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Irrational Numbers on the Number Line
Task
Without using your calculator, label approximate locations for the following numbers on the number line.

$\pi$

$(\frac12 \times \pi)$

$2\sqrt2$

$\sqrt{17}$
IM Commentary
When students plot irrational numbers on the number line, it helps reinforce the idea that they fit into a number system that includes the more familiar integer and rational numbers. This is a good time for teachers to start using the term "real number line" to emphasize the fact that the number system represented by the number line is the real numbers. When students begin to study complex numbers in high school, they will encounter numbers that are not on the real number line (and are, in fact, on a "number plane"). This task could be used for assessment, or if elaborated a bit, could be used in an instructional setting.
Solution

$\pi$ is slightly greater than 3.
 $(\frac12 \times \pi)$ is slightly less than $1.5$.
 $(2\sqrt2)^2 = 4\cdot2=8$ and $3^2=9$, so $2\sqrt2$ is slightly less than 3.
 $\sqrt{16} = 4$, so $\sqrt{17}$ is slightly greater than 4.
Irrational Numbers on the Number Line
Without using your calculator, label approximate locations for the following numbers on the number line.

$\pi$

$(\frac12 \times \pi)$

$2\sqrt2$

$\sqrt{17}$
Comments
Log in to commentCam says:
over 3 yearsA nice alternative approach  thanks!
Richard Askey says:
over 3 yearsThere should be a second solution to both parts c and d. Consider the number line as the horizontal axis of a plane, with the vertical axis labeled the same way. For part c, 8 = 2^2 + 2^2, so by the converse of the Pythagorean theorem, an 8th grade topic, there is a right triangle with sides 2,2, and 2*square root of 2. Draw this right triangle with vertices at (0,0) and (2,0) and (2,2), and then use a compass to draw the circle with (0,0) as the center and containing the point (2,2). The point where the circle hits the positive initial number line is the point (2*square root 2,0). 17=16+1 can be used in a similar way. Some eighth grade students are ready to see a solution like this, and even a few are likely to find it themselves.
Debbie says:
over 4 yearsGreat nocalculator problem for my Test tomorrow! Thanks