Engaging math activities for the summer break - Day 8
The setup A student is presented with a small pile of objects.
The task To split a pile into two; count the number of objects in each, and compute the product of the two numbers. This step is repeated with any of the present piles until the only remaining piles each contain a single object. The computed products are then summed up.
What's to observe? Regardless of how and which piles are split the result is always the same.
For example, starting with the pile of 5 objects, one can proceed as in
5 → 2, 3 → 1, 1, 3 → 1, 1, 1, 2 → 1, 1, 1, 1, 1,
with the sum 2·3 + 1·1 + 1·2 + 1·1 = 10.
Another possibility would be
5 → 1, 4 → 1, 2, 2 → 1, 1, 1, 2 → 1, 1, 1, 1, 1,
with the sum 1·4 + 2·2 + 1·1 + 1·1 = 10.
Even when starting with such a small number as 5 there are several ways to perform the task.
What's to prove? One must prove that the resulting sum is independent of the manner the task is performed.
There is a beautifully intuitive proof that is accessible in early grades. It only requires the knowledge of the basic principles of counting, most essentially the product rule.
Imagine tying each object to every other object by a thread. In all, if there are N objects to start with, there are going to be (N - 1)N/2 threads.
To split a pile into two parts of x and y objects respectively, one has to cut xy threads connecting chips from the two parts. Which is exactly the number to be added to the total. To reach the stage of no nontrivial piles one will have to cut all the original threads. Which mean that the running total of the intermediate products will always at the end be equal to (N - 1)N/2.
Related posts: