Free Math Worksheets

- Compare the left and right sides of each equation. - Did only the order of the numbers change? - Did the grouping change while the numbers stayed in the same order? - Was a factor multiplied by every term inside parentheses?

1. In a), the addends stay in the same order, but the grouping changes. This illustrates the associative property of addition. 2. In b), the order of the factors is reversed without changing the product. This illustrates the commutative property of multiplication. 3. In c), the factor \(8\) is multiplied by each addend inside the parentheses. This illustrates the distributive property.

a) Associative property b) Commutative property c) Distributive property

- Evaluate both quotients. - Write the second quotient as a fraction. - Compare the two values.

1. \(24 \div 6 = 4\). 2. \(6 \div 24 = \frac{1}{4}\). 3. Since \(4 \ne \frac{1}{4}\), changing the order changes the quotient. Division is not commutative.

Lukas is not correct. \(24 \div 6 = 4\), but \(6 \div 24 = \frac{1}{4}\), so division is not commutative.

- Look for two factors whose product is especially easy to compute. - Which property allows you to change the order of factors? - Which property allows you to change how factors are grouped?

1. Use the commutative property to reorder the factors and the associative property to regroup them: \((2.5 \cdot 4) \cdot 13\). 2. Compute the first product: \(2.5 \cdot 4 = 10\). 3. Multiply by the remaining factor: \(10 \cdot 13 = 130\). 4. Reordering used the commutative property, and regrouping used the associative property.

The value is \(130\). The properties used are the commutative and associative properties of multiplication.

- Evaluate the grouped operation first on each side. - Compare the left and right sides separately for each operation. - Recall the name of the property that changes grouping but not order.

1. With subtraction, the left side is \(100 - (20 - 5) = 85\), while the right side is \((100 - 20) - 5 = 75\). The equation is false. 2. With division, the left side is \(100 \div (20 \div 5) = 25\), while the right side is \((100 \div 20) \div 5 = 1\). The equation is false. 3. The property that permits regrouping in addition and multiplication is the associative property.

The equation is false for both subtraction, because \(85 \ne 75\), and division, because \(25 \ne 1\). The property is the associative property.

- Simplify each expression before classifying it. - Recall how the natural, integer, rational, and real number sets are nested. - A square root of a non-perfect-square integer is irrational.

1. \(\sqrt{2.25} = 1.5\), so the smallest set is \(\mathbb{Q}\). 2. \(\sqrt{12} = 2\sqrt{3}\), which is irrational, so the smallest listed set is \(\mathbb{R}\). 3. The difference of the irrational number \(\pi\) and the rational number \(3\) is irrational, so the smallest listed set is \(\mathbb{R}\). 4. \(\frac{\sqrt{50}}{\sqrt{2}} = \sqrt{25} = 5\), so the smallest set is \(\mathbb{N}\). 5. The terminating decimal \(-3.14\) is rational but not an integer, so the smallest set is \(\mathbb{Q}\).

a) \(\mathbb{Q}\) b) \(\mathbb{R}\) c) \(\mathbb{R}\) d) \(\mathbb{N}\) e) \(\mathbb{Q}\)

- Evaluate \(0^2\) first. - Use the zero-product property. - Check which possible solution belongs to the natural numbers.

1. Evaluate the right side: \(0^2 = 0\). 2. A product equals \(0\) when at least one factor equals \(0\). 3. From \(25 - x = 0\), we get \(x = 25\). 4. From \(25 + x = 0\), we get \(x = -25\), which is not a natural number. 5. Therefore, the only natural-number solution is \(x = 25\).

\(x = 25\)

- Identify whether each statement changes order, grouping, or distribution. - Use the same variables on both sides of each equation. - Preserve the operation described in the statement.

1. The commutative property of multiplication is \(ab = ba\). 2. The associative property of addition is \((a + b) + c = a + (b + c)\). 3. The distributive property is \(a(b + c) = ab + ac\).

a) \(ab = ba\) b) \((a + b) + c = a + (b + c)\) c) \(a(b + c) = ab + ac\)

- Compare each line with the line before it. - Decide whether the order or the grouping changed. - Match each change to its general property equation.

1. The factors \(9\) and \(4\) are reversed, using the commutative property: \(ab = ba\). 2. The factors are regrouped so \(25\) and \(4\) are multiplied first, using the associative property: \((ab)c = a(bc)\).

1) Commutative property: \(ab = ba\) 2) Associative property: \((ab)c = a(bc)\)

- Evaluate both sides when needed. - Reversing order tests a commutative property. - Changing grouping tests an associative property.

1. For a), both sides equal \(103\). The statement is true by the commutative property of addition. 2. For b), both sides equal \(1300\). The statement is true by the associative property of multiplication. 3. For c), \(100 - 30 = 70\), but \(30 - 100 = -70\). The statement is false; subtraction is not commutative.

a) True; commutative property b) True; associative property c) False

- Substitute the given values carefully. - Compare both the magnitude and sign of each pair of results. - A commutative operation gives the same result after reversing the order.

1. For addition, \(a + b = -4.5 + 1.5 = -3\) and \(b + a = 1.5 + (-4.5) = -3\). 2. For subtraction, \(a - b = -4.5 - 1.5 = -6\) and \(b - a = 1.5 - (-4.5) = 6\). 3. Addition is commutative because changing the order does not change the sum. Subtraction is not commutative because changing the order changes the difference.

1) \(a + b = -3\) and \(b + a = -3\) 2) \(a - b = -6\) and \(b - a = 6\) Addition is commutative, but subtraction is not.

- On the left, evaluate \(a + b\) first. - On the right, evaluate \(b + c\) first. - The property changes grouping without changing order.

1. For the first set, the left side is \((-15.4 + 6.7) + 3.3 = -8.7 + 3.3 = -5.4\). The right side is \(-15.4 + (6.7 + 3.3) = -15.4 + 10 = -5.4\). 2. For the second set, the left side is \(\left(\frac{5}{9} + \frac{2}{9}\right) - \frac{2}{9} = \frac{5}{9}\). The right side is \(\frac{5}{9} + \left(\frac{2}{9} - \frac{2}{9}\right) = \frac{5}{9}\). 3. Both cases illustrate the associative property of addition.

1) Both sides equal \(-5.4\). 2) Both sides equal \(\frac{5}{9}\). The equation illustrates the associative property of addition.

- Evaluate each grouped expression separately. - Compare the two final values. - A single counterexample is enough to show that an operation is not associative.

1. Evaluate the left side: \((10 - 5) - 2 = 5 - 2 = 3\). 2. Evaluate the right side: \(10 - (5 - 2) = 10 - 3 = 7\). 3. Since \(3 \ne 7\), the equation is false. 4. Subtraction is not associative, so changing grouping symbols can change the value.

The equation is false because \(3 \ne 7\). Subtraction is not associative.

- Evaluate the grouped product first in each expression. - Track the signs of the factors. - Compare the intermediate values created by the two groupings.

1. For a), \((-2.5 \cdot 4)(-1.8) = (-10)(-1.8) = 18\). 2. For b), \(-2.5(4 \cdot -1.8) = (-2.5)(-7.2) = 18\). 3. Both results are equal, illustrating the associative property. The first grouping is easier because it creates the intermediate product \(-10\).

a) \(18\) b) \(18\) c) Both results are equal. The first grouping is easier because \(-2.5 \cdot 4 = -10\).

- Solve each equation before classifying its solution or solutions. - Recall how \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), and \(\mathbb{R}\) are nested. - Distinguish between a rational value that is an integer and one that is not. - A square root of a non-perfect-square natural number is irrational.

1. For a), \(x = -4\). The smallest listed set containing \(-4\) is \(\mathbb{Z}\). 2. For b), \(x = \frac{6}{8} = \frac{3}{4}\). The smallest listed set containing this value is \(\mathbb{Q}\). 3. For c), \(x = \sqrt{7}\) or \(x = -\sqrt{7}\). Because \(7\) is not a perfect square, both solutions are irrational. The smallest listed set containing both solutions is \(\mathbb{R}\). 4. For d), \(x = 1.1\) or \(x = -1.1\). Both terminating decimals are rational, so the smallest listed set is \(\mathbb{Q}\). 5. For e), \(x = 30\). The smallest listed set containing \(30\) is \(\mathbb{N}\).

a) \(x = -4\); smallest set: \(\mathbb{Z}\) b) \(x = \frac{3}{4}\); smallest set: \(\mathbb{Q}\) c) \(x = \sqrt{7}\) or \(x = -\sqrt{7}\); smallest set: \(\mathbb{R}\), because both solutions are irrational. d) \(x = 1.1\) or \(x = -1.1\); smallest set: \(\mathbb{Q}\) e) \(x = 30\); smallest set: \(\mathbb{N}\)

- Evaluate the expression in parentheses first on the left side. - On the right side, be careful when subtracting a negative number. - Which property multiplies one factor by each term in a sum or difference?

1. Evaluate the left side: \((-5)(20 - 4) = (-5)(16) = -80\). 2. Evaluate the right side: \((-5) \cdot 20 - (-5) \cdot 4 = -100 - (-20) = -80\). 3. Both sides equal \(-80\), so the equation is true. 4. The factor \(-5\) is multiplied by each term in the difference, so the equation illustrates the distributive property.

The equation is true because both sides equal \(-80\). It illustrates the distributive property.

- Look for two factors whose product is \(100\). - One property changes order; another changes grouping. - For the division example, evaluate each grouping separately.

1. For a), one method is \((25 \cdot 17) \cdot 4 = 425 \cdot 4 = 1700\). 2. An efficient method is \((25 \cdot 4) \cdot 17 = 100 \cdot 17 = 1700\). 3. Reordering factors uses the commutative property, and regrouping factors uses the associative property. 4. For c), \(12 \div (6 \div 2) = 12 \div 3 = 4\), but \((12 \div 6) \div 2 = 2 \div 2 = 1\). Since \(4 \ne 1\), division is not associative.

a) \(1700\) b) Commutative property and associative property c) \(12 \div (6 \div 2) = 4\), but \((12 \div 6) \div 2 = 1\).

- Test a positive number, a negative number, and \(0\) for \(x\). - Recall that the principal square root is never negative. - When is a square root defined in the real number system?

1. The principal square root is always nonnegative, so \(\sqrt{x^2} = |x|\). Therefore, \(\sqrt{x^2} = x\) only when \(x \geq 0\). 2. The expression \(\sqrt{x}\) is real only when \(x \geq 0\). For every such value, \((\sqrt{x})^2 = x\). Therefore, the statement is true only when \(x \geq 0\). 3. For \(\sqrt{-x}\) to be real, the radicand must satisfy \(-x \geq 0\). This is equivalent to \(x \leq 0\).

a) True only for \(x \geq 0\). b) True only for \(x \geq 0\). c) True only for \(x \leq 0\).

- A single counterexample disproves an “always” statement. - Recall what repeating decimals represent. - What does it mean for two sets to be disjoint? - How do you calculate the midpoint of two numbers?

1. Statement 1 is false. For example, \(\sqrt{2} + (-\sqrt{2}) = 0\), which is rational. 2. Statement 2 is false. A repeating decimal such as \(0.\overline{3} = \frac{1}{3}\) has infinitely many digits but is rational. 3. Statement 3 is true. Rational and irrational numbers are disjoint by definition. 4. Statement 4 is true. If \(a\) and \(b\) are distinct rational numbers, their midpoint \(\frac{a+b}{2}\) is rational and lies strictly between them.

1. False; \(\sqrt{2} + (-\sqrt{2}) = 0\). 2. False; \(0.\overline{3} = \frac{1}{3}\). 3. True. 4. True; the midpoint is another rational number.

- How do you calculate the midpoint of two numbers? - After finding one midpoint, apply the same process to one of the smaller intervals. - What happens when a positive rational number is divided by \(2\)?

1. The midpoint of \(a = \frac{3}{4}\) and \(b = 1\) is \(m_1 = \frac{\frac{3}{4}+1}{2} = \frac{7}{8}\). 2. The midpoint of \(a = \frac{3}{4}\) and \(m_1 = \frac{7}{8}\) is \(m_2 = \frac{\frac{3}{4}+\frac{7}{8}}{2} = \frac{13}{16}\). 3. The midpoint of \(m_1 = \frac{7}{8}\) and \(b = 1\) is \(m_3 = \frac{\frac{7}{8}+1}{2} = \frac{15}{16}\). Thus, three possible numbers are \(\frac{13}{16}\), \(\frac{7}{8}\), and \(\frac{15}{16}\). 4. Suppose a smallest positive rational number \(r\) existed. The number \(\frac{r}{2}\) is rational and satisfies \(0 < \frac{r}{2} < r\), contradicting the claim that \(r\) is the smallest positive rational number.

a) One possible set is \(\frac{13}{16}\), \(\frac{7}{8}\), and \(\frac{15}{16}\). b) For every positive rational number \(r\), the rational number \(\frac{r}{2}\) is positive and smaller than \(r\). Therefore, no smallest positive rational number exists.

- Approximate the square roots with a calculator. - To find \(\sqrt{n}\) in an interval, square the interval endpoints. - The definition of dense must hold for every pair of different numbers in the set.

1. Since \(\sqrt{5} \approx 2.236\) and \(\sqrt{6} \approx 2.449\), a rational number such as \(q = 2.3\) lies between them. 2. Squaring the interval endpoints gives \(2.5^2 < n < 3^2\), or \(6.25 < n < 9\). The natural numbers \(7\) and \(8\) work, so either \(\sqrt{7}\) or \(\sqrt{8}\) is a valid irrational number. 3. The set is not dense. Between \(\sqrt{1} = 1\) and \(\sqrt{2}\), there is no other number of the form \(\sqrt{n}\) because there is no natural number strictly between \(1\) and \(2\).

a) For example, \(q = 2.3\) b) For example, \(\sqrt{7}\) or \(\sqrt{8}\) c) No. There is no number of the form \(\sqrt{n}\) strictly between \(\sqrt{1}\) and \(\sqrt{2}\).

- How do you find the midpoint of two fractions? - Use a common denominator to compare the three fractions. - Can the midpoint process be repeated without ending?

1. The midpoint is \(m = \frac{a+b}{2} = \frac{\frac{2}{3}+\frac{3}{4}}{2}\). Using a common denominator, \(\frac{2}{3} = \frac{8}{12}\) and \(\frac{3}{4} = \frac{9}{12}\), so \(m = \frac{\frac{17}{12}}{2} = \frac{17}{24}\). Because \(m\) is a ratio of integers, it is rational. 2. Using denominator \(24\), \(a = \frac{16}{24}\), \(m = \frac{17}{24}\), and \(b = \frac{18}{24}\). Therefore, \(a < m < b\). 3. Each new midpoint is rational and lies strictly between \(a\) and the previous midpoint. The process can continue indefinitely, so there are infinitely many rational numbers between \(a\) and \(b\).

a) \(m = \frac{17}{24}\), which is rational. b) \(\frac{16}{24} < \frac{17}{24} < \frac{18}{24}\), so \(a < m < b\). c) There are infinitely many rational numbers between \(a\) and \(b\).

- A terminating decimal is rational. - Scale and shift a familiar irrational number so that it lies in the interval. - What number is exactly halfway between two different real numbers?

1. One possible rational number is \(q = 3.05\). 2. One possible irrational number is \(i = 3 + \frac{\sqrt{2}}{20}\). Since \(1.4 < \sqrt{2} < 1.5\), it follows that \(3.07 < 3 + \frac{\sqrt{2}}{20} < 3.075\), so \(i\) lies between \(3\) and \(3.1\). The number is irrational because \(\frac{\sqrt{2}}{20}\) is irrational, and adding the rational number \(3\) preserves irrationality. 3. If \(x < y\), their midpoint \(m = \frac{x+y}{2}\) is real. Also, \(x < \frac{x+y}{2} < y\), so another real number always lies between two different real numbers.

a) For example, \(q = 3.05\) b) For example, \(i = 3 + \frac{\sqrt{2}}{20}\) c) For \(x < y\), the midpoint \(\frac{x+y}{2}\) is real and satisfies \(x < \frac{x+y}{2} < y\).

- Assume the sum is rational and look for a contradiction. - Rearrange the equation so the irrational number is alone. - Recall that rational numbers are closed under subtraction. - For a counterexample, look for two irrational numbers that cancel.

1. Suppose, for contradiction, that \(q = r + s\) is rational. 2. Then \(s = q - r\). Because the difference of two rational numbers is rational, this would make \(s\) rational. 3. That contradicts the assumption that \(s\) is irrational. Therefore, \(r + s\) must be irrational. 4. For part b), choose \(\sqrt{2}\) and \(-\sqrt{2}\). Both are irrational, but their sum is \(0\), which is rational.

a) If \(r+s\) were rational, then \(s = (r+s)-r\) would be rational, a contradiction. Therefore, \(r+s\) is irrational. b) No. For example, \(\sqrt{2} + (-\sqrt{2}) = 0\).

- Try a natural number smaller than \(4\) whose square is divisible by \(4\). - Identify the exact step in the contradiction argument that is no longer valid. - Compare the reduced fraction for \(\sqrt{4}\) with the assumed fraction.

1. The statement is false. Let \(p=2\). Then \(p^2=4\) is divisible by \(4\), but \(p=2\) is not divisible by \(4\). 2. An attempted contradiction proof would begin with \(\sqrt{4}=\frac{p}{q}\) in lowest terms and obtain \(p^2=4q^2\). 3. This equation implies that \(p\) is even, so write \(p=2k\). Substitution gives \(4k^2=4q^2\), or \(k^2=q^2\). It does not imply that \(q\) is even. 4. Therefore, no contradiction to the fraction being in lowest terms appears. In fact, \(p=2\) and \(q=1\) give the reduced representation \(\sqrt{4}=2\).

The statement is false. For \(p=2\), \(p^2=4\) is divisible by \(4\), but \(p\) is not divisible by \(4\). In the attempted proof, \(p^2=4q^2\) shows only that \(p\) is even. Writing \(p=2k\) gives \(k^2=q^2\), not that \(q\) is even, so no contradiction occurs. Indeed, \(\sqrt{4}=\frac{2}{1}=2\).

- Assume the sum is rational. - Isolate the known irrational number. - What is true about the difference of two rational numbers?

1. Assume the opposite: suppose \(x=3+\sqrt{2}\) is rational. 2. Because \(3\) is rational, the difference \(x-3\) would also be rational. 3. But \(x-3=(3+\sqrt{2})-3=\sqrt{2}\). This would make \(\sqrt{2}\) rational, contradicting the given fact. 4. Therefore, the assumption is false, and \(3+\sqrt{2}\) is irrational.

Assume \(3+\sqrt{2}\) is rational. Subtracting the rational number \(3\) would make \(\sqrt{2}\) rational, contradicting the given fact. Therefore, \(3+\sqrt{2}\) is irrational.

- What number lies exactly halfway between two numbers? - Assume the sum is rational and solve for the irrational addend. - Can an interval be made so small that irrational numbers no longer occur in it?

1. Statement a) is true. If \(a\) and \(b\) are different rational numbers, their midpoint \(\frac{a+b}{2}\) is rational and lies strictly between them. 2. Statement b) is true. Suppose a rational number \(r\) and an irrational number \(i\) had a rational sum \(s\). Then \(i=s-r\) would be rational because the difference of two rational numbers is rational. This contradicts the assumption that \(i\) is irrational. 3. Statement c) is false. Every interval with two different endpoints contains irrational numbers, so no nonempty interval can contain only rational numbers.

a) True b) True c) False

- Add the two numbers, then divide by \(2\) to find a midpoint. - For part c), rewrite both endpoints with denominator \(80\). - Do not reuse either midpoint.

1. The first midpoint is \(m_1 = \frac{\frac{1}{5}+\frac{1}{4}}{2} = \frac{\frac{9}{20}}{2} = \frac{9}{40}\). 2. The second midpoint is \(m_2 = \frac{\frac{1}{5}+\frac{9}{40}}{2} = \frac{\frac{17}{40}}{2} = \frac{17}{80}\). 3. With denominator \(80\), the endpoints are \(a = \frac{16}{80}\) and \(b = \frac{20}{80}\). The fractions strictly between them are \(\frac{17}{80}\), \(\frac{18}{80}\), and \(\frac{19}{80}\). Since \(\frac{17}{80}=m_2\) and \(\frac{18}{80}=m_1\), a valid additional number is \(\frac{19}{80}\).

a) \(m_1 = \frac{9}{40}\) b) \(m_2 = \frac{17}{80}\) c) \(\frac{19}{80}\)

- Rewrite the quotients as fractions. - Clear the denominators by multiplying both sides by \(ab\). - Use the fact that \(a\) and \(b\) are positive.

1. Since \(a\) and \(b\) are positive, multiply both sides by \(ab\): \(a^2 = b^2\). 2. Positive numbers with equal squares are equal, so \(a = b\). 3. Conversely, if \(a = b\), then both quotients equal \(1\). For example, \(5 \div 5 = 1\) and \(12 \div 12 = 1\).

The quotient is unchanged exactly when \(a = b\). Examples include \(5 \div 5 = 1\) and \(12 \div 12 = 1\).

- Evaluate both forms in each comparison. - A property fails if one counterexample gives different results. - Recall which operations allow their inputs to be reversed without changing the result.

1. For a), \(50 - 20 = 30\), while \(20 - 50 = -30\). Since the results differ, subtraction is not commutative. 2. For b), \((80 \div 8) \div 2 = 10 \div 2 = 5\), while \(80 \div (8 \div 2) = 80 \div 4 = 20\). Since the results differ, division is not associative. 3. Addition and multiplication are commutative.

a) Subtraction is not commutative. b) Division is not associative. c) Addition or multiplication

- Write each rational number as a ratio of integers. - Compare square roots of perfect squares with square roots of non-perfect squares. - Look for irrational square roots whose product simplifies to a perfect square.

1. Let \(\frac{a}{b}\) and \(\frac{c}{d}\) be rational numbers, with \(b\), \(c\), and \(d\) nonzero. Then \(\frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc}\), which is another ratio of integers. Therefore, \(\mathbb{Q}\) is closed under division by a nonzero rational number. 2. The rational numbers are not closed under taking square roots because a rational input can have an irrational square root. For example, \(2\) is rational, but \(\sqrt{2}\) is irrational. 3. For part b), \(x^2=4\) has rational solutions \(x=\pm2\), while \(x^2=2\) has real but irrational solutions \(x=\pm\sqrt{2}\). 4. The claim in part c) is false. Both factors in \(\sqrt{2}\cdot\sqrt{2}\) are irrational, but their product is \(2\), which is rational.

a) Dividing two rational numbers, with a nonzero divisor, produces another ratio of integers. However, a rational number such as \(2\) can have an irrational square root. b) For example, \(x^2=4\) has rational solutions, and \(x^2=2\) has real but irrational solutions. c) False; \(\sqrt{2}\cdot\sqrt{2}=2\) is a rational product of two irrational numbers.

- Does being dense mean that a set contains every real number? - Which real numbers cannot be written as ratios of integers? - What would follow about \(\sqrt{3}\) if \(2+\sqrt{3}\) were rational?

1. The rational numbers are dense: another rational number lies between any two different rational numbers. However, density does not mean that the set contains every point. Numbers such as \(\sqrt{2}\) and \(\pi\) are points on the number line but are not rational. 2. The irrational numbers must be added. Together, the rational and irrational numbers form the real numbers \(\mathbb{R}\). 3. The number \(x=2+\sqrt{3}\) represents an exact real number and therefore an exact point on the number line. It is irrational. If \(x\) were rational, then \(x-2=\sqrt{3}\) would also be rational, which is false.

1) The claim is false. The rational numbers are dense but do not include irrational points such as \(\sqrt{2}\). 2) The irrational numbers 3) \(x\) is an exact point on the number line and is irrational.

- Solve each equation for \(x\) first. - Test small natural-number values to see whether negative numbers, noninteger fractions, or irrational roots can occur. - Recall the definitions of the number sets before choosing the smallest one. - The chosen set must contain every possible solution for every allowed choice of \(a\) and \(b\).

1. Solving gives \(x=a+ab=a(1+b)\). This is always a natural number, so the smallest set is \(\mathbb{N}\). 2. Solving gives \(x=b-a^2\). This is always an integer, and it can be negative; for example, \(a=2\) and \(b=1\) give \(x=-3\). Therefore, the smallest set is \(\mathbb{Z}\). 3. Solving gives \(x=\frac{b+1}{a}\). This is always rational, but it need not be an integer; for example, \(a=3\) and \(b=1\) give \(x=\frac{2}{3}\). Therefore, the smallest set is \(\mathbb{Q}\). 4. Solving gives \(x=\pm\sqrt{ab+1}\). The radicand is always positive, so the solutions are real. They can be irrational; for example, \(a=b=1\) gives \(x=\pm\sqrt{2}\). Therefore, the smallest set is \(\mathbb{R}\).

1) \(\mathbb{N}\) 2) \(\mathbb{Z}\) 3) \(\mathbb{Q}\) 4) \(\mathbb{R}\)

- Isolate \(x^2\). - Check when the quantity under a square root is nonnegative. - Choose \(\frac{b}{a}\) to be the square of a noninteger fraction. - In part c), consider whether the two prime numbers must be different.

1. Dividing by \(a\) gives \(x^2=\frac{b}{a}\). Because \(a\) and \(b\) are positive, \(\frac{b}{a}>0\). Therefore, the two real solutions are \(x=\sqrt{\frac{b}{a}}\) and \(x=-\sqrt{\frac{b}{a}}\). 2. For example, let \(a=4\) and \(b=1\). Then \(x^2=\frac{1}{4}\), so \(x=\pm\frac{1}{2}\). Both solutions are rational but not integers. 3. The claim is false because the two primes are not required to be different. If \(a=b=2\), then \(x^2=\frac{2}{2}=1\), so \(x=\pm1\), which are rational. When \(a\) and \(b\) are distinct primes, the solutions are irrational, but that additional condition was not stated.

a) \(x=\pm\sqrt{\frac{b}{a}}\), and \(\frac{b}{a}>0\), so there are exactly two real solutions. b) For example, \(a=4\) and \(b=1\) give \(x=\pm\frac{1}{2}\). c) False. For example, \(a=b=2\) gives \(x=\pm1\), which are rational.

- For (1), test two simple nonzero perfect squares. - For (2), consider separately what happens when \(x\) is negative, zero, or positive. - For (3), determine when the negative of a square can be nonnegative.

1. Statement (1) is false. For example, let \(a = 9\) and \(b = 16\). Then \(\sqrt{9} + \sqrt{16} = 3 + 4 = 7\), but \(\sqrt{9 + 16} = \sqrt{25} = 5\). 2. Since \(\sqrt{x^2} = |x|\), the equation is \(|x| = -x\). This is true for every \(x \leq 0\), so there are infinitely many solutions. Statement (2) is true. 3. The radicand must be nonnegative: \(-(x-5)^2 \geq 0\). Since \((x-5)^2 \geq 0\), its opposite is nonnegative only when \((x-5)^2 = 0\). Thus, \(x = 5\) is the only value in the domain. Statement (3) is true.

(1) False. For example, \(\sqrt{9} + \sqrt{16} = 7\), while \(\sqrt{25} = 5\). (2) True. Every \(x \leq 0\) is a solution. (3) True. The expression is defined only for \(x = 5\).

- For a), consider negative integers. - For b), write the rational number as a fraction in lowest terms. - For c), include both square-root solutions before applying the sign condition. - For d), choose a non-perfect-square integer between \(16\) and \(25\).

1. For a), any negative integer works; for example, \(x = -3\). 2. For b), no such number exists. If \(x = \frac{p}{q}\) is in lowest terms and \(x^2\) is an integer, then \(q^2\) must divide \(p^2\). Since \(p\) and \(q\) have no common factor, this is possible only when \(q = 1\), which would make \(x\) an integer. 3. For c), the two solutions of \(x^2 = 5\) are \(\sqrt{5}\) and \(-\sqrt{5}\). The negative solution is \(x = -\sqrt{5}\). 4. For d), one example is \(x = \sqrt{17}\). Since \(16 < 17 < 25\), \(4 < \sqrt{17} < 5\), and \(\sqrt{17}\) is irrational.

a) For example, \(x = -3\) b) No such number exists. c) \(x = -\sqrt{5}\) d) For example, \(x = \sqrt{17}\)

- Can a decimal with at most two digits after the decimal point lie strictly between \(0.01\) and \(0.02\)? - Consider the midpoint of two terminating decimals. - Does a set of fixed ruler marks have a smallest spacing?

1. The set \(D\) is not dense. There is no number in \(D\) strictly between \(0.01\) and \(0.02\), because any such number would need more than two digits after the decimal point. 2. The set of all terminating decimals is dense. If \(x < y\) are terminating decimals, then their midpoint \(\frac{x+y}{2}\) is also a terminating decimal and satisfies \(x < \frac{x+y}{2} < y\). 3. In a dense set, every interval between two different elements contains another element of the set, no matter how far the number line is magnified. A set of equally spaced marks has a fixed positive gap between neighboring points, even when that gap is small.

a) No. There is no element of \(D\) strictly between \(0.01\) and \(0.02\). b) Yes. The midpoint of any two different terminating decimals is another terminating decimal between them. c) A dense set has another element between every two different elements. Closely spaced ruler marks still have a fixed gap between neighboring marks.

- What does it mean for a fraction to be in lowest terms? - Square the assumed fraction equation to remove the radical. - Write a number divisible by \(5\) as \(5k\). - Look for a contradiction to the lowest-terms assumption.

1. Assume the opposite: suppose \(\sqrt{5}\) is rational. Then \(\sqrt{5}=\frac{p}{q}\), where \(p\) and \(q\) are natural numbers and the fraction is in lowest terms. 2. Squaring gives \(5=\frac{p^2}{q^2}\), so \(p^2=5q^2\). 3. Thus, \(p^2\) is divisible by \(5\). By the given fact, \(p\) is divisible by \(5\), so \(p=5k\) for some natural number \(k\). 4. Substituting gives \((5k)^2=5q^2\), so \(25k^2=5q^2\) and \(5k^2=q^2\). 5. Therefore, \(q^2\) is divisible by \(5\), so \(q\) is also divisible by \(5\). 6. Both \(p\) and \(q\) are divisible by \(5\), contradicting the assumption that \(\frac{p}{q}\) is in lowest terms. Therefore, \(\sqrt{5}\) is irrational.

Assuming \(\sqrt{5}=\frac{p}{q}\) in lowest terms leads to \(p^2=5q^2\). The given divisibility fact then shows that both \(p\) and \(q\) are divisible by \(5\), contradicting that the fraction is in lowest terms. Therefore, \(\sqrt{5}\) is irrational.

- Determine whether each square root is an integer. - What does lying strictly between consecutive integers imply about the denominator of a reduced fraction? - Check whether the condition \(b>1\) holds in each case. - When can a reduced fraction equal an integer?

1. Suppose \(\sqrt{7}=\frac{a}{b}\) in lowest terms. Since \(2^2<7<3^2\), \(\sqrt{7}\) is not an integer, so \(b>1\). 2. Squaring gives \(7=\frac{a^2}{b^2}\). Because \(a\) and \(b\) are relatively prime, \(a^2\) and \(b^2\) are also relatively prime. Thus, \(\frac{a^2}{b^2}\) is in lowest terms with denominator greater than \(1\), so it cannot equal the integer \(7\). This is a contradiction. 3. For \(\sqrt{16}\), the value is \(4\), which has the reduced representation \(\frac{4}{1}\). Here the denominator is \(1\), so the step requiring \(b>1\) does not apply. 4. The key difference is that \(\sqrt{7}\) lies strictly between consecutive integers, while \(\sqrt{16}\) is an integer.

For \(\sqrt{7}\), any reduced rational representation would have denominator \(b>1\). After squaring, the reduced fraction \(\frac{a^2}{b^2}\) could not equal the integer \(7\), producing a contradiction. For \(\sqrt{16}=4=\frac{4}{1}\), the denominator can be \(1\), so no contradiction occurs. The argument differs at the step where \(b>1\) is established.

- Begin by assuming the opposite of what you want to prove. - Locate \(\sqrt{24}\) between consecutive integers. - Use prime factors to explain why squaring preserves relative primality. - Can a reduced fraction with denominator greater than \(1\) equal an integer?

1. Assume \(\sqrt{24}\) is rational, so \(\sqrt{24}=\frac{a}{b}\), where \(a\) and \(b\) are natural numbers and \(\frac{a}{b}\) is in lowest terms. 2. Since \(4^2<24<5^2\), \(4<\sqrt{24}<5\). Therefore, \(\sqrt{24}\) is not an integer, so a reduced fraction representing it must have \(b>1\). 3. Squaring gives \(24=\frac{a^2}{b^2}\). Because \(a\) and \(b\) have no common prime factor, squaring them creates no new prime factors. Thus, \(a^2\) and \(b^2\) are also relatively prime. 4. The fraction \(\frac{a^2}{b^2}\) is therefore in lowest terms and has denominator \(b^2>1\). Such a fraction cannot equal the integer \(24\). This contradiction shows that \(\sqrt{24}\) is irrational.

Assume \(\sqrt{24}=\frac{a}{b}\) in lowest terms. Since \(4<\sqrt{24}<5\), the denominator must satisfy \(b>1\). Squaring gives \(24=\frac{a^2}{b^2}\). Relatively prime \(a\) and \(b\) have relatively prime squares, so this is a reduced fraction with denominator greater than \(1\), which cannot equal the integer \(24\). Therefore, \(\sqrt{24}\) is irrational.