Connecting arithmetic and geometry

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I’ve been kicking around a few ideas about connecting arithmetic and geometry. My first thoughts were revisiting an idea we’ve played with a few times before:

So, today I decided to look at these two sums with the boys:

1 + 2 + 3 + 4 + \ldots + n, and

1^2 + 2^2 + 3^3 + \ldots + n^2.

We got off to a slower than expected start because my younger son didn’t remember the formula for the first sum correctly. I’m trying hard to break through the idea of relying on remembering formulas, so I was actually happy to review where the formula came from, though.

After the introduction we moved on to studying the first series using snap cubes. What is the geometry hiding behind the formula.

This part of the project to a totally unexpected turn, though:

I decided to keep going with the new sum that added up to n^2 to see if we could make another connection. The boys did remember that the sum of odd integers connects to perfect squares, so I challenged them to find the connection between that formula and the new one they just stumbled on.

Finally we moved on to the sum of squares formula. Lots of fun questions from the boys here, including if the idea extended to 4 dimensions!

The shape here is more difficult to build that it initially seems, but they got through it and now hopefully have a better idea of where the formula comes from.

We wrapped up by looking very briefly at pyramids.

I’d like to do more projects like this one and develop a bunch of different ways to share connections between arithmetic and geometry with kids.

Published by mjlawler

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2 thoughts on “Connecting arithmetic and geometry”

  1. If you break up an nxn square {(a,b)} into (ab), you can recombine the first two sets into (a<=b), and get the equation n^2 = (n+1 choose 2) + (n choose 2).

    I feel like you're doing something close to that, in 3-d, with your sums of squares.
    Now I want to take the
    (a<b<c), (a<c<b), (b<a<c), (b<c<a), (c<a<b), (c<b<a) n! of these in general
    and the
    (a=bc), (a=c    b), (b=ca) (n choose 2)*(n-1)! of these
    and the
    (a=b=c)
    and recombine somehow.
    Maybe the best thing is to give up on the (a=b=c) part, and attach the
    others? If there’s a way to do it it’s special to 3 dimensions. But that’s the kind of printer you have anyway.
    So I’m looking for a matching in this bipartite graph, and since it’s actually a wheel there are exactly two matchings! One is
    (a<=b<c), (a<c<=b), (b<a<=c), (b<=c<a), (c<=a<b), (c<b<=a)
    What I'm getting at with this is that if you made three copies each of these two slightly different half-pyramids, they'd fit together into a cube minus its main diagonal.

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