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Deriving Properties of Fractions

Date: 08/10/2003 at 17:02:28
From: Andrew
Subject: Deriving properties of fractions

I am wondering how to derive the rule for multiplying fractions,
namely that a/b x c/d = ac/bd, using lower-level properties of
multiplication and rational numbers.

I find myself going in circles. For instance, starting with a/b x c/d, 
I think "Multiplying q by r is the same as dividing q by 1/r," so I 
transform it into a/b / d/c. But the reason I know d/c is the 
reciprocal of c/d is that I know from rote memorization of the rules 
that c/d x d/c = 1, but that is using the very rule I'm trying to 
prove.

I  found this entry in the archives that deals with my problem on
an intuitive basis:

   When to Add or Multiply Denominators?
   http://mathforum.org/library/drmath/view/58177.html 

What I'm looking for is what Dr. Peterson calls a "deeper (algebraic)"
explanation.

I think I derived the rule for adding fractions with the same
denominator as follows:
Starting with a/c + b/c

= (a x 1/c) + (b x 1/c)   because x divided by y is the same as
multiplying x by the reciprocal of y

= (1/c)(a + b)   by the distributive property of multiplication

= (a + b) / c    because any x multiplied by the reciprocal of y is
the same as x divided by y


Date: 08/10/2003 at 22:03:42
From: Doctor Rick
Subject: Re: Deriving properties of fractions

Hi, Andrew.

That's correct for addition of fractions.

Let's see what we can do for multiplication of fractions, using only 
the definition of division as multiplication by the multiplicative 
inverse, plus the commutative and associative properties of 
multiplication. I'll include EVERY step, something we don't often do. 
We have

  (a/b) * (c/d) = (a * 1/b) * (c * 1/d)
                = ((a * 1/b) * c) * 1/d (associativity)
                = (a * (1/b * c)) * 1/d (associativity)
                = (a * (c * 1/b)) * 1/d (commutativity)
                = ((a * c) * 1/b) * 1/d (associativity)
                = (a * c) * (1/b * 1/d) (associativity)

I need one more thing in order to finish. I need a theorem that the 
multiplicative inverse of a product is the product of the 
multiplicative inverses (inverse distributes over multiplication):

  1/(b*d) = 1/b * 1/d

Knowing this, I can complete the proof I started:

            ... = (a * c) * 1/(b * d)   (distributivity of inverse)
                = (a * c) / (b * d)     (definition of division)

and that's it. So, how can I prove my theorem? By definition, a number 
times the multiplicative inverse of the number is 1. If we multiply 
(b*d) by (1/b * 1/d) and get 1, then by definition, (1/b * 1/d) is the 
multiplicative inverse of (b*d). (I am glossing over one point: 
showing that the multiplicative inverse is unique.) Let's see:

  (b*d) * (1/b * 1/d) = ((b*d) * 1/b) * 1/d (associativity)
                      = ((d*b) * 1/b) * 1/d (commutativity)
                      = (d * (b * 1/b)) * 1/d (associativity)
                      = (d * 1) * 1/d       (multiplicative inverse)
                      = d * 1/d             (multiplicative identity)
                      = 1                   (multiplicative inverse)

The distributivity of inversion over multiplication is proved, so our 
theorem about multiplication of fractions is complete.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 08/10/2003 at 23:43:48
From: Andrew
Subject: Thank you (Deriving properties of fractions)

Hey Doctor Rick,

Thanks for your detailed answer!  Not only did that set my mind at 
ease, I learned something valuable about looking ahead in my proofs.

Thanks again,

- Andrew
Associated Topics:
High School Logic
High School Number Theory

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