Proof of (a/b).(c/d)-1 = (a.d)/(b.c) | Better Method for Homework

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The discussion centers on proving the equation (a/b)/(c/d) = (a/b).(c/d)-1 using two different methods. One method simplifies the expression step-by-step, while the other uses a fraction manipulation approach. Participants express preferences for the elegance of each proof, with some favoring the first method for its clarity. There is a light-hearted exchange about the correctness of the approaches, with acknowledgments of corrections made during the conversation. Ultimately, both methods are validated, but opinions on which is superior vary among the contributors.
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Homework Statement


proof (a/b)/(c/d) = (a/b).(c/d)-1 = (a.b-1).(c.d-1)-1 = (a.b-1)(d.c-1)=(a.d).(b-1.c-1) = (a.d).(b.c)-1 = (a.d)/(b.c)



Homework Equations





The Attempt at a Solution



the other way to proof (a/b).(c/d)-1 = [(a.b-1)/(c.d-1)].(d/d-1) = (a.d.b-1)/(c.d.d-1) =(a.d.b-1)/c = (a.d)/(b.c)

which one is better proof ?
 
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i mean in more elegant way ..oh yea thanks before ..
 
dianzz said:

Homework Statement


proof (a/b)/(c/d) = (a/b).(c/d)-1 = (a.b-1).(c.d-1)-1 = (a.b-1)(d.c-1)=(a.d).(b-1.c-1) = (a.d).(b.c)-1 = (a.d)/(b.c)



Homework Equations





The Attempt at a Solution



the other way to proof (a/b).(c/d)-1 = [(a.b-1)/(c.d-1)].(d/d-1) = (a.d.b-1)/(c.d.d-1) =(a.d.b-1)/c = (a.d)/(b.c)

which one is better proof ?

you mean d/d:wink:


all works well, though I prefer the first
 
drizzle said:
you mean d/d:wink:


all works well, though I prefer the first

yea you right ..thanx for correcting:wink:

mm..you prefer the first ,but i dontnow I am suite with the last :biggrin:..
 

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