Approximate irrational numbers with rationals to compare, locate and estimate them

Learning goal: Use rational approximations of irrational numbers to compare their size, locate them approximately on a number line diagram, and estimate the value of expressions such as π² by truncating decimal expansions and explaining how to continue to better approximations.

An irrational number such as √2 cannot be written exactly as a fraction or as a terminating decimal, but it can be trapped between two rational numbers as tightly as we like. That trapping is what "rational approximation" means: a rational number that is close to the irrational one, together with a statement of how close. The approximation 1.4 for √2 is useful only when we also know that √2 lies between 1.4 and 1.5, because then we know the approximation is off by less than 0.1. This lesson builds such approximations, uses them to compare irrational numbers with one another and with rational numbers, and places them on a number line.

Trapping √2 between consecutive integers. The number √2 is the positive number whose square is 2. Because 1² = 1 is less than 2 and 2² = 4 is greater than 2, and because squaring makes positive numbers larger in the same order, √2 must be between 1 and 2. This step uses one fact repeatedly: for positive numbers a and b, if a² < b² then a < b. The comparison of squares tells us the comparison of the numbers themselves.

Refining to tenths, then hundredths. Now test tenths. 1.4² = 1.96, which is less than 2, and 1.5² = 2.25, which is greater than 2, so 1.4 < √2 < 1.5. Test hundredths inside that interval: 1.41² = 1.9881 and 1.42² = 2.0164, so 1.41 < √2 < 1.42. Test thousandths: 1.414² = 1.999396 and 1.415² = 2.002225, so 1.414 < √2 < 1.415. Each stage truncates the decimal expansion one digit further and squeezes √2 into an interval one tenth as wide. To continue, square the candidates 1.4140, 1.4141, …, 1.4149 and find the two whose squares straddle 2. The process never ends, because √2 is irrational and no finite decimal equals it, but it can be carried to any precision that a problem needs. The number line diagram in this lesson shows the interval from 1.4 to 1.5 with the point 1.414 inside it, and also marks √3 ≈ 1.732, which is located later in the lesson.

Comparing irrational numbers. To compare √10 and π, approximate each to the same place. Since 3.1² = 9.61 and 3.2² = 10.24, √10 lies between 3.1 and 3.2; since 3.16² = 9.9856 and 3.17² = 10.0489, √10 lies between 3.16 and 3.17. The expansion of π begins 3.14159…, so π lies between 3.14 and 3.15. Because every number between 3.16 and 3.17 is larger than every number between 3.14 and 3.15, √10 > π. The key move is to approximate until the intervals no longer overlap; comparing 3.1 with 3.1 to one decimal place decides nothing, so the precision must increase until it does.

Comparing an irrational with a rational. Is √5 larger or smaller than 2.2? Square the rational: 2.2² = 4.84, which is less than 5, so 2.2 < √5. Also 2.3² = 5.29 > 5, so √5 is between 2.2 and 2.3. There is no need to find digits of √5; squaring the rational candidate and comparing with 5 settles it. Likewise, to compare 3√2 with 4, note 3√2 is between 3 × 1.414 = 4.242 and 3 × 1.415 = 4.245, so 3√2 > 4. Multiplying an interval by a positive number keeps its endpoints in order.

Locating on a number line. To locate √3 approximately, note 1.7² = 2.89 and 1.8² = 3.24, so √3 is between 1.7 and 1.8, and 1.73² = 2.9929 while 1.74² = 3.0276, so √3 is between 1.73 and 1.74. On a number line marked in tenths, place √3 a little past 1.7, about three tenths of the way toward 1.8. Always say what precision the placement uses, and know the difference between trapping and rounding: 1.73 < √3 < 1.74 traps √3 between hundredths, but to round it to the nearest hundredth you must also test the midpoint, and 1.735² = 3.010225 > 3 shows √3 is below 1.735. So "√3 to the nearest hundredth is 1.73" is a claim that can be checked, while "√3 is about 1.7" is a coarser claim that is also true. Both are rational approximations; they differ in how tightly they trap the number.

Estimating expressions. To estimate π², use an interval for π and square its endpoints. From 3.1 < π < 3.2 we get 9.61 < π² < 10.24, so π² is between 9.6 and 10.3. From 3.14 < π < 3.15 we get 9.8596 < π² < 9.9225, so π² is about 9.9. A tighter input interval gives a tighter output interval. For √2 + √3, add the lower endpoints and the upper endpoints: 1.414 + 1.732 = 3.146 and 1.415 + 1.733 = 3.148, so √2 + √3 is between 3.146 and 3.148, roughly 3.15. When you add or multiply intervals of positive numbers, the lowest result comes from the smallest inputs and the highest from the largest, which is why endpoint-by-endpoint calculation gives a valid trap.

A worked model, in full. Estimate 2π − √10 to one decimal place. Take 3.14 < π < 3.15, so 6.28 < 2π < 6.30. Take 3.16 < √10 < 3.17. Subtracting an interval is the one place where care is needed: the smallest difference is the smallest 2π minus the largest √10, which is 6.28 − 3.17 = 3.11, and the largest difference is 6.30 − 3.16 = 3.14. So 2π − √10 is between 3.11 and 3.14, and to one decimal place it is 3.1. A student who subtracts lower from lower and upper from upper would get 3.12 and 3.13, an interval that happens to sit inside the correct one here but that is not guaranteed to contain the true value in general. The safe rule is: for a difference, pair the low end of the first with the high end of the second, and vice versa.

Common misconceptions. First, "√2 = 1.414" is false; 1.414 is a rational approximation whose square is 1.999396, not 2. Write ≈, not =. Second, a calculator value such as 1.4142135624 is also only an approximation, rounded to the display. Third, the decimal expansion of an irrational number does not "eventually stop" if you compute far enough; irrational numbers are exactly the real numbers whose expansions are non-repeating and non-terminating, so refining an approximation is a process that can always be continued.

Worked interval for √50: Since 7² = 49 < 50 and 8² = 64 > 50, √50 is between 7 and 8. Test tenths: 7.0² = 49 and 7.1² = 50.41, so 7.0 < √50 < 7.1. Test hundredths: 7.07² = 49.9849 and 7.08² = 50.1264, so 7.07 < √50 < 7.08. Trapping is not yet rounding, so test the midpoints: 7.05² = 49.7025 < 50 puts √50 above 7.05, so to the nearest tenth √50 ≈ 7.1; 7.075² = 50.055625 > 50 puts it below 7.075, so to the nearest hundredth √50 ≈ 7.07. To continue, square 7.070, 7.071, … until the squares straddle 50.

Worked comparison of three numbers: Order √15, 3.9 and π + 0.7 from least to greatest. Square 3.9: 15.21 > 15, so √15 < 3.9; also 3.8² = 14.44 < 15, so √15 is between 3.8 and 3.9. For π + 0.7, use 3.14 < π < 3.15 to get 3.84 < π + 0.7 < 3.85. Now √15 needs more precision: 3.87² = 14.9769 < 15 and 3.88² = 15.0544 > 15, so 3.87 < √15 < 3.88. The intervals for π + 0.7 and √15 no longer overlap, so the order is π + 0.7 < √15 < 3.9.

Worked estimate of an expression: Estimate the area of a circle of radius 2 cm, which is 4π square centimeters, to the nearest tenth. Using 3.14 < π < 3.15, 12.56 < 4π < 12.60. Both endpoints round to 12.6, so the area is about 12.6 cm². Checking with a coarser interval 3.1 < π < 3.2 gives 12.4 < 4π < 12.8, which is consistent but not precise enough to decide the tenths digit; the hundredths interval was needed.

Worked cube root: Locate ∛20 between consecutive tenths. 2³ = 8 and 3³ = 27, so 2 < ∛20 < 3. Then 2.7³ = 19.683 < 20 and 2.8³ = 21.952 > 20, so 2.7 < ∛20 < 2.8. On a number line from 2 to 3 marked in tenths, ∛20 sits just past 2.7.

Number line construction: Draw a number line from 1 to 4 marked in tenths. Locate √2, √3, √5, √7 and √10 by trapping each between consecutive tenths with squared candidates, and record beside the line the two squares you compared for each number; locate π using its given expansion 3.14159…, which places it between 3.1 and 3.2. Then draw a zoomed line from 1.4 to 1.5 marked in hundredths and place √2 using 1.41² and 1.42².

Independent performance: Choose two irrational numbers of your own, one a square root of a non-square whole number and one an expression involving π. Trap each between hundredths, showing every square or product you computed; if the two intervals overlap, refine to thousandths (or further) until they do not. Then decide which is larger with a sentence that names the two non-overlapping intervals. Estimate the sum and the difference of your two numbers using endpoint arithmetic, stating which endpoints you paired for the difference and why. Success criteria: every interval is justified by explicit calculations, the comparison cites non-overlapping intervals, and the final estimates are stated with ≈ and a named precision, never with =.