Picturing irrational numbers Students often meet irrational numbers for the first time as they begin working with the Pythagorean theorem. I’d like students to understand that irrational numbers are just another type of number (like fractions were when they were in 2nd or 3rd grade) that are kind of difficult to evaluate and place by size on a number line. I can come close to naming an irrational number but all of my standard formed numbers (like decimal approximations) are slightly inaccurate. But … I’m really good at drawing them. As you begin working with the Pythagorean theorem why not let your students create a little art project that might help them actually see a segment of length of 2. Using my drawing tools, I’ve created a 1-inch by 1-inch square. I feel pretty confident that the sides of my square are in fact 1 inch long. 1. Measure my square’s sides with a ruler and see if I’ve done a pretty accurate job. Is my square about right? 2. Now draw the diagonal of my square and use your ruler to measure the length of the diagonal that you just created. Hard to precisely measure isn’t it? Make a rough estimate of the diagonal’s length. Translate whatever measurement you found to a decimal number. Please show all of your work here. 3. With the Pythagorean theorem you can calculate very precisely what the diagonal of this square must measure. Do that math and show your work below. 4. On your calculator see what value your answer to problem #3 is approximately and write that number here. 5. Is that about what you got when you measured the length? 1 inch Let’s look at more irrational numbers. Instead of making a square with sides = 1 inch, I’m going to draw only ½ of a square … just two adjacent sides of a square. The angle between those two sides will have to be a right angle because squares have only right angles. 1 inch For your final project you will draw this little half-square in the center of a plain piece of paper. (It might be handy to create a little cardboard 1-inch square to use as a template.) Next you will need to make a triangle from this little half-square. It will be an isosceles right triangle and you should label all of its sides with their lengths. 1 inch 1 inch To find the hypotenuse value of this triangle I’ll use the Pythagorean theorem. This time I’ll do the math for you but for your project you will need to find all of the hypotenuse values yourself. 1! + 1! = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ! 2 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ! 2 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 √2 1 in c 1 inch Now the fun begins. I’m going to add a new right-angled 1-inch leg to the base of the hypotenuse of that first right triangle. And I’ll also add a new hypotenuse. h 1 inch 6. Use the Pythagorean theorem again to calculate the length of this new hypotenuse. Show your work. 7. Now measure the length of this hypotenuse with your ruler and change your measurement to decimal form. 8. Is the decimal form close to the Pythagorean solution? Project Continue to add one-inch legs (at right angles) to your last hypotenuse and label all of each new triangle’s sides using the Pythagorean theorem. Add on at least 15 new triangles. If you are enjoying this you can certainly add more than 15 triangles. (In all the example pictures that we’ve shown at the end of this activity the students’ new right triangle’s started to overlap their early right triangles. It looks best if you make those last, overlapping right triangles appear to be behind the first part of the spiral. Just always connect your next hypotenuse to the same middle point in the drawing but don’t draw in the line on top of the previous pictures.) Decide how you might like to color or animate your drawing and create something that your classmates and your teacher might enjoy. I’ve given some examples of other students work at the end of this activity sheet. Enjoy! This is a famous spiral called the Wheel or Spiral of Theodorus. Theodorus was a Greek mathematician in the 5th century B.C. He was a Pythagorean and one of Plato’s teachers. Plato gave him credit for proving that the square roots of 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, and 17 were irrational numbers. 1 inch 1 in 1 in ch ch 1 √9 h ch 1 in 3 0 √1 √5 √11 1 inch 1 inch in c √12 √2 h 1 inch Brought to you by Yummymath.com 1 inch c in = ch 2 √3 1 1 in ch √8 √7 √6 1 in √4 =