Proof for Dividing Fractions
Date: 01/18/2002 at 11:35:12 From: Cathy Subject: Proof - dividing fractions I am sure there is a proof for dividing fractions. Could you please send it to me?
Date: 01/18/2002 at 12:06:45 From: Doctor Peterson Subject: Re: Proof - dividing fractions Hi, Cathy. Do you mean an explanation as to why we divide fractions the way we do? Try the Dr. Math FAQ: Dividing Fractions http://mathforum.org/dr.math/faq/faq.divide.fractions.html An actual proof will depend somewhat on what definitions and axioms you want to base the proof on. Several of the links at the bottom of the FAQ give informal demonstrations that could be turned into proofs. But let's try doing it algebraically. Just to make sure I don't accidentally assume what I'm proving, I'll write "DIV" for the division sign rather than "/" as usual, which sort of implies the connection between fractions and division. I'll suppose that we have defined a fraction as a pair "a/b" for which equality, addition, and multiplication have been defined in the standard ways. We want to determine the value of a/b DIV c/d. First, we have to know how we are defining division. I would define it as the inverse of multiplication, so that x DIV y = z if and only if x = y * z So our answer, a fraction x/y, must satisfy a/b DIV c/d = x/y if and only if a/b = c/d * x/y By the definition of fraction multiplication, c/d * x/y = (cx)/(dy) Therefore, a/b DIV c/d = x/y if and only if a/b = (cx)/(dy) Now, how do we define equality of fractions? x/y = u/v if and only if xv = uy So we can say a/b DIV c/d = x/y if and only if ady = bcx and again, reinterpreting it as a different pair of fractions, a/b DIV c/d = x/y if and only if x/y = (ad)/(bc) But the right side, by the definition of multiplication, is (a/b)(d/c); so we've found that a/b DIV c/d = (ad)/(bc) = (a/b)(d/c) But this means that we divide by c/d by multiplying by d/c. Is that the sort of proof you want? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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