Today’s opener was chosen for totally selfish reasons, which I will explain below. We’re starting our statistics unit today, so we’ll be a doing a lot of review of means, meadians, quatiles, and graphical displays – all with an eye towards interpretation. But only after I have students respond to a prompt for me:

What is a least common denominator (LCD)? Provide directions for finding an LCD to someone who may not know how to find one.

This Friday, I will be out of school for the Association of Math Teachers of New Jersey conference, where I am looking forward to participating in an Ignite session, hosted by my friends from the Drexel Math Forum. In these talks, a speaker has 5 minutes and 20 slides to share their idea – mine is on the importance of language skills in math classrooms.

So, today’s opener was entirely selfish, as I was looking for examples to share during the Ignite. The LCD problem is one I have given before during rational expressions units. – try it with your classes and watch the misconceptions fly! Responses to this prompt can often be pigeonholed:

**THE “EXPLANATION BY EXAMPLE” CROWD**

**THE “NOT QUITE COMPLETE” CROWD**

**THE “MINIMALISTS”**

To be fair, I gave this prompt out of context, as fractions aren’t on our radar now. But it’s fascinating to see what built-in ideas students come to the high school with regarding a task they have now done for many years.

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## 10 replies on “Class Opener – Day 32 – What the Heck is an LCD, Anway?”

Bob

I like this opener

Ralph

What was your feedback to the second student in the “Not quite complete” category?

Usually I do this activity with Algebra 2 classes as we approach a rational expressions unit. In that setting, I would have groups share their definitions and try to “crowd source” a working definition, with peers providing feedback. If conversation wanes or we reach a stalemate, I provide some examples which cause their working definitions to work and not work.

Whatever happened to the Euclidean Algorithm?

This get you the HCF, and with a teeny bit of thinking gets the LCD as well.

have a look at this:

http://en.wikipedia.org/wiki/Euclidean_algorithm

Interesting algorithm, and would be see how students repond to it and/or synthesize the steps. Usually when I do this activity, I am about to start adding/subtraction rational expressions, and think about LCD’s of pairs like 6x^2y^3 and 8xy^4. As such, I want to get them thinking about factors and the need to select the highest power of each factor.

I figured that out, but why not dive straight in with some simple examples like

a/x+2/xy or 3x+1/x+4/x^2

There is no need for LCD arithmetically.

Clearly if, for example, there is a quadratic in one of the denominators then it might be a good idea to factorize it!

Good suggestions all. But I am curious how you approach equations which contain ration expressions, where often multiplying both sides by a common factor simplifies the equation.

for integer denominators a & b, it seems to be true that

(1 / remainder(b/a))*b = LCM

Do you find this to be the case? Might be an interesting thing to go through.

I have never encountered that formula before, but I think it would be interesting to have kids think about it and prove/disprove/find counter-examples.

for 4 and for 15 for example. key thing is you represent the remainder as a fraction so 15/4 leaves you with 3/4 left over.

I think I miswrote the equation which should be seen as

[1 / {1 – rem(b/a)}]*b = LCM

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60

15, 30, 45, 60

[1 / {1 – rem(15/4)}]*15 = [1 / {1 – (3/4)}] * 15 = {1 / (1/4)} * 15 = 4 * 15 = 60

say 7 and 20

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140

[1 / {1 – rem(20/7)}]*20 = [1 / {1 – (6/7)}]*20 = {1 / (1/7)} * 20 = 7 * 20 = 140

Seems to work, at least for the few cases I have tried. I can’t remember how I came up with the formula yesterday… but.

(1 / (1 – rem(b/a))) = y

started by thinking that with two numbers a and b you know that a*x = b*y where ax and by are the LCM. So x = (b / a) y where x & y are integers with no common factors. Not really sure how I wound up with the rest…