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8.NS Comparing Rational and Irrational Numbers
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8.NS Comparing Rational and Irrational Numbers

Alignments to Content Standards

  • Grade 8
    Domain
    NS: The Number System
    Cluster
    Know that there are numbers that are not rational, and approximate them by rational numbers.
    Standard
    Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., $\pi^2$). For example, by truncating the decimal expansion of $\sqrt{2}$, show that $\sqrt{2}$ is between $1$ and $2$, then between $1.4$ and $1.5$, and explain how to continue on to get better approximations.

Tags

Tags: MP 7

For each pair of numbers, decide which is larger without using a calculator. Explain your choices.

  1. $\pi^2$ or $9$

  2. $ \sqrt{50}$ or $\sqrt{51}$

  3. $\sqrt{50}$ or $8$

  4. $-2\pi$ or $-6$


Commentary

This task can be used to either build or assess initial understandings related to rational approximations of irrational numbers.



Solutions

Solution: Solution
  1. $\pi \gt 3$ so $\pi^2 \gt 9$.

  2. $ \sqrt{50} \lt \sqrt{51}$ because $50 = (\sqrt{50})^2 \lt (\sqrt{51})^2 = 51.$

  3. $7^2 = 49$ and $8^2 = 64$. Thus we have that $\sqrt{49} \lt \sqrt{50} \lt \sqrt{64}$. So $\sqrt{50} \lt 8$.

  4. $\pi \gt 3$ so $2\pi \gt 2\cdot3$. If you look at these numbers on the number line, that means that $2\pi$ is farther to the right than $6$. When you look at their opposites, $-2\pi$ will be farther to the left than $-6$, so $-2\pi \lt -6$.



Public Comments
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  • Ken wrote this public comment over 1 year ago

    Some of these solutions seem more about properties of square roots than about the concept of irrational numbers.

    • Cam McLeman wrote this public reply over 1 year ago

      Well, square roots of integers certainly make up a large proportion of the irrational numbers we expect middle school students to be conscious of. That said, I don't disagree with your general point. While it's true that $\sqrt{50}\approx \sqrt{49}=7$, and this is pretty convincing that $\sqrt{50}<8$, I don't think this (or the more careful solution provided for this fact) is the type of reasoning being requested under the heading "rational approximation" in the standard. If the task disappears for a while, it's because we've pulled it off-line for fixing up.

      • Kristin Umland wrote this public reply over 1 year ago

        It seems to me that noting that $\sqrt{50}\approx \sqrt{49} = 7$ is equivalent to saying that 7 is a decent rational approximation of $\sqrt{50}$ in the right context, so in that sense, I think it does fit the standard nicely. Perhaps part (b) is less clearly connected to the standard, but I think it helps bridge the perspectives on square roots encapsulated in 8.EE.A.2 and 8.NS.A.2. In general, I think this task is about helping students develop/demonstrate "irrational number sense" which is a main reason to find/recognize rational approximations of irrational numbers. I think this task an appropriate member of a set illustrating the standard, although we need more. Any suggestions?