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

$\sqrt{2}$ is between 1 and 2, because squaring the endpoints 1 and 2 gives values that bracket 2.

$\sqrt{2}$ equals 1.5, the number exactly halfway between 1 and 2.

$\sqrt{2}$ is between 1 and 4, matching the two squared values directly.

This tells us nothing about $\sqrt{2}$ without a calculator.

Yes — since we cannot write the exact decimal, the number does not have one specific location and cannot be plotted accurately.

No — $\sqrt{2}$ has a precise, exact location on the number line (e.g., the diagonal of a unit square); we just approximate that location using rational bounds like 1.41 to 1.42, which is precise enough to plot.

No — but only whole numbers and simple fractions can be plotted accurately; $\sqrt{2}$ can only be plotted roughly, in the wrong place.

Yes — irrational numbers are not real numbers, so they don't belong on the real number line at all.

The error stays exactly the same after the first two decimal places.

The error eventually grows again, because rounding mistakes accumulate the more digits you add, like a copying error compounding.

The error keeps shrinking — each additional digit makes the approximation roughly 10 times more accurate, and the process always converges, never gets worse.

The error becomes 0 once enough digits are added, since $\sqrt{2}$ can eventually be written exactly.

It should use the approximately-equal symbol ($\sqrt{2} \approx 1.41$), not equals — since $1.41^2 = 1.9881 \neq 2$, $1.41$ is close to $\sqrt{2}$ but not equal to it.

There is no problem — $1.41$ is the standard rounded value of $\sqrt{2}$, so writing $=$ is acceptable shorthand in any context.

The problem is that 1.41 has too few digits; using more digits (e.g., 1.41421356) would make the equals sign correct.

The problem is that $\sqrt{2}$ should be written as a fraction before comparing it to a decimal at all.

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.