Back to Use rational approximations of irrational numbers — Problem 3 · Task Set 3

8.NS.A.2 Tutor Intake -- Rational Approximations of Irrational Numbers

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Grade 8·10 problems·~11 min·Common Core Math - Grade 8·container·8-ns-a-2

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Procedures

Which is greater: $\sqrt{5}$ or $\sqrt{7}$ ? Use that
$\sqrt{5}$ is between 2.2 and 2.3 (since $2.2^2 = 4.84$ and
$2.3^2 = 5.29$ ), and $\sqrt{7}$ is between 2.6 and 2.7 (since
$2.6^2 = 6.76$ and $2.7^2 = 7.29$ ).

It cannot be determined without refining both approximations further.

$\sqrt{5} > \sqrt{7}$ , because 5 is closer to a perfect square than 7 is.

They are equal, since both are irrational numbers.

$\sqrt{7} > \sqrt{5}$ , because the interval for $\sqrt{7}$ (2.6 to 2.7) lies entirely above the interval for $\sqrt{5}$ (2.2 to 2.3), so no further refinement is needed.