Halving & Doubling: Very Fun to Play With

On last week’s Last Week Tonight with John Oliver, John Oliver used the mental math/computation strategy of halving and doubling as a punchline to a news story on nuclear waste.

The graphics nicely–and quickly!–illustrate why this strategy works. Starting with 1 × 20 (one football field twenty feet tall), if we double the first factor (area in football fields) and halve the second factor (height in feet), the product (volume in piles of nuclear waste), expressed as 2 × 10, remains the same. Similarly, we can halve and double to visualize that 1 × 20 is equivalent to ½ × 40. (Oliver also throws in the commutative property at the end–twenty football fields one foot tall.)

This reminded me of a video clip from Sherry Parrish’s Number Talks. In it, the teacher poses the problem 16 × 35. The fifth graders share several strategies: partial products (10 × 30 + 10 × 5 + 6 × 30 + 6 × 5); making friendly numbers (20 × 35 − 4 × 35); halving and doubling (8 × 70); and prime factors (ultimately unhelpful here).

I’ve probably shared this video in about a dozen workshops. There are some predictable responses from attendees. Often “not my kids” is the first reaction. I remind teachers that the teacher in this video has implemented this routine three to five times a week in her classroom. This isn’t her kids’ first number talk. Pose 16 × 35 in your fifth–or ninth!–grade classroom tomorrow and, yeah, the conversation will probably fall flat. Also, this teacher is part of a schoolwide effort (seen in other videos shared at these workshops).

Teachers are always amazed by Molly’s halving and doubling strategy. Every. Single. Time. I ask attendees to anticipate strategies but they don’t see this one coming. I note that doubling and halving wasn’t introduced through 16 × 35. I would introduce this through a string of computation problems (e.g., 1 × 12,  2 × 6, 4 × 3). “What do you notice? What patterns do you see? Does it always work? Why?” We can answer this by calling on the associative property: 16 × 35 = (8 × 2) × 35 = 8 × (2 × 35) = 8 × 70 above. Better yet, having students play with cutting and rearranging arrays provides another (connected) explanation.

Rather than playing with virtual piles of nuclear waste, I had fun with arrays of candy buttons:

Number Talks (pdf)

 

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Building Capacity

This week, we spent one day with 15 math teams (almost 50 teachers and administrators) from 15 elementary schools in Surrey. (I’ll blog about this project soon.) Part of this day was  devoted to having teachers work together to solve problems. These problems help set the stage for some of the important themes schools will be exploring by participating in this project over three years. These include:

  • conceptual understanding
  • concrete, pictorial, and symbolic representations
  • use of manipulatives
  • communication
  • connections between mathematical ideas
  • learning and teaching through problem-solving
  • multiple solutions
  • reasoning
  • attitudes and self-confidence

We gave the following problem, from Figure This!:

Take two identical sheets of paper (8½ inches by 11 inches). Roll one sheet into a short cylinder and the other into a tall cylinder. Does one hold more than the other?

A common misconception is that the two cylinders hold the same because the two pieces of paper are the same size. Teachers use a variety of strategies to explore the relationship between surface area and volume.

The first approach most teams take is to calculate the volume of each cylinder. They’ll ask for (or google!) formulas. Even after determining the volume of each cylinder, many will remain unconvinced. Prompted with “How might young children solve this?” teachers will fill each cylinder with manipulatives available at their tables and compare the results, similar to the third act of Dan Meyer’s Popcorn Picker or John Scammell’s Surface Area vs. Volume.

One of the things that I enjoy most about posing these problems to teachers is that each time someone will come up with a solution that I haven’t seen before. For example, this week one team solved this problem by solving a simpler problem. That is, they compared rectangular prisms. (Is ‘squarular prism’ a thing? It should be.)

They argued that the conclusion would be the same but the calculations would be easier. They found a way around the formulas V = πr²h and C = 2πr. True problem-solving!

Part of me geeks out at seeing innovative solutions. The other part of me kicks himself for not making this a bigger part of my own classroom. A lot of missed opportunities– maybe one day I’ll get a do-over.