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
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