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Approximate irrational numbers

Learning goals

  • Bounds an irrational number between nearby rational numbers.
  • Uses perfect squares, perfect cubes, or decimal benchmarks to refine the estimate.
  • Approximates to a requested precision.
  • Locates the approximation correctly on a number line.

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Approximate irrational numbers 1
Without using a calculator, find two consecutive whole numbers between which \(\sqrt{27}\) lies. Explain using perfect squares.

Hints

- Find a perfect square just smaller than the number under the root. - Find the next larger perfect square. - Compare their positive square roots.

Solution

1. The neighboring perfect squares are \(25=5^2\) and \(36=6^2\). 2. Since \(25<27<36\), taking positive square roots gives \(5<\sqrt{27}<6\).

Answer

Since \(5^2=25<27<36=6^2\), \(5<\sqrt{27}<6\).
Approximate irrational numbers 2
Between which two consecutive integers does \(\sqrt[3]{-50}\) lie? Use two nearby perfect cubes to explain.

Hints

- Try small negative integers and cube them. - What happens to the sign when you cube a negative integer? - Put the two cube values in the correct order on the number line.

Solution

1. The neighboring perfect cubes are \((-4)^3=-64\) and \((-3)^3=-27\). 2. Because \(-64<-50<-27\), the cube root is between \(-4\) and \(-3\).

Answer

Since \((-4)^3=-64<-50<-27=(-3)^3\), \(-4<\sqrt[3]{-50}<-3\).
Approximate irrational numbers 3
Let \(a=-\sqrt{13}\) and \(b=-3.5\). Which number lies farther to the right on the number line? Justify your answer by comparing squares without using a calculator.

Hints

- Compare the magnitudes by squaring the positive values. - Recall how the order changes when positive numbers are replaced by their opposites. - Compute \(3.5^2\) exactly.

Solution

1. Compare the positive magnitudes. Since \((\sqrt{13})^2=13\) and \(3.5^2=12.25\), \(\sqrt{13}>3.5\). 2. For negative numbers, the number with the smaller magnitude is greater. Therefore, \(-3.5>-\sqrt{13}\), so \(b\) lies farther to the right.

Answer

\(b=-3.5\) lies farther to the right because \(-3.5>-\sqrt{13}\).
Approximate irrational numbers 4
For each number \(150\), \(180\), and \(210\), decide whether its square root lies between \(13\) and \(14\). Justify each decision by calculating \(13^2\) and \(14^2\) and comparing.

Hints

- How does comparing radicands to perfect squares help compare their square roots? - If a square root lies between two numbers, where must its radicand lie relative to their squares? - Calculate the two boundary squares first.

Solution

1. The boundary squares are \(13^2 = 169\) and \(14^2 = 196\). 2. Since \(150 < 169\), \(\sqrt{150} < 13\). 3. Since \(169 < 180 < 196\), \(13 < \sqrt{180} < 14\). 4. Since \(210 > 196\), \(\sqrt{210} > 14\).

Answer

Only \(\sqrt{180}\) lies between \(13\) and \(14\), because \(169 < 180 < 196\).
Approximate irrational numbers 5
Use \(\sqrt{15}\approx3.8729833\ldots\). a) Give lower and upper decimal bounds for \(\sqrt{15}\) at increments of \(0.1\), \(0.01\), and \(0.0001\). b) Which interval is the tightest, and what is its width? c) Round \(\sqrt{15}\) to the nearest thousandth.

Hints

- Truncate the decimal at the requested place for the lower bound. - The upper bound is the next decimal at that increment. - Use the digit immediately after the thousandths place when rounding.

Solution

1. At increments of \(0.1\), \(3.8<\sqrt{15}<3.9\). 2. At increments of \(0.01\), \(3.87<\sqrt{15}<3.88\). 3. At increments of \(0.0001\), \(3.8729<\sqrt{15}<3.8730\). 4. The tightest interval is \((3.8729,3.8730)\), and its width is \(3.8730-3.8729=0.0001\). 5. The thousandths digit is \(2\), and the next digit is \(9\), so \(\sqrt{15}\approx3.873\) to the nearest thousandth.

Answer

a) \(3.8<\sqrt{15}<3.9\); \(3.87<\sqrt{15}<3.88\); \(3.8729<\sqrt{15}<3.8730\) b) The tightest interval has width \(0.0001\). c) \(\sqrt{15}\approx3.873\)
Approximate irrational numbers 6
Use \(\sqrt{13}\approx3.606\) and \(\sqrt{2}\approx1.414\). Estimate \(A=2\sqrt{13}\) and \(B=5\sqrt{2}\) to the nearest tenth. Which value is greater?

Hints

- Use the supplied square-root approximations. - Multiply before rounding the final results. - Round both values to the same place before comparing.

Solution

1. \(A\approx2\cdot3.606=7.212\), so \(A\approx7.2\). 2. \(B\approx5\cdot1.414=7.070\), so \(B\approx7.1\). 3. Since \(7.2>7.1\), \(A>B\).

Answer

\(A \approx 7.2\) and \(B \approx 7.1\). Therefore, \(A > B\).
Approximate irrational numbers 7
Use \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\) to evaluate each expression. Round the result to the nearest hundredth. a) \(1.5\sqrt{2} + 2\) b) \(2\sqrt{3} - \sqrt{2}\) c) \(\frac{\sqrt{3} + \sqrt{2}}{2}\)

Hints

- Substitute the given square-root approximations first. - Follow the order of operations carefully. - Round only after completing the calculation. - Use the thousandths digit to round to the nearest hundredth.

Solution

1. For a), \(1.5 \cdot 1.414 + 2 = 2.121 + 2 = 4.121\), so the result is approximately \(4.12\). 2. For b), \(2 \cdot 1.732 - 1.414 = 3.464 - 1.414 = 2.050\), so the result is approximately \(2.05\). 3. For c), \(\frac{1.732 + 1.414}{2} = \frac{3.146}{2} = 1.573\), so the result is approximately \(1.57\).

Answer

a) \(\approx 4.12\) b) \(\approx 2.05\) c) \(\approx 1.57\)
Approximate irrational numbers 8
Use \(\sqrt{3}\approx1.7321\), \(\sqrt{10}\approx3.1623\), \(\sqrt{8}\approx2.8284\), \(\sqrt{2}\approx1.4142\), and \(\sqrt{5}\approx2.2361\). Estimate each expression and round to the nearest hundredth. 1) \(2.45+\sqrt{3}\) 2) \(7.18-\sqrt{10}\) 3) \(\sqrt{8}+\sqrt{2}\) 4) \(3\sqrt{5}-4\)

Hints

- Substitute the supplied approximation for each square root. - Complete all addition, subtraction, and multiplication before rounding. - Use the thousandths digit to round to the nearest hundredth.

Solution

1. Using \(\sqrt{3} \approx 1.7321\), \(2.45 + \sqrt{3} \approx 4.1821\), which rounds to \(4.18\). 2. Using \(\sqrt{10} \approx 3.1623\), \(7.18 - \sqrt{10} \approx 4.0177\), which rounds to \(4.02\). 3. Using \(\sqrt{8} \approx 2.8284\) and \(\sqrt{2} \approx 1.4142\), the sum is approximately \(4.2426\), which rounds to \(4.24\). 4. Using \(\sqrt{5} \approx 2.2361\), \(3\sqrt{5} - 4 \approx 6.7083 - 4 = 2.7083\), which rounds to \(2.71\).

Answer

1) \(\approx 4.18\) 2) \(\approx 4.02\) 3) \(\approx 4.24\) 4) \(\approx 2.71\)
Approximate irrational numbers 9
Use \(\sqrt{7}\approx2.64575\), \(\sqrt{2}\approx1.41421\), and \(\sqrt{15}\approx3.87298\). Evaluate each expression and round to the nearest hundredth. 1) \(4\sqrt{7}\) 2) \(9\sqrt{2}\) 3) \(2\sqrt{15}\)

Hints

- Substitute the supplied approximation for each square root. - Multiply before rounding. - Use the thousandths digit to round to the nearest hundredth.

Solution

1. \(4\sqrt{7} \approx 4 \cdot 2.64575 = 10.583\), so the result is approximately \(10.58\). 2. \(9\sqrt{2} \approx 9 \cdot 1.41421 = 12.72789\), so the result is approximately \(12.73\). 3. \(2\sqrt{15} \approx 2 \cdot 3.87298 = 7.74596\), so the result is approximately \(7.75\).

Answer

1) \(\approx 10.58\) 2) \(\approx 12.73\) 3) \(\approx 7.75\)
Approximate irrational numbers 10
Two intervals are marked on the number line. Which interval contains \(\sqrt{13}\)? Justify the choice by squaring the endpoints.
Figure for problem 549857

Hints

- Identify the endpoints of each marked interval. - Compare the squares of nearby decimals with the radicand. - The square root lies between endpoints whose squares surround the radicand.

Solution

1. \(3.6^2=12.96<13\). 2. \(3.7^2=13.69>13\). 3. Therefore, \(3.6<\sqrt{13}<3.7\). 4. Interval B is the interval from \(3.6\) to \(3.7\).

Answer

Interval B
Approximate irrational numbers 11
The number line is marked in tenths. Between which two consecutive tenths does \(\sqrt{29}\) lie? Justify your answer by squaring both endpoints.
Figure for problem 550158

Hints

- First identify the consecutive whole numbers that contain the square root. - Test neighboring tenths by squaring them. - The lower endpoint must have a square below \(29\), and the upper endpoint must have a square above \(29\).

Solution

1. Since \(5^2=25<29<36=6^2\), \(\sqrt{29}\) lies between \(5\) and \(6\). 2. \(5.3^2=28.09<29\). 3. \(5.4^2=29.16>29\). 4. Therefore, \(5.3<\sqrt{29}<5.4\).

Answer

\(5.3<\sqrt{29}<5.4\)
Approximate irrational numbers 12
Show that \(\sqrt{32}\) lies between \(5.6\) and \(5.7\), without finding its decimal value on a calculator.

Hints

- Squaring a positive guess tells you which side of the root it lies on. - Calculate both squares carefully to two decimal places. - Turn the squared comparisons back into a statement about the root.

Solution

1. Square the lower guess: \(5.6^2=31.36\), which is below \(32\). 2. Square the upper guess: \(5.7^2=32.49\), which is above \(32\). 3. Since both guesses are positive, \(5.6<\sqrt{32}<5.7\).

Answer

\(5.6^2=31.36<32<32.49=5.7^2\), so \(5.6<\sqrt{32}<5.7\).
Approximate irrational numbers 13
Where does \(\sqrt{2}\) belong relative to the marked points: left of \(A\), between \(A\) and \(B\), or right of \(B\)? Explain using squares.
Figure for problem 559906

Hints

- Read both marked positions from the number line. - Compare the square of each positive endpoint with \(2\). - If an endpoint's square is too small, where must the root lie?

Solution

1. The marked positions are \(A=1.4\) and \(B=1.5\). 2. Their squares are \(1.4^2=1.96\) and \(1.5^2=2.25\). 3. Since \(1.96<2<2.25\), \(1.4<\sqrt{2}<1.5\). It belongs between the markers.

Answer

Between \(A\) and \(B\), because \(1.4^2=1.96<2<2.25=1.5^2\).
Approximate irrational numbers 14
A square garden has area \(70\,\text{ft}^2\). Without a calculator, between which two whole numbers of feet is its side length? Explain.

Hints

- The area of a square is its side length squared. - Find two nearby whole-number sides and compare their areas. - Use feet for the final side-length estimate, not square feet.

Solution

1. A square with side \(8\,\text{ft}\) has area \(64\,\text{ft}^2\). 2. A square with side \(9\,\text{ft}\) has area \(81\,\text{ft}^2\). 3. Since \(64<70<81\), the garden's side is between \(8\) and \(9\) feet.

Answer

Since \(8^2=64<70<81=9^2\), the side length is between \(8\,\text{ft}\) and \(9\,\text{ft}\).
Approximate irrational numbers 15
Mei estimates \(\sqrt{27}\) by dividing \(27\) by a guess, then averaging the guess and that quotient. Starting with \(9\), she writes \(9+27\div9=12\) as her new guess. Find her mistake and the correct new guess. Square the correct guess to decide whether it is exact, too high, or too low.

Hints

- Work out the quotient Mei used before checking her new guess. - Does her written calculation follow every step of the stated rule? - What can the square of a corrected guess tell you about its size?

Solution

1. Mei added \(9\) and \(27\div9\) but forgot to divide their sum by \(2\) to find the average. 2. Since \(27\div9=3\), the correct guess is \((9+3)\div2=6\). 3. Its square is \(6^2=36>27\). The new guess is too high, not exact.

Answer

Mei forgot to divide by \(2\). The correct guess is \(6\); it is too high because \(6^2=36>27\).
Approximate irrational numbers 16
A circle of radius \(r>0\) has area \(\pi r^2\). A square drawn inside the circle has area \(2r^2\). A square drawn around the circle has area \(4r^2\). a) Use the two squares to find lower and upper bounds for \(\pi\). b) A 12-sided polygon drawn inside the circle has area \(3r^2\). Another 12-sided polygon surrounds the circle and has area **less than** \(3.216r^2\). Find new bounds for \(\pi\) and the difference between those bounds. c) Which shapes give a narrower interval for \(\pi\)? Compare the two interval widths.

Hints

- Which shape has the smallest area? Which has the largest? - Divide each inequality by the same positive quantity. - A narrower interval gives more precise bounds.

Solution

1. The circle lies between the two squares: \(2r^2<\pi r^2<4r^2\). Since \(r^2>0\), divide to get \(2<\pi<4\). 2. The new area bounds give \(3r^2<\pi r^2<3.216r^2\), so \(3<\pi<3.216\). 3. The new interval has width \(3.216-3=0.216\), while the square interval has width \(4-2=2\). 4. Because \(0.216<2\), the 12-sided polygons give tighter bounds.

Answer

a) \(2<\pi<4\) b) \(3<\pi<3.216\); width \(0.216\) c) The 12-sided polygons, since \(0.216<2\).
Approximate irrational numbers 17
Order the numbers from least to greatest without using a calculator: \(\sqrt{18},\ 4.1,\ \sqrt{16},\ 4,\ \sqrt{20}\)

Hints

- Evaluate any square roots of perfect squares first. - To compare a positive decimal with a square root, compare their squares. - Use the order of the radicands to compare positive square roots.

Solution

1. Since \(\sqrt{16}=4\), those two values are equal. 2. Because \(4.1^2=16.81<18\), \(4.1<\sqrt{18}\). 3. Since \(18<20\), \(\sqrt{18}<\sqrt{20}\). 4. Therefore, \(4=\sqrt{16}<4.1<\sqrt{18}<\sqrt{20}\).

Answer

\(4=\sqrt{16}<4.1<\sqrt{18}<\sqrt{20}\)
Approximate irrational numbers 18
Without using a calculator, insert \(<\), \(>\), or \(=\) to make each statement true. Briefly justify each answer. a) \(\sqrt{0.09}\) ___ \(0.09\) b) \(\sqrt{1.21}\) ___ \(1.1\) c) \(\sqrt{20}\) ___ \(4.5\) d) \(\sqrt{\frac{1}{2}}\) ___ \(\frac{1}{2}\)

Hints

- Evaluate square roots of perfect-square decimals when possible. - When both values are nonnegative, compare their squares. - Remember how squaring affects numbers between \(0\) and \(1\).

Solution

1. In a), \(\sqrt{0.09}=0.3\), and \(0.3>0.09\). 2. In b), \(1.1^2=1.21\), so \(\sqrt{1.21}=1.1\). 3. In c), \(4.5^2=20.25>20\), so \(\sqrt{20}<4.5\). 4. In d), both quantities are positive, and \(\left(\frac{1}{2}\right)^2=\frac{1}{4}<\frac{1}{2}\). Therefore, \(\sqrt{\frac{1}{2}}>\frac{1}{2}\).

Answer

a) \(\sqrt{0.09}>0.09\) b) \(\sqrt{1.21}=1.1\) c) \(\sqrt{20}<4.5\) d) \(\sqrt{\frac{1}{2}}>\frac{1}{2}\)
Approximate irrational numbers 19
Estimate each square root without using a calculator. For every part, first locate the root between the two given benchmarks. Then find the midpoint of those benchmarks, square that midpoint, and use the comparison to decide the requested rounding. Your work must include the midpoint and its square. a) Estimate \(\sqrt{500}\) to the nearest whole number. Use \(22^2=484\) and \(23^2=529\). b) Estimate \(\sqrt{0.05}\) to the nearest tenth. Use \(0.2^2=0.04\) and \(0.3^2=0.09\). c) Estimate \(\sqrt{50{,}000}\) to the nearest ten. Use \(220^2=48{,}400\) and \(230^2=52{,}900\).

Hints

- Use the two given square benchmarks to establish an interval for each root. - The rounding boundary is the midpoint of the two candidate rounded values. - Square that midpoint and compare its square with the radicand before choosing the rounded value.

Solution

1. Since \(484<500<529\), \(22<\sqrt{500}<23\). The midpoint is \(22.5\), and \(22.5^2=506.25>500\), so \(\sqrt{500}<22.5\). Therefore, \(\sqrt{500}\approx22\) to the nearest whole number. 2. Since \(0.04<0.05<0.09\), \(0.2<\sqrt{0.05}<0.3\). The midpoint is \(0.25\), and \(0.25^2=0.0625>0.05\), so \(\sqrt{0.05}<0.25\). Therefore, \(\sqrt{0.05}\approx0.2\) to the nearest tenth. 3. Since \(48{,}400<50{,}000<52{,}900\), \(220<\sqrt{50{,}000}<230\). The midpoint is \(225\), and \(225^2=50{,}625>50{,}000\), so \(\sqrt{50{,}000}<225\). Therefore, \(\sqrt{50{,}000}\approx220\) to the nearest ten.

Answer

a) Midpoint \(22.5\); \(22.5^2=506.25>500\), so \(\sqrt{500}\approx22\). b) Midpoint \(0.25\); \(0.25^2=0.0625>0.05\), so \(\sqrt{0.05}\approx0.2\). c) Midpoint \(225\); \(225^2=50{,}625>50{,}000\), so \(\sqrt{50{,}000}\approx220\).
Approximate irrational numbers 20
A square storage room has a floor area of \(32\,\text{m}^2\). a) Find the side length \(s\). Give the exact value as a square root and an approximation rounded to the nearest hundredth. b) Between which two consecutive whole numbers of meters does the side length lie? Justify your answer by comparing perfect squares. c) If the side length were doubled, what would the new floor area be?
Figure for problem 514354

Hints

- Is \(32\) a perfect square? - Which familiar perfect squares are closest to \(32\)? - How does a square’s area change when its side length is multiplied by a factor?

Solution

1. The side length is \(s = \sqrt{32}\,\text{m} \approx 5.66\,\text{m}\). 2. Since \(5^2 = 25\) and \(6^2 = 36\), and \(25 < 32 < 36\), \(\sqrt{32}\) lies between \(5\) and \(6\). 3. Doubling the side length multiplies the area by \(2^2 = 4\). The new area is \(4 \cdot 32\,\text{m}^2 = 128\,\text{m}^2\).

Answer

a) \(s = \sqrt{32}\,\text{m} \approx 5.66\,\text{m}\) b) The side length lies between \(5\,\text{m}\) and \(6\,\text{m}\). c) The new floor area is \(128\,\text{m}^2\).
Approximate irrational numbers 21
Consider \(\sqrt{7}\). a) Find two rational numbers with exactly one decimal place between which \(\sqrt{7}\) lies. b) Sushma claims, “If I keep narrowing rational intervals, eventually I will find two rational endpoints so close together that \(\sqrt{7}\) is exactly one of them.” Explain why this claim contradicts the definition of an irrational number.

Hints

- Square one-decimal-place candidates to compare them with \(7\). - Recall the defining difference between rational and irrational decimal expansions. - Distinguish approximating a number arbitrarily closely from equaling it exactly.

Solution

1. Since \(2.6^2 = 6.76\) and \(2.7^2 = 7.29\), \(2.6 < \sqrt{7} < 2.7\). 2. Every terminating decimal is rational because it can be written as a fraction of integers. 3. If the interval process ended with \(\sqrt{7}\) equal to a rational endpoint, then \(\sqrt{7}\) would be rational. 4. But \(\sqrt{7}\) is irrational. Its decimal expansion is nonterminating and nonrepeating. Rational intervals can approximate it as closely as desired, but no terminating rational endpoint equals it exactly.

Answer

a) \(2.6 < \sqrt{7} < 2.7\) b) The claim is false. Every terminating decimal endpoint is rational, but \(\sqrt{7}\) is irrational, so no such endpoint can equal \(\sqrt{7}\) exactly.
Approximate irrational numbers 22
Let \(x = \sqrt{45}\) and \(y = 6.71\). a) Find an interval of width \(0.01\) that contains \(x\). b) Compare \(x\) and \(y\). Which number is greater? Justify your answer using the interval from part a).

Hints

- Square nearby hundredths to determine whether they are below or above \(45\). - What does the upper endpoint of the interval tell you when comparing \(x\) with \(6.71\)?

Solution

1. Since \(6.70^2 = 44.89\) and \(6.71^2 = 45.0241\), \(44.89 < 45 < 45.0241\). 2. Therefore, \(6.70 < \sqrt{45} < 6.71\), so an interval of width \(0.01\) containing \(x\) is \([6.70, 6.71]\). 3. Because \(x < 6.71 = y\), \(y\) is greater than \(x\).

Answer

a) \(x \in [6.70, 6.71]\) b) \(y > x\), because \(6.71^2 = 45.0241 > 45\), so \(\sqrt{45} < 6.71\).
Approximate irrational numbers 23
A real number \(z\) lies in the interval \(I_3 = [5.38, 5.39]\). Determine whether each number could be the value of \(z\). Justify each answer with a calculation. a) \(\sqrt{29}\) b) \(\frac{43}{8}\) c) \(5.382\)

Hints

- A number is in \([a, b]\) when it is at least \(a\) and at most \(b\). - Compare a square root with positive bounds by squaring the bounds. - Convert the fraction to a decimal for a direct comparison.

Solution

1. For a), square the positive interval endpoints: \(5.38^2 = 28.9444\) and \(5.39^2 = 29.0521\). Since \(28.9444 < 29 < 29.0521\), it follows that \(5.38 < \sqrt{29} < 5.39\). Therefore, \(\sqrt{29}\) is in the interval. 2. For b), \(\frac{43}{8} = 5.375\). Since \(5.375 < 5.38\), the number is not in the interval. 3. For c), \(5.38 < 5.382 < 5.39\), so \(5.382\) is in the interval.

Answer

a) Yes, because \(5.38^2 < 29 < 5.39^2\). b) No, because \(\frac{43}{8} = 5.375 < 5.38\). c) Yes, because \(5.38 < 5.382 < 5.39\).
Approximate irrational numbers 24
The Babylonian method for approximating square roots can be viewed geometrically as repeatedly changing a rectangle into a more square-like rectangle with the same area. A rectangle has area \(18\,\text{cm}^2\) and one side length \(a_0=6\,\text{cm}\). a) Find the other side length \(b_0\). b) The next approximation \(a_1\) is the mean of \(a_0\) and \(b_0\). Find \(a_1\). c) Find the new side length \(b_1\) that keeps the area equal to \(18\,\text{cm}^2\). d) Calculate \(b_1^2\) and \(a_1^2\). Use those comparisons with \(18\,\text{cm}^2\) to place the square's side length \(\sqrt{18}\,\text{cm}\) between \(b_1\) and \(a_1\). Then find the absolute square error \(\left|a_1^2-18\,\text{cm}^2\right|\).

Hints

- Relate the side lengths of a rectangle to its area. - The next Babylonian approximation uses the mean of the two current side lengths. - After finding \(a_1\) and \(b_1\), compare their squares with the target area. - Use the order of the positive squares to infer the order of the corresponding positive side lengths and \(\sqrt{18}\).

Solution

1. Since \(A=a_0b_0\), \(b_0=18\div6=3\,\text{cm}\). 2. The mean is \(a_1=\frac{6+3}{2}=4.5\,\text{cm}\). 3. To preserve the area, \(b_1=18\div4.5=4\,\text{cm}\). 4. \(b_1^2=(4\,\text{cm})^2=16\,\text{cm}^2\), while \(a_1^2=(4.5\,\text{cm})^2=20.25\,\text{cm}^2\). 5. Since \(16\,\text{cm}^2<18\,\text{cm}^2<20.25\,\text{cm}^2\) and \(b_1\) and \(a_1\) are positive, \(4\,\text{cm}<\sqrt{18}\,\text{cm}<4.5\,\text{cm}\). 6. The absolute square error of \(a_1\) is \(\left|20.25-18\right|\,\text{cm}^2=2.25\,\text{cm}^2\).

Answer

a) \(b_0=3\,\text{cm}\) b) \(a_1=4.5\,\text{cm}\) c) \(b_1=4\,\text{cm}\) d) \(b_1^2=16\,\text{cm}^2<18\,\text{cm}^2<20.25\,\text{cm}^2=a_1^2\), so \(4\,\text{cm}<\sqrt{18}\,\text{cm}<4.5\,\text{cm}\). The absolute square error of \(a_1\) is \(2.25\,\text{cm}^2\).
Approximate irrational numbers 25
Two students estimate \(\sqrt{2}\). Student A uses \(1.41\), and Student B uses \(1.42\). a) Use \(1.41^2=1.9881\) and \(1.42^2=2.0164\) to show that \(\sqrt{2}\) lies between the two estimates. b) The midpoint of the estimates is \(1.415\), and \(1.415^2=2.002225\). Which student has the closer estimate? Explain.

Hints

- Compare the given squares with \(2\) to locate the square root. - The midpoint separates values closer to \(1.41\) from values closer to \(1.42\). - Square the midpoint to decide on which side of it \(\sqrt{2}\) lies.

Solution

1. Since \(1.9881<2<2.0164\), it follows that \(1.41<\sqrt{2}<1.42\). 2. Because \(1.415^2=2.002225>2\), \(\sqrt{2}<1.415\). The exact value lies below the midpoint between \(1.41\) and \(1.42\), so it is closer to \(1.41\). Student A has the closer estimate.

Answer

a) \(1.41<\sqrt{2}<1.42\) b) Student A; \(1.41\) is closer to \(\sqrt{2}\).
Approximate irrational numbers 26
The number line shows an interval \(I\). a) Show that \(\sqrt{10}\) lies in \(I\) by squaring the endpoints. b) Compare \(\sqrt{10}\) with \(3.165\). c) Round \(\sqrt{10}\) to the nearest hundredth.
Figure for problem 514417

Hints

- Square the two endpoints of the displayed interval. - Use \(3.165\) as the midpoint between the hundredths. - The side of the midpoint determines the nearest hundredth.

Solution

1. The interval is \(I=[3.16,3.17]\). Since \(3.16^2=9.9856<10\) and \(3.17^2=10.0489>10\), \(3.16<\sqrt{10}<3.17\). 2. Since \(3.165^2=10.017225>10\), \(\sqrt{10}<3.165\). 3. The value lies below the midpoint between \(3.16\) and \(3.17\), so \(\sqrt{10}\approx3.16\) to the nearest hundredth.

Answer

a) \(3.16<\sqrt{10}<3.17\) b) \(\sqrt{10}<3.165\) c) \(\sqrt{10}\approx3.16\)
Approximate irrational numbers 27
Find three nested decimal intervals that contain \(\sqrt{11}\): first an interval with consecutive whole-number endpoints, then one with consecutive tenth endpoints, and then one with consecutive hundredth endpoints.

Hints

- Which perfect squares are closest to \(11\)? - Square consecutive tenths to narrow the interval. - Continue by testing consecutive hundredths.

Solution

1. Since \(3^2 = 9\) and \(4^2 = 16\), \(3 < \sqrt{11} < 4\). The first interval is \([3, 4]\). 2. Since \(3.3^2 = 10.89\) and \(3.4^2 = 11.56\), \(3.3 < \sqrt{11} < 3.4\). The second interval is \([3.3, 3.4]\). 3. Since \(3.31^2 = 10.9561\) and \(3.32^2 = 11.0224\), \(3.31 < \sqrt{11} < 3.32\). The third interval is \([3.31, 3.32]\).

Answer

1. \([3, 4]\) 2. \([3.3, 3.4]\) 3. \([3.31, 3.32]\)
Approximate irrational numbers 28
A nested-interval approximation for \(\sqrt{2}\) begins with \(I_1 = [1, 2]\). Each new interval is formed by dividing the previous interval into \(10\) equal parts and selecting the part that contains \(\sqrt{2}\). a) What is the width of the fourth interval \(I_4\)? b) Use squaring to determine \(I_4\), and decide whether \(1.414\) belongs to it.

Hints

- What happens to an interval’s width when it is divided into \(10\) equal parts? - Square the proposed endpoints to check whether \(\sqrt{2}\) lies between them. - A closed interval includes both endpoints.

Solution

1. The width of \(I_1\) is \(1\). Each subdivision divides the width by \(10\), so the widths of \(I_2\), \(I_3\), and \(I_4\) are \(0.1\), \(0.01\), and \(0.001\), respectively. 2. Since \(1.4^2 = 1.96\) and \(1.5^2 = 2.25\), \(I_2 = [1.4, 1.5]\). 3. Since \(1.41^2 = 1.9881\) and \(1.42^2 = 2.0164\), \(I_3 = [1.41, 1.42]\). 4. Since \(1.414^2 = 1.999396\) and \(1.415^2 = 2.002225\), \(I_4 = [1.414, 1.415]\). The number \(1.414\) is the lower endpoint, so it belongs to this closed interval.

Answer

a) The width of \(I_4\) is \(0.001\). b) \(I_4 = [1.414, 1.415]\), and \(1.414\) belongs to the interval.
Approximate irrational numbers 29
Consider \(\sqrt{18}\). a) Find an interval of the form \([n, n+1]\), where \(n\) is a natural number, that contains \(\sqrt{18}\). b) Without directly evaluating the square root on a calculator, decide whether \(\sqrt{18}\) is closer to \(4.2\) or \(4.3\). Justify your answer by comparing squares.

Hints

- Begin with neighboring perfect squares. - The midpoint determines which of two approximations is closer. - Find the midpoint of \(4.2\) and \(4.3\), then compare its square with \(18\).

Solution

1. Since \(4^2 = 16 < 18 < 25 = 5^2\), \(4 < \sqrt{18} < 5\). Thus, \(\sqrt{18}\) lies in \([4, 5]\). 2. The midpoint of \(4.2\) and \(4.3\) is \(4.25\). Since \(4.2^2 = 17.64 < 18\) and \(4.25^2 = 18.0625 > 18\), \(4.2 < \sqrt{18} < 4.25\). 3. Because \(\sqrt{18}\) is below the midpoint, it is closer to \(4.2\) than to \(4.3\).

Answer

a) \(\sqrt{18} \in [4, 5]\) b) \(\sqrt{18}\) is closer to \(4.2\), because \(4.2 < \sqrt{18} < 4.25\), and \(4.25\) is the midpoint of \(4.2\) and \(4.3\).
Approximate irrational numbers 30
Consider \(\sqrt{13}\). a) Find the two consecutive natural numbers between which \(\sqrt{13}\) lies. b) Use squaring to determine whether \(\sqrt{13}\) is closer to \(3.6\) or \(3.61\). c) Without a calculator, explain why \(\frac{11}{3}\) must be greater than \(\sqrt{13}\).

Hints

- Use neighboring perfect squares. - To compare \(3.6\) and \(3.61\), square their midpoint. - A positive number is greater than \(\sqrt{13}\) when its square is greater than \(13\).

Solution

1. Since \(3^2 = 9 < 13 < 16 = 4^2\), \(3 < \sqrt{13} < 4\). 2. The midpoint of \(3.6\) and \(3.61\) is \(3.605\). Since \(3.605^2 = 12.996025 < 13\), \(\sqrt{13}\) lies above the midpoint. Therefore, it is closer to \(3.61\). 3. \(\left(\frac{11}{3}\right)^2 = \frac{121}{9} > 13\). Since \(\frac{11}{3}\) is positive, \(\frac{11}{3} > \sqrt{13}\).

Answer

a) \(3 < \sqrt{13} < 4\) b) \(\sqrt{13}\) is closer to \(3.61\). c) Since \(\left(\frac{11}{3}\right)^2 > 13\), \(\frac{11}{3} > \sqrt{13}\).
Approximate irrational numbers 31
A square lot has an area of \(18\,\text{m}^2\), so its side length is \(s = \sqrt{18}\,\text{m}\). a) Between which two consecutive whole numbers does \(s\) lie? b) Determine the tenths interval containing \(s\) by comparing the squares of \(4.1\), \(4.2\), and \(4.3\). c) Briefly explain why this process never produces a terminating decimal equal to \(\sqrt{18}\).

Hints

- Which familiar perfect squares are nearest to \(18\)? - Compare how the square changes as the decimal input increases by tenths. - Recall the difference between rational and irrational numbers.

Solution

1. Since \(4^2 = 16\) and \(5^2 = 25\), \(4 < \sqrt{18} < 5\). 2. \(4.1^2 = 16.81\), \(4.2^2 = 17.64\), and \(4.3^2 = 18.49\). Therefore, \(4.2 < \sqrt{18} < 4.3\), so the tenths digit is \(2\). 3. The integer \(18\) is not a perfect square, so \(\sqrt{18}\) is irrational. Its decimal expansion is nonterminating and nonrepeating, so increasingly narrow decimal intervals never end at an exact terminating decimal.

Answer

a) \(4\,\text{m} < s < 5\,\text{m}\) b) \(4.2 < \sqrt{18} < 4.3\), so the tenths digit is \(2\). c) \(\sqrt{18}\) is irrational, so its decimal expansion is nonterminating and nonrepeating.
Approximate irrational numbers 32
Let \(A=\sqrt{0.36}+\sqrt{0.25}\) and \(B=\sqrt{0.36+0.25}\). Which expression has the greater value? Justify your answer without approximating the square root in \(B\).

Hints

- Find the exact value of \(A\). - Combine the numbers inside \(B\). - Compare two positive values by comparing their squares.

Solution

1. Evaluate \(A\): \(A=0.6+0.5=1.1\). 2. Combine the numbers under the radical in \(B\): \(B=\sqrt{0.61}\). 3. Both values are positive. Since \(1.1^2=1.21>0.61\), \(1.1>\sqrt{0.61}\). Therefore, \(A>B\).

Answer

\(A>B\)
Approximate irrational numbers 33
Order the real numbers from least to greatest without using a calculator: \(\sqrt{5},\ 2.23,\ 2.\overline{2},\ \frac{9}{4},\ 2.24\)

Hints

- Convert the repeating decimal and fraction to comparable decimal forms. - Compare a positive decimal with \(\sqrt{5}\) by squaring the decimal. - Place the two decimal bounds on opposite sides of \(\sqrt{5}\).

Solution

1. Write the rational values in comparable form: \(2.\overline{2}=2.2222\ldots\) and \(\frac{9}{4}=2.25\). 2. Since \(2.23^2=4.9729<5\), \(2.23<\sqrt{5}\). 3. Since \(2.24^2=5.0176>5\), \(\sqrt{5}<2.24\). 4. Therefore, \(2.\overline{2}<2.23<\sqrt{5}<2.24<\frac{9}{4}\).

Answer

\(2.\overline{2}<2.23<\sqrt{5}<2.24<\frac{9}{4}\)
Approximate irrational numbers 34
The area \(A\) of a circle is given. Use a calculator to find the radius \(r\) and round to the nearest hundredth. a) \(A = 10\,\text{cm}^2\) b) \(A = 35.5\,\text{m}^2\) c) \(A = 0.8\,\text{dm}^2\)

Hints

- Which formula relates a circle’s area and radius? - Rearrange the formula so the squared radius is isolated before taking a square root. - Use the third decimal place to round to the nearest hundredth. - Include the correct length unit in each answer.

Solution

1. From \(A = \pi r^2\), solve for the radius: \(r = \sqrt{\frac{A}{\pi}}\). 2. For a), \(r = \sqrt{\frac{10}{\pi}}\,\text{cm} \approx 1.78\,\text{cm}\). 3. For b), \(r = \sqrt{\frac{35.5}{\pi}}\,\text{m} \approx 3.36\,\text{m}\). 4. For c), \(r = \sqrt{\frac{0.8}{\pi}}\,\text{dm} \approx 0.50\,\text{dm}\).

Answer

a) \(r \approx 1.78\,\text{cm}\) b) \(r \approx 3.36\,\text{m}\) c) \(r \approx 0.50\,\text{dm}\)
Approximate irrational numbers 35
The area of a circle is \(A = \pi r^2\). a) Use a calculator to find the radius of a circle with area \(50\,\text{cm}^2\). Round to the nearest hundredth. b) Find the new area \(A_{\text{new}}\) needed to make the radius exactly twice the original circle’s radius. c) In general, how does a circle’s radius change when its area is multiplied by \(4\)? Justify your answer using the formula.

Hints

- Solve the circle-area formula for the positive radius in part a). - Substitute \(2r\) for the radius to see how the area changes. - Use the same relationship in reverse when the area is multiplied by \(4\).

Solution

1. For part a), \(r=\sqrt{\frac{50}{\pi}}\,\text{cm}\approx3.99\,\text{cm}\). 2. If the radius doubles, \(A_{\text{new}}=\pi(2r)^2=4\pi r^2=4A\). Therefore, \(A_{\text{new}}=4\cdot50\,\text{cm}^2=200\,\text{cm}^2\). 3. Conversely, a circle with radius \(2r\) has area \(4A\). Because a radius is positive, multiplying a circle's area by \(4\) doubles its radius.

Answer

a) \(r \approx 3.99\,\text{cm}\) b) \(A_{\text{new}} = 200\,\text{cm}^2\) c) The radius doubles when the area is multiplied by \(4\).
Approximate irrational numbers 36
Estimate \(15\sqrt{2}\) to the nearest tenth using two methods. a) First round \(\sqrt{2}\) to the nearest tenth, and then multiply by \(15\). b) Instead, use \(\sqrt{2}\approx1.414\), multiply first, and round only the final product to the nearest tenth. c) The correctly rounded value is \(21.2\). Which method is more accurate, and why?

Hints

- Follow the stated order of operations in each method. - Keep all three given decimal places in method b). - Consider what happens to an early approximation error when it is multiplied by \(15\).

Solution

1. Method a): \(\sqrt{2}\approx1.4\), so \(15\cdot1.4=21.0\). 2. Method b): \(15\cdot1.414=21.21\), which rounds to \(21.2\). 3. Method b) is more accurate. Method a) rounds before multiplying, so the initial rounding error is also multiplied by \(15\). Method b) keeps more information and rounds only once, at the end.

Answer

a) \(21.0\) b) \(21.2\) c) Method b) is more accurate because it delays rounding until the final result.
Approximate irrational numbers 37
Consider \(4\sqrt{5}\). a) Compute \((4\sqrt{5})^2\). Use the result to find two consecutive whole numbers between which \(4\sqrt{5}\) lies. b) Use \(\sqrt{5}\approx2.236\) to estimate \(4\sqrt{5}\) to the nearest tenth.

Hints

- Square the entire positive expression rather than rewriting the radical. - Compare the resulting value with neighboring perfect squares. - Multiply the supplied decimal approximation before rounding.

Solution

1. \((4\sqrt{5})^2=16\cdot5=80\). 2. Since \(8^2=64<80<81=9^2\) and \(4\sqrt{5}\) is positive, \(8<4\sqrt{5}<9\). 3. Using \(\sqrt{5}\approx2.236\), \(4\sqrt{5}\approx4\cdot2.236=8.944\), which rounds to \(8.9\).

Answer

a) \(8<4\sqrt{5}<9\) b) \(4\sqrt{5}\approx8.9\)
Approximate irrational numbers 38
A square has area \(A=7\,\text{cm}^2\). a) Find the side length \(s\) and perimeter \(P\). Round both values to the nearest hundredth. b) Use \(d^2=s^2+s^2\) to find the diagonal length \(d\). Round to the nearest hundredth.

Hints

- Use the area to determine the positive side length. - A square has four equal sides. - For the diagonal, substitute \(s^2=7\) into the given relationship and then take the positive square root.

Solution

1. The side length is \(s=\sqrt{7}\,\text{cm}\approx2.65\,\text{cm}\). 2. The perimeter is \(P=4s=4\sqrt{7}\,\text{cm}\approx10.58\,\text{cm}\). 3. For the diagonal, \(d^2=s^2+s^2=7+7=14\). Since a length is positive, \(d=\sqrt{14}\,\text{cm}\approx3.74\,\text{cm}\).

Answer

a) \(s \approx 2.65\,\text{cm}\); \(P \approx 10.58\,\text{cm}\) b) \(d \approx 3.74\,\text{cm}\)
Approximate irrational numbers 39
Find all approximate solutions to each equation. Round to the nearest hundredth. 1) \(x^2 - 13 = 0\) 2) \(0.5x^2 = 11\) 3) \(4x^2 - 30 = 50\)

Hints

- Isolate \(x^2\) before taking a square root. - Include both the positive and negative solutions. - Use the thousandths digit to round to the nearest hundredth.

Solution

1. \(x^2 = 13\), so \(x = \pm\sqrt{13} \approx \pm 3.61\). 2. Multiply by \(2\): \(x^2 = 22\). Therefore, \(x = \pm\sqrt{22} \approx \pm 4.69\). 3. Add \(30\) and divide by \(4\): \(x^2 = 20\). Therefore, \(x = \pm\sqrt{20} \approx \pm 4.47\).

Answer

1) \(x \approx 3.61\) or \(x \approx -3.61\) 2) \(x \approx 4.69\) or \(x \approx -4.69\) 3) \(x \approx 4.47\) or \(x \approx -4.47\)
Approximate irrational numbers 40
Consider the equation \(x^2=40\). a) Without using a calculator’s square-root key, find two consecutive whole numbers between which the positive solution lies. Explain. b) Use a calculator to find both solutions to the nearest hundredth.

Hints

- Which familiar perfect squares are nearest to \(40\)? - Compare \(40\) with those squares. - If a positive number has square \(40\), what can you say about its opposite?

Solution

1. Since \(6^2 = 36\) and \(7^2 = 49\), and \(36 < 40 < 49\), the positive solution \(\sqrt{40}\) lies between \(6\) and \(7\). 2. \(\sqrt{40} \approx 6.32455\ldots\), so the positive solution is approximately \(6.32\). 3. The negative solution is its opposite, approximately \(-6.32\).

Answer

a) The positive solution lies between \(6\) and \(7\). b) \(x \approx 6.32\) or \(x \approx -6.32\)
Approximate irrational numbers 41
Every number in the list \(1.4,\ 1.41,\ 1.414,\ 1.4142,\ldots\) is a terminating decimal approximation to \(\sqrt{2}\). Ioana argues that \(\sqrt{2}\) must therefore be rational because all the approximations are rational. Explain why the conclusion does not follow.

Hints

- Separate “close to” from “equal to.” - Ask whether any finite truncation contains the complete decimal expansion. - Classification applies to the exact number, not to its approximations.

Solution

1. Each listed approximation is rational because it terminates. 2. None of the listed decimals equals \(\sqrt{2}\) exactly. 3. A sequence of rational approximations can approach an irrational number arbitrarily closely. 4. The exact decimal expansion of \(\sqrt{2}\) is nonterminating and nonrepeating, so \(\sqrt{2}\) is irrational.

Answer

The argument is invalid. Rational approximations can approach an irrational number without ever equaling it; \(\sqrt{2}\) remains irrational.
Approximate irrational numbers 42
The number line shows three labeled points. Which labeled point is closest to \(\sqrt{5}\)? Justify your choice without using a calculator’s square-root key.
Figure for problem 549856

Hints

- Square the two neighboring decimal values. - Decide which interval contains the irrational number. - Compare with the midpoint of that interval.

Solution

1. \(2.2^2=4.84\) and \(2.3^2=5.29\), so \(2.2<\sqrt{5}<2.3\). 2. The midpoint is \(2.25\), and \(2.25^2=5.0625>5\). 3. Therefore, \(\sqrt{5}<2.25\), so it is closer to \(2.2\) than to \(2.3\). 4. Point A marks \(2.2\), so A is closest.

Answer

Point A
Approximate irrational numbers 43
The number line shows points P and Q. Which point gives the better nearest-tenth approximation to \(\sqrt{70}\)? Use the midpoint to justify your decision.
Figure for problem 549858

Hints

- Find the value halfway between the two candidate tenths. - Compare the square of that midpoint with the radicand. - Use which side of the midpoint contains the root.

Solution

1. The midpoint between \(8.3\) and \(8.4\) is \(8.35\). 2. \(8.35^2=69.7225<70\), so \(\sqrt{70}>8.35\). 3. Also, \(8.4^2=70.56>70\), so \(\sqrt{70}<8.4\). 4. Therefore, \(\sqrt{70}\) is closer to \(8.4\), which is point Q.

Answer

Point Q; \(\sqrt{70}\approx8.4\) to the nearest tenth.
Approximate irrational numbers 44
Use the bound \(3.14<\pi<3.15\) to estimate \(\pi^2\) to the nearest tenth. Show how squaring the bounds supports the rounded value.

Hints

- Apply the same increasing operation to both positive bounds. - Calculate the two endpoint values accurately. - Check whether the entire resulting interval rounds to one tenth value.

Solution

1. Squaring the positive bounds gives \(3.14^2<\pi^2<3.15^2\). 2. \(3.14^2=9.8596\) and \(3.15^2=9.9225\). 3. Every value in this interval rounds to \(9.9\) to the nearest tenth. 4. Therefore, \(\pi^2\approx9.9\).

Answer

\(\pi^2\approx9.9\)
Approximate irrational numbers 45
The number line shows a marked interval and its midpoint for \(\sqrt[3]{30}\). First cube the two endpoints to verify that \(\sqrt[3]{30}\) lies in the marked interval. Then cube the midpoint shown on the number line and use that comparison to round \(\sqrt[3]{30}\) to the nearest hundredth.
Figure for problem 549860

Hints

- Use cubes, not squares, to verify the endpoints of the marked interval. - The midpoint shown on the number line is the rounding boundary between the two hundredths. - Cube that midpoint and compare its cube with \(30\).

Solution

1. \(3.10^3=29.791<30\). 2. \(3.11^3=30.080231>30\). 3. Therefore, \(3.10<\sqrt[3]{30}<3.11\). 4. The midpoint is \(3.105\), and \(3.105^3=29.935382625<30\). 5. Since cubing is increasing on the real numbers, \(\sqrt[3]{30}>3.105\), so it rounds to \(3.11\) to the nearest hundredth.

Answer

\(3.10^3=29.791<30<30.080231=3.11^3\), so \(3.10<\sqrt[3]{30}<3.11\). The midpoint satisfies \(3.105^3=29.935382625<30\), so \(\sqrt[3]{30}>3.105\) and \(\sqrt[3]{30}\approx3.11\) to the nearest hundredth.
Approximate irrational numbers 46
The number line shows a highlighted interval. Use squaring to show that \(\sqrt{27}\) lies in this interval. Then decide whether \(\sqrt{27}\) or \(5.20\) is greater, and state how far apart they are at most.
Figure for problem 549861

Hints

- Square both endpoints of the highlighted interval. - Compare the results with the radicand. - Use the interval width to bound the difference.

Solution

1. \(5.19^2=26.9361<27\). 2. \(5.20^2=27.04>27\). 3. Therefore, \(5.19<\sqrt{27}<5.20\). 4. Thus, \(5.20\) is greater than \(\sqrt{27}\), and their difference is less than \(0.01\).

Answer

\(5.19<\sqrt{27}<5.20\). The number \(5.20\) is greater, and the difference is less than \(0.01\).
Approximate irrational numbers 47
The number line shows two candidate tenths and their midpoint for \(\sqrt[3]{-20}\). First cube the two candidate tenths to verify that \(\sqrt[3]{-20}\) lies between them. Then cube the displayed midpoint and use that comparison to decide which tenth is closer. Your work must include the midpoint and its cube. Round \(\sqrt[3]{-20}\) to the nearest tenth.
Figure for problem 554573

Hints

- Cube the two candidate tenths shown on the number line and compare those cubes with the radicand. - Remember that the real cube function preserves order, including for negative inputs. - Cube the displayed midpoint and use that comparison as the rounding test rather than comparing distances between the endpoint cubes and the radicand.

Solution

1. Cube the candidate tenths: \((-2.8)^3=-21.952\) and \((-2.7)^3=-19.683\). 2. Since \(-21.952<-20<-19.683\) and cubing is increasing on the real numbers, \(-2.8<\sqrt[3]{-20}<-2.7\). 3. Cube the midpoint: \((-2.75)^3=-20.796875\). 4. Since \(-20>-20.796875\), \(\sqrt[3]{-20}>-2.75\), so it is closer to \(-2.7\). 5. Therefore, \(\sqrt[3]{-20}\approx-2.7\) to the nearest tenth.

Answer

\((-2.8)^3=-21.952<-20<-19.683=(-2.7)^3\), so \(-2.8<\sqrt[3]{-20}<-2.7\). The midpoint satisfies \((-2.75)^3=-20.796875<-20\), so \(\sqrt[3]{-20}>-2.75\) and \(\sqrt[3]{-20}\approx-2.7\) to the nearest tenth.
Approximate irrational numbers 48
Use the bounds \(3.141<\pi<3.142\) and \(1.414<\sqrt{2}<1.415\) to bound \(2\pi-\sqrt{2}\). Then round \(2\pi-\sqrt{2}\) to the nearest hundredth. Do not use a calculator decimal for \(\pi\) or \(\sqrt{2}\).

Hints

- For the smallest possible difference, pair the smaller possible value of the first term with the larger possible value of the quantity being subtracted. - Reverse those choices to obtain the largest possible difference. - Check whether the entire resulting interval rounds to the same hundredth.

Solution

1. For a lower bound, use the lower bound for \(2\pi\) and the upper bound for \(\sqrt{2}\): \(2\cdot3.141-1.415=4.867\). 2. For an upper bound, use the upper bound for \(2\pi\) and the lower bound for \(\sqrt{2}\): \(2\cdot3.142-1.414=4.870\). 3. Therefore, \(4.867<2\pi-\sqrt{2}<4.870\). 4. Every number in this interval rounds to \(4.87\) to the nearest hundredth.

Answer

\(4.867<2\pi-\sqrt{2}<4.870\), so \(2\pi-\sqrt{2}\approx4.87\) to the nearest hundredth.
Approximate irrational numbers 49
Riley says \(\sqrt[3]{70}\approx8.4\) because \(8.4^2\) is close to \(70\). Explain why Riley's method is invalid. Then show that \(\sqrt[3]{70}\) lies between \(4.1\) and \(4.2\). Finally, cube the midpoint \(4.15\) and use that result to decide which tenth is closer.

Hints

- Match the inverse operation to the root index before checking any decimal estimate. - Test the two proposed tenths using the appropriate power. - Use their midpoint to decide the nearest tenth without needing the exact cube root.

Solution

1. Riley used squares, but a cube root must be checked with cubes. 2. \(4.1^3=68.921\) and \(4.2^3=74.088\), so \(68.921<70<74.088\). Therefore, \(4.1<\sqrt[3]{70}<4.2\). 3. The midpoint is \(4.15\), and \(4.15^3=71.473375\). 4. Since \(70<71.473375\), \(\sqrt[3]{70}<4.15\), so it is closer to \(4.1\). 5. Thus, \(\sqrt[3]{70}\approx4.1\) to the nearest tenth.

Answer

Riley used squares instead of cubes. \(4.1^3=68.921<70<74.088=4.2^3\), so \(4.1<\sqrt[3]{70}<4.2\). Since \(4.15^3=71.473375>70\), \(\sqrt[3]{70}<4.15\) and \(\sqrt[3]{70}\approx4.1\) to the nearest tenth.
Approximate irrational numbers 50
Which point, \(P\) or \(Q\), is closer to \(-\sqrt{5}\)? Decide without using a calculator.
Figure for problem 559907

Hints

- Find the midpoint of the two marked positions. - Compare the square of the midpoint's positive distance from zero with \(5\). - Remember to reverse the position when taking the negative of a positive root.

Solution

1. The markers are \(P=-2.3\) and \(Q=-2.2\), with midpoint \(-2.25\). 2. Since \(2.25^2=5.0625>5\), we have \(\sqrt{5}<2.25\), so \(-\sqrt{5}>-2.25\). 3. Also, \(2.2^2<5<2.3^2\), so \(-\sqrt{5}\) is between the marked values. It lies to the right of their midpoint and is closer to \(Q\).

Answer

\(Q\) is closer to \(-\sqrt{5}\). Since \(2.2^2<5<2.3^2\), the root lies between the points. Also, \(2.25^2=5.0625>5\), so \(-\sqrt{5}>-2.25\), placing it nearer \(Q\).
Approximate irrational numbers 51
A rectangle has area \(20\,\text{cm}^2\) and one side is \(5\,\text{cm}\) long. To estimate the side of a square with the same area, take the average of the rectangle's two side lengths. Square that average. Then compare it with \(5^2\). Which gives an area closer to \(20\,\text{cm}^2\)?

Hints

- Which calculation undoes multiplying two side lengths to get an area? - An average lies between the two values being averaged. Why might that help here? - A change in side length changes square area. Is it enough just to compare the side lengths?

Solution

1. The other side of the rectangle is \(20\div5=4\,\text{cm}\). 2. The average of the two side lengths is \((5+4)\div2=4.5\,\text{cm}\). 3. A square with side \(4.5\,\text{cm}\) has area \(4.5^2=20.25\,\text{cm}^2\), which is \(0.25\,\text{cm}^2\) above the rectangle's area. 4. Using the original side of \(5\,\text{cm}\) gives \(5^2=25\,\text{cm}^2\), an error of \(5\,\text{cm}^2\). The average gives a smaller area error, so it is the better side-length estimate.

Answer

Average: \(4.5\,\text{cm}\). Its square has area \(20.25\,\text{cm}^2\), an error of \(0.25\,\text{cm}^2\). The original \(5\,\text{cm}\) guess gives \(25\,\text{cm}^2\), an error of \(5\,\text{cm}^2\). The average is better because its squared area is closer to \(20\,\text{cm}^2\).
Approximate irrational numbers 52
Without a calculator, find two consecutive natural numbers \(n\) and \(n+1\) such that \(T = \sqrt{40 + \sqrt{125}}\) lies between them. Show your reasoning.

Hints

- Begin with the inner square root and bound it between nearby whole numbers. - Use those bounds to estimate the entire radicand of the outer square root. - Which perfect squares surround that resulting interval?

Solution

1. Since \(11^2 = 121\) and \(12^2 = 144\), \(11 < \sqrt{125} < 12\). 2. Add \(40\) to each part: \(51 < 40 + \sqrt{125} < 52\). 3. Because \(7^2 = 49\) and \(8^2 = 64\), \(49 < 40 + \sqrt{125} < 64\). 4. Taking square roots gives \(7 < \sqrt{40 + \sqrt{125}} < 8\). Therefore, \(T\) lies between \(7\) and \(8\).

Answer

The value of \(T\) lies between \(7\) and \(8\).
Approximate irrational numbers 53
The Babylonian method approximates \(\sqrt{A}\) using \(x_{n+1}=\frac{1}{2}\left(x_n+\frac{A}{x_n}\right)\). Use the rule for \(A=2\), starting with \(x_1=1\). Find \(x_2\) and \(x_3\) as reduced fractions. Then calculate the exact difference \(x_3^2-2\).

Hints

- Apply the given averaging rule twice. - Add fractions using common denominators. - Keep your fractions exact until you subtract from the squared value.

Solution

1. With \(x_1=1\), \(x_2=\frac{1}{2}\left(1+\frac{2}{1}\right)=\frac{3}{2}\). 2. Next, \(x_3=\frac{1}{2}\left(\frac{3}{2}+\frac{4}{3}\right)=\frac{17}{12}\). 3. Square the third approximation: \(x_3^2=\frac{289}{144}\). 4. Subtract: \(x_3^2-2=\frac{289-288}{144}=\frac{1}{144}\).

Answer

\(x_2=\frac{3}{2}\), \(x_3=\frac{17}{12}\), and \(x_3^2-2=\frac{1}{144}\)
Approximate irrational numbers 54
Use one Babylonian step to improve the estimate of \(\sqrt{13}\), starting with \(x_1=3.5\). Use \(x_2=\frac{1}{2}\left(x_1+\frac{13}{x_1}\right)\). a) Calculate \(x_2\) as a fraction and round it to four decimal places. b) Compare \(x_2^2\) with \(13\). Is \(x_2\) above or below \(\sqrt{13}\)? c) Use \(3.60^2\) and \(3.61^2\) to show that \(x_2\) is within \(0.01\) of \(\sqrt{13}\).

Hints

- Substitute \(x_1=3.5\) into the given iteration formula and keep an exact fraction. - Compare the square of the approximation with the radicand. - Locate both values inside the same interval of width \(0.01\).

Solution

1. \(x_2=\frac{1}{2}\left(3.5+\frac{13}{3.5}\right)=\frac{1}{2}\left(\frac{7}{2}+\frac{26}{7}\right)=\frac{101}{28}\approx3.6071\). 2. \(x_2^2=\frac{10201}{784}\), while \(13=\frac{10192}{784}\). Therefore, \(x_2^2>13\), so \(x_2>\sqrt{13}\). 3. Since \(3.60^2=12.96<13\) and \(3.61^2=13.0321>13\), \(3.60<\sqrt{13}<3.61\). Also, \(3.60<x_2<3.61\). Two numbers in this interval differ by less than \(0.01\), so \(x_2\) is within \(0.01\) of \(\sqrt{13}\).

Answer

a) \(x_2=\frac{101}{28}\approx3.6071\) b) \(x_2>\sqrt{13}\) c) \(x_2\) is within \(0.01\) of \(\sqrt{13}\).
Approximate irrational numbers 55
Álvaro proposes \(\frac{161}{72}\) as a rational approximation to \(\sqrt{5}\). a) Compare \(\left(\frac{161}{72}\right)^2\) with \(5\). Is the approximation above or below \(\sqrt{5}\)? b) Use \(2.236^2=4.999696\) and \(2.237^2=5.004169\) to locate \(\sqrt{5}\) in an interval of width \(0.001\). c) Show that \(\frac{161}{72}\) is in the same interval, and give an upper bound for its absolute error.

Hints

- For positive quantities, compare squares to determine their order. - Use the given endpoint squares to locate \(\sqrt{5}\). - Relate the width of a common interval to the maximum possible distance between two points inside it.

Solution

1. \(\left(\frac{161}{72}\right)^2=\frac{25921}{5184}\), while \(5=\frac{25920}{5184}\). The squared approximation is greater than \(5\), so \(\frac{161}{72}>\sqrt{5}\). 2. Since \(4.999696<5<5.004169\), \(2.236<\sqrt{5}<2.237\). 3. \(\frac{161}{72}=2.236111\ldots\), so it also lies between \(2.236\) and \(2.237\). Two numbers in the same interval of width \(0.001\) differ by less than \(0.001\). Therefore, \(\left|\frac{161}{72}-\sqrt{5}\right|<0.001\).

Answer

a) \(\frac{161}{72}>\sqrt{5}\) b) \(2.236<\sqrt{5}<2.237\) c) \(\left|\frac{161}{72}-\sqrt{5}\right|<0.001\)
Approximate irrational numbers 56
Two students propose rational approximations for \(\sqrt{10}\). Lucas proposes \(\frac{19}{6}\), and Sofia proposes \(\frac{22}{7}\). a) Without a calculator, use squaring to decide which approximation is closer to \(\sqrt{10}\). b) Use a calculator to find the absolute error of each approximation. Who is correct?

Hints

- First square each fraction to determine which side of \(\sqrt{10}\) it lies on. - For two values on opposite sides, compare \(\sqrt{10}\) with their midpoint. - Square the midpoint rather than directly evaluating \(\sqrt{10}\).

Solution

1. The approximations lie on opposite sides of \(\sqrt{10}\): \(\left(\frac{22}{7}\right)^2 = \frac{484}{49} < 10\), while \(\left(\frac{19}{6}\right)^2 = \frac{361}{36} > 10\). 2. Their midpoint is \(m = \frac{1}{2}\left(\frac{22}{7} + \frac{19}{6}\right) = \frac{265}{84}\). Since \(m^2 = \frac{70225}{7056} < 10\), the midpoint is below \(\sqrt{10}\). Therefore, \(\sqrt{10}\) is closer to the larger approximation, \(\frac{19}{6}\). 3. Using \(\sqrt{10} \approx 3.162278\), the errors are \(\left|\frac{19}{6} - \sqrt{10}\right| \approx 0.004389\) and \(\left|\frac{22}{7} - \sqrt{10}\right| \approx 0.019421\). Lucas is correct.

Answer

a) \(\frac{19}{6}\) is closer to \(\sqrt{10}\). b) Lucas’s absolute error is approximately \(0.004389\), and Sofia’s is approximately \(0.019421\). Lucas is correct.
Approximate irrational numbers 57
Two irrational numbers are \(a = \sqrt{2}\) and \(b = 2-\sqrt{2}\). a) Determine whether the sum \(s = a+b\) is rational or irrational. b) Find the first three digits after the decimal point of \(d = a-b\) by bounding \(\sqrt{2}\) to four decimal places.

Hints

- Simplify the sum before classifying it. - Simplify the difference before substituting bounds. - Track how multiplying and subtracting change both interval endpoints. - Which decimal digits are fixed throughout the final interval?

Solution

1. The sum is \(s = \sqrt{2} + (2-\sqrt{2}) = 2\), which is rational. 2. The difference is \(d = \sqrt{2}-(2-\sqrt{2}) = 2\sqrt{2}-2\). 3. Since \(1.4142 < \sqrt{2} < 1.4143\), multiplying by \(2\) gives \(2.8284 < 2\sqrt{2} < 2.8286\). 4. Subtracting \(2\) gives \(0.8284 < d < 0.8286\). Throughout this interval, the first three digits after the decimal point are \(828\).

Answer

a) \(s = 2\), so the sum is rational. b) The first three digits after the decimal point are \(828\).
Approximate irrational numbers 58
The Babylonian method uses \(x_{n+1}=\frac{1}{2}\left(x_n+\frac{A}{x_n}\right)\) to approximate \(\sqrt{A}\). Use it to approximate \(\sqrt{10}\), starting with \(x_0=3\). a) Calculate \(x_1\) and \(x_2\) as reduced fractions. b) Round \(x_2\) to six decimal places. c) Find the exact absolute difference between \(x_2^2\) and \(10\).

Hints

- Use the starting value in the stated iteration rule. - Keep the steps as fractions so rounding does not affect the last part. - When dividing by a fraction, use its reciprocal. - Compare the square of your second approximation with the number under the root.

Solution

1. Substitute \(A=10\) and \(x_0=3\): \(x_1=\frac{1}{2}\left(3+\frac{10}{3}\right)=\frac{19}{6}\). 2. Use the rule again: \(x_2=\frac{1}{2}\left(\frac{19}{6}+\frac{60}{19}\right)=\frac{721}{228}\). 3. Convert to a decimal: \(\frac{721}{228}\approx3.162281\). 4. The squared error is \(\left|\frac{519841}{51984}-10\right|=\frac{1}{51984}\).

Answer

a) \(x_1=\frac{19}{6}\), \(x_2=\frac{721}{228}\) b) \(x_2\approx3.162281\) c) \(\frac{1}{51984}\)

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