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

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In summary, the proof shows that (a/b)/(c/d) is equal to (a.d)/(b.c) through the manipulation of algebraic expressions. There is another way to prove this, but the first method is considered more elegant. Both methods yield the same result.
  • #1
dianzz
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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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  • #2
i mean in more elegant way ..oh yea thanks before ..
 
  • #3
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
 
  • #4
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:..
 

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

What is the concept of "Proof of (a/b).(c/d)-1 = (a.d)/(b.c) | Better Method for Homework"?

The concept of this statement is to prove that the expression (a/b).(c/d)-1 is equal to (a.d)/(b.c) through a better and more efficient method for solving homework problems. This involves using algebraic manipulation and properties of fractions.

How is this proof different from traditional methods?

This proof uses the distributive property and simplification of fractions to directly show the equality of the two expressions, instead of using substitution or cross-multiplication.

What are the benefits of using this method for solving homework problems?

This method is more efficient and can save time when solving problems involving fractions. It also helps to develop a better understanding of algebraic concepts and properties of fractions.

Can this proof be applied to other similar expressions?

Yes, this proof can be applied to other expressions involving fractions and the distributive property. It is a general method that can be used to prove various equations and identities.

How can I use this proof in my own homework or studies?

You can use this proof as a guide for solving problems involving fractions and to develop a better understanding of algebraic concepts. It can also be used as a reference for similar proofs and equations.

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