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Topic: To meet a challenge
Replies: 18   Last Post: Oct 6, 2012 10:30 AM

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Paul A. Tanner III

Posts: 5,920
Registered: 12/6/04
Re: To meet a challenge
Posted: Sep 28, 2012 2:36 PM
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On Fri, Sep 28, 2012 at 12:03 PM, kirby urner <> wrote:
> This is how we teach teachers about fractions in the geek world, which is
> tending either towards home schooling or in some cases revamping older
> buildings. **
> The above URL links to properly formatted code in the computer language used
> in the Litvins' text.
> GCD(a,b) is computed using Euclid's Method, taken up by Knuth as probably
> one of the most important algorithms of all time.
> If a K16 curriculum does not mention Euclid's Method say by grade 10, it's
> likely not worth the trees that were mowed to truck it to your state.
> If you're a parent with a pitch fork, flip to the index and take a look. If
> Euclid's Method is not in Common Core Standards, write to your state
> representative and think about finding a more serious school perhaps based
> overseas in a less backward nation.
> Typical of the 1900s was to factor both a and b into unique primes using the
> "factor tree" method and then taking the product of all primes in common as
> the GCD. This method does not scale well as it requires a complete
> factorization, which is both time consuming and impractical in many cases.
> Assume rational number p/q is represented as Q(p, q).
> Also, we know:
> LCM(a,b) = (a*b)/GCD(a,b) # Knuth, using greatest common divisor derived by
> Euclid's Method
> So if adding Q(m, n) + Q(a, b) you may choose to go:
> LCM(n, b) = c # to get a common denominator (lowest common multiple)
> k = c/n is how many times n goes into c (no remainder) so (m/n)(k/k) =
> (mk)/(nk) = (mk)/c
> j = c/b is how many times b goes into c (no remainder) so (a/b)(j/j) =
> (aj)/(bj) = (aj)/c
> Q(m, n) + Q(a, b) = Q(mk, c) + Q(aj, c) where c = LCM(n, b) and k = c/n, j =
> c/b

This equation is essentially this one (using different variables),
translated into standard hand-written notation (but of course with the
ascii limitations here) and using the commutative property on the
products of each addend on the right side:

a/b +- c/d = [(m/b)a]/m +- [(m/d)c]/m

for m being the least common multiple of the denominators b and d
(LCD, least common denominator).

But when I said that we teach fraction addition/subtraction using the
LCD such that the algorithm is too long and complicated and clunky to
be written elegantly and concisely as a single equation, this single
equation above is not the single equation I had in mind. In fact, this
equation above is only part of this long and complicated and clunky
algorithm. That is, on the right side we are not yet at an expression
that is a single noncomplex fraction.

When I'm talking about being able to write the algorithm elegantly and
concisely as a single equation, I mean such that the left side is
written as the sum/difference of the fractions to be transformed into
a single noncomplex fraction and the right side is that single
fraction, like this equation (and where I use the "/" symbol
surrounded by a space to be the main dividing line):

a/b +- c/d = (m/b)a +- (m/d)c / m.

Side Note 1: Each of (m/b) and (m/d) is an integer when each of the
variables is an integer - when we are in the rationals. And in the
algebraic context of adding rational functions, each of (m/b) and
(m/d) is a polynomial. That is, even though this single fractional
expression in abstract form on the right is written here in a way that
can be viewed as a complex fraction in that the numerator at least
contains fractional expressions, by what I just said, it is not a
complex fraction when it is applied in the specific contexts. If one
wishes to replace each of (m/b) and (m/d) with some expression that is
not a fractional expression to make what I just said explicit in the
abstract formula, then one is free to do so. (One could use notation
like m_b and m_d, or (m,b) and (m,d), or other ways to avoid actually
writing a fractional form if one wishes to avoid that.)

This generalizes to any number of fractions elegantly and concisely,
letting n be this any number of fractions:

a_1/b_1 +- ... +- a_n/b_n = (m/b_1)a_1 +- ... +- (m/b_n)a_n / m.

How to teach this equality as an algorithm to students? I use the
verbalization "m over the bottom times the top" for each addend. And
since this method generalizes so elegantly and concisely to any number
of fractions, a number of times I actually introduce the method on
three fractions to show how easy it is - if they see that it's easy on
three or more fractions, it of course it is easy on just two. Here's a
favorite set of three fractions I like to use, one with an easy LCD:

3/4 + 5/6 - 7/8.

To show how quick and easy we can go in one written step from a sum of
any number of fractions to one non-complex fraction, I write this out
and then on the right side make a long dividing line with LCM 24 on
the bottom. And since we now have m and all the "bottoms" and the
"tops" written out at this point,

3/4 + 5/6 - 7/8 = / 24,

to fill out the numerator on the right I then perform "m over the
bottom times the top" in our heads three times and copy the sign
twice, getting

3/4 + 5/6 - 7/8 = 18 + 20 - 21 / 24.

The rest is just any needed simplification of the single non-complex
fraction on the right side, which is what we have to consider with
fraction multiplication and fraction division when we get to a single
noncomplex fraction there.

If we wanted to write out the pattern that they would use in algebra
in the adding/subtracting of rational functions, we would do
everything exactly the same, keeping the same pattern, but doing only
"m over the bottom" in our heads, resulting in

3/4 + 5/6 - 7/8 = (6)(3) + (4)(5) - (3)(7) / 24.

Again: Where we replace each of these integers with polynomials, this
last pattern is exactly what we do with this method on rational

Side note 2: I say this last part because part of the reason that we
teach fraction addition/subtraction with the LCD is because it is an
anticipation of rational function addition/subtraction.

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