新<( ̄. ̄)/)》之銘言:
: E:
: 請嚴格証明合成函數不滿足交換律
: 如果有人會寫,
: 請幫我寫出詳細過程,
: 謝謝唷~~
Suppose f:A->B g:B->A be two functions ,then by the definition of
comosition of functions we have :
f(g(.)):g(B)∩A->B and g(f(.)):f(A)∩B->A .
Since the domains of f(g(.)) and g(f(.)) are not generally equal ,
so these two functions are not generally equal .
This completes the proof.
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BTW, you can also make a counterexample to proof this .
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08/02 16:41, , 1F
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