## 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 know where we're at: my 6th graders are working on adding and subtracting fractions, algebra kids are solving compound inequalities, geometry group just took a test on chapter 6 (not doing SBG with this class). And I have no clue what chapter 7 is about.

Yet I'm well aware that the state test is in the first week of May when we still have almost a full quarter left of school. No one has given me an intelligent explanation as to why a multiple-choice test administered in the name of "Algebra 1 Standards Test" does not allow students to get a full year's worth of algebra and why it takes 3 months to get the results.

They might as well rename the test as "We-Hate-Students-So-Here's-Your-Algebra-1-Standards-Test-Version-A," and "We-Hate-Teachers-Too-So-Let's-See-How-Poorly-Your-Students-Do-On-This-Algebra-1-Standards-Test-Version-B."

Anyway, 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.

**********

*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:

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

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:

I say nothing more and continue to walk around. Now I'm seeing more on their whiteboards.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 TVisthe number line!

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.

**1**. Draw an arc of radius 3 units.

**2**. Draw an arc of radius 4 units.

**3**. Draw the line

*y*= -3.

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

**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."

I agree with everything you said about state testing. My 7th and 8th graders test in MARCH!!! And I KNOW I'm at least a month behind on the pacing guide. My belief is that they learn the concepts...radical thought, I know!

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March?? That's criminal. Sorry to hear that, Betty. I'll stop whining then. Thanks for dropping in.

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Awesome.

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Thanks, Kate!

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Boom! Math bomb. Wicked.

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Thanks, Matt aka Mr.MarriedtoHotWifeandDaddyofCutestBaby. (Did I get that right?)

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You got one thing wrong - that gene is not defective. Maybe it's an adaptive mutation?

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True true. Or I'm counting on it being a recessive gene. Thanks, Sue.

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I’ve never seen a student approach this type of problem using this method. It got me thinking though, “Will this always work?” Specifically, I mean will this method always produce a leg and hypotenuse that are integers. If they aren’t integers then you will have to plot the points for two irrational numbers, which is pretty challenging to do accurately on graph paper.

So, I started out by trying sqrt(10) and sqrt(11) as you suggested. sqrt(10) did not seem to work as there would be another leg/hypotenuse that was irrational, but sqrt(11) did produce an integer leg and hypotenuse. Trying more values, a pattern emerged: this technique only works with the square root of odd numbers. For example:

sqrt(7) gives sides of 3 and 4

sqrt(9) gives sides of 4 and 5

sqrt(11) gives sides of 5 and 6

sqrt(13) gives sides of 6 and 7

I realized the reason why was because you were subtracting one leg from the hypotenuse and the difference between any two consecutive perfect squares was an odd number. This led to a pattern of:

sqrt(x + y) gives sides of x and y

(where x is the length of a leg, y is the length of the hypotenuse, and x and y are consecutive whole numbers)

Thanks for sharing this Fawn.

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Your comment is exactly what I asked the kids today. After Gabe showed his sqrt(7), I asked everyone for sqrt(11). They were able to duplicate the process using 5 and 6 as the other two sides. Then I asked for sqrt(10) and sure enough they were stuck. (Gabe's group got it though because they went ahead of the class.) Then I asked, "Why did you assume that sqrt(10) had to be one of the legs? Couldn't it be another side, like the hypotenuse maybe?" So with this hint, it didn't take them long to figure out 1 and 3 for the legs to produce a hypotenuse of sqrt(10).

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It's like you live in a strange far away Utopia where kids actually care a little bit about math class.

I'm sure your excellence and enthusiasm have something to do with that.

Coolest thing I've seen today.

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Ha! We refer to this campus as the "little school among the lemon orchards," so a bit Utopian in setting at least. I make these kids love math. Dammit, I love it, so will they. It kinda works. Thanks for the kind words, Marshall.

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Beautiful!! You, him, the problem, the solution, all of it....

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So so kind, thanks, Dan. I'm trying to have more of the "yum mantra."

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I can't think of anything new to say, but I just want to agree with all of the commenters above. This is a great story, and paints a nice picture of the Math Atmosphere in your classroom.

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Sometimes I wish I could videotape them. They are a good bunch of kids who have a lot of respect for each other's thinking. We laugh a lot. Thanks, Steve.

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If we make the hypotenuse an even number

say H = (N+1)/2, with the opposite side say Y = (N-1)/2 will get the adjacent side X = sqrt(N) where N is odd.

Note that Y = H -1 so Y will also be odd.

Im not certain that this will be irrational in all cases but interesting!

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Thanks, tomctutor. I think this is what Robert K is saying above?

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If we now make hypotenuse an odd number (H =2N+1 say) and opposite Y =2N (even) so that H =Y+1, we get the interesting fact that adjacent X =sqrt(H+Y)which is a generalisation of Robert K's idea.

If X irrational then H+Y (=4N+1) must be prime, which of course is not always the case. If H+Y not prime,e.g. N =2, X is rational.

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What I meant to say was if H+Y (=4N+1)is prime then X must be irrational - not the other way about (sorry)!

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So the converse isn't true.

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Wonderful! (I think you just made many tear up!)

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Sweet of you to drop me a note, thank you, Shecky. I always enjoy reading your tidbits and summaries at Math-Frolic!

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