The question was mine, but the answer was all his.

I carry a defective teacher gene that makes me intolerant of pacing guides and curriculum guidelines.

I’m sure I’m a month behind in the pacing guide, and I’m pretty sure no part of today’s lesson will appear on any state test.

But I did it anyway.

The algebra kids have been doing a lot of graphing inequalities on the number line, and the geometry kids have been doing a bunch of a-squared-plus-b-squared-equals-c-squared for their Viewmongus lesson.

So I’m looking at my geometry kids and thinking: Hmmm… I wonder if you know how to graph an irrational number on the number line.

I ask them to locate sqrt(7) on the number line. Two students pull out their calculators. The rest are scribbling and erasing, doodling too I suppose. They know sqrt(7) is somewhere between 2 and 3.

Five minutes pass quietly.

Another five minutes, their faces grow longer. Gabe brings his journal up to me. He writes:

You could create a right triangle with a vertex at 0 and the side being sqrt(7) going along the number line. The other vertex would be the point where the sqrt(7) is located.

I nod my head but ask, “Want to help me understand what you wrote with a picture?”

He comes back with this sketch. I tell him, “I see. Something to do with right triangles. Okay. But you know that I still don’t see how you arrive at sqrt(7) right there, right? I think you’re on to something, Gabemeister.”

Enough individual think/stuck time has passed, I put them into random groups of three. I walk around, they’re talking about the problem, but I’m not seeing much on the whiteboards. Maia [bottom right] thinks she’s funny. Gabe gets himself a compass.

I’m watching the clock. Jack wonders if he could just make a number line for irrational numbers only. You know, 0 here, then sqrt(1) here, then sqrt(2) here, then oh never mind. Gabe looks busy with his compass. His group mates are observing him carefully and listening to him mumble.

Time to nudge them along. I ask, “This question that I’m asking you of where is sqrt(7) or sqrt(any irrational number) on the number line came from my watching you in the last activity. What were you doing in last two days?”

They burst out:

Aspect ratio thingies.

The diagonal of the TV!

Mathalicious!

Right triangles!

Pythagorean theorems — but this [number line] is not a right triangle!

Viewmongus!

The diagonal of the TV is the number line!

I say nothing more and continue to walk around. Now I’m seeing more on their whiteboards.

All the groups are working with right triangles, but I’m unable to take more pictures of their work as I’m now with Gabe’s group and listening to him explain how he thinks he has the answer.

He goes over his drawing above step-by-step by drawing another similar one. I’m retelling his story using Geometer’s Sketchpad.

Step 1: Draw an arc of radius 3 units.

Step 2: Draw an arc of radius 4 units.

Step 3: Draw the line y = -3.

Step 4: Draw a perpendicular line (pink) to the red line where it intersects blue arc.

Step 5: Sharing this final step right now with you makes me tear up.

Tomorrow Gabe will get to share his way of finding sqrt(7) on the number line with his class.

Then maybe I’ll ask the class to try finding sqrt(11), sqrt(10), sqrt(bring-it-on-Mrs.Win-we-got-this). Maybe they’ll have fun doing this.

I’d posed a question that wasn’t lined up with the curriculum. It just came to me as I was watching my kids do math. But the question lined up with my gut, and my heart went along and said, Do this because it’ll remind you of why you love this job.

1. Miss H
Posted March 24, 2017 at 5:23 am | Permalink

I know this was a very long time ago, but my maths club did this yesterday evening and had a brilliant time, so thank you! They came up with this solution, and another, where they constructed triangles with hypotenuse root 2, then root 3, then root7.

2. Dave Smith
Posted April 9, 2019 at 9:32 am | Permalink

In general this should work for integer circles (a,b) and root(N) where N is also integer when N=(a + b)/mod(a – b).

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