Free math worksheets

- Recall the definition of a rational number in terms of two integers. - Check the numerator, denominator, and the condition on the denominator.

1. A rational number can be written as a quotient of two integers with a nonzero denominator. 2. In \(-\frac{11}{6}\), both \(-11\) and \(6\) are integers and the denominator is nonzero. 3. Therefore, \(-\frac{11}{6}\) is rational.

Rational, because \(-\frac{11}{6}\) is a quotient of two integers with a nonzero denominator.

- Evaluate square roots of perfect squares before classifying them. - A terminating or repeating decimal represents a rational number. - A fraction of two integers is rational when its denominator is not zero.

1. \(\sqrt{121}=11\), so it is rational. 2. The decimal \(0.\overline{45}\) repeats, so it is rational. In fact, \(0.\overline{45}=\frac{45}{99}=\frac{5}{11}\). 3. The number \(\pi\) is irrational. 4. The number \(18\) is not a perfect square, so \(\sqrt{18}\) is irrational. 5. The number \(\frac{22}{7}\) is a quotient of two integers with a nonzero denominator, so it is rational.

a) Rational; \(\sqrt{121}=11\). b) Rational; its decimal expansion repeats. c) Irrational. d) Irrational; \(18\) is not a perfect square. e) Rational; it is a quotient of integers.

- Reduce each fraction before judging its square root. - For a reduced fraction, check the numerator and denominator separately. - Its square root is rational only when both parts are perfect squares.

1. Reduce the fraction: \(\frac{18}{50}=\frac{9}{25}\). Both \(9\) and \(25\) are perfect squares, so \(\sqrt{\frac{18}{50}}=\frac{3}{5}\), which is rational. 2. Reduce the fraction: \(\frac{45}{80}=\frac{9}{16}\). Both \(9\) and \(16\) are perfect squares, so \(\sqrt{\frac{45}{80}}=\frac{3}{4}\), which is rational. 3. Reduce the fraction: \(\frac{14}{63}=\frac{2}{9}\). The numerator \(2\) is not a perfect square, so \(\sqrt{\frac{14}{63}}=\sqrt{\frac{2}{9}}\) is irrational.

a) Rational; \(\frac{3}{5}\) b) Rational; \(\frac{3}{4}\) c) Irrational; \(\sqrt{\frac{2}{9}}\)

- Replace repeated occurrences of the same irrational number with one symbol. - Look for terms that cancel exactly. - Use the defining relationship between a square root and its square.

1. \(r+5-r=5\), so the value is rational. 2. \(3+r=3+\sqrt{13}\). If this sum were rational, subtracting the rational number \(3\) would make \(\sqrt{13}\) rational. Therefore, the value is irrational. 3. Since \(r^2=13\), \(\frac{r^2}{13}=1\), so the value is rational. 4. \(r\cdot r=r^2=13\), so the value is rational.

a) Rational b) Irrational c) Rational d) Rational

- Decide whether a screen can show infinitely many digits. - Separate an exact value from a rounded value. - Check whether the radicand is a perfect square.

1. The displayed decimal \(4.3589\) is a rounded approximation, not the complete decimal expansion. 2. Since \(19\) is not a perfect square, \(\sqrt{19}\) is irrational. 3. Its exact decimal expansion is nonterminating and nonrepeating, even though a calculator shows only finitely many digits.

Marisol’s conclusion is incorrect. The display is a finite rounded approximation; the exact value \(\sqrt{19}\) is irrational.

- Substitute each listed denominator separately. - Simplify the quotient before judging its square root. - Look for whole-number perfect squares.

1. If \(m=2\), then \(\sqrt{\frac{72}{2}}=\sqrt{36}=6\), which is rational. 2. If \(m=3\), then \(\sqrt{\frac{72}{3}}=\sqrt{24}\), which is irrational because \(24\) is not a perfect square. 3. If \(m=8\), then \(\sqrt{\frac{72}{8}}=\sqrt{9}=3\), which is rational. 4. If \(m=18\), then \(\sqrt{\frac{72}{18}}=\sqrt{4}=2\), which is rational.

\(m=2\), \(m=8\), and \(m=18\)

- Determine the value or type of each square root before considering the outside sign. - Compare the effect of negation with the classification of the positive square root. - For the squared radicand, simplify the inside expression first.

1. a) \(-\sqrt{36}=-6\), which is rational. 2. b) \(\sqrt{18}\) is irrational because \(18\) is not a perfect square, so its negative is also irrational. 3. c) \(-\sqrt{(-7)^2}=-\sqrt{49}=-7\), which is rational. 4. Negating a real number changes its sign but not whether it is rational or irrational.

a) Rational b) Irrational c) Rational A negative sign outside a square root changes the sign of the value, but it does not change whether the value is rational or irrational.

- Translate each verbal description into a known number form. - Simplify any operation that reverses a square root. - Use decimal behavior to classify the last two labels.

1. An integer decreased by \(\sqrt{2}\) is irrational. 2. \((\sqrt{23})^2=23\), so the square is rational. 3. An eventually repeating decimal is rational. 4. A nonterminating, nonrepeating decimal is irrational.

a) Irrational b) Rational c) Rational d) Irrational

- Convert the terminating decimal to a fraction. - Check the numerator and denominator separately for perfect-square structure. - Simplify before classifying.

1. Rewrite \(0.0049\) as \(\frac{49}{10000}\). 2. Both \(49\) and \(10000\) are perfect squares. 3. \(\sqrt{0.0049}=\sqrt{\frac{49}{10000}}=\frac{7}{100}=0.07\). 4. Therefore, Lina is correct and the value is rational.

Lina is correct. \(\sqrt{0.0049}=0.07\), so the value is rational.

- Identify which digits repeat forever. - Separate the finite prefix from the infinite tail. - Classification depends on the long-term decimal behavior.

1. The decimal has a finite initial part \(31415926535\). 2. After that initial part, the block \(27\) repeats forever. 3. A decimal that eventually repeats is rational, regardless of its finite beginning.

\(x\) is rational because its decimal expansion eventually repeats the block \(27\).

- Focus on what the machine does after the repeated block stops. - Decide whether infinitely many nonzero digits remain. - Classify the final decimal behavior, not the temporary pattern.

1. After \(150\) decimal digits, every later digit is \(0\). 2. Therefore, the decimal terminates when unnecessary trailing zeros are removed. 3. Every terminating decimal is rational. 4. The finite repetition of \(314\) does not determine the classification; the eventual all-zero tail does.

The number is rational because its decimal expansion terminates after \(150\) decimal places.

- Reduce the fraction inside each square root first. - For a whole-number radicand, check whether it is a perfect square. - For a reduced fractional radicand, check whether both numerator and denominator are perfect squares.

1. Reduce the radicand: \(\frac{48}{3}=16\). Therefore, \(\sqrt{\frac{48}{3}}=\sqrt{16}=4\), which is rational. 2. Reduce the radicand: \(\frac{15}{3}=5\). Since \(5\) is not a perfect square, \(\sqrt{\frac{15}{3}}=\sqrt{5}\) is irrational. 3. Reduce the radicand: \(\frac{14}{56}=\frac{1}{4}\). Therefore, \(\sqrt{\frac{14}{56}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\), which is rational.

a) Rational; \(4\) b) Irrational; \(\sqrt{5}\) c) Rational; \(\frac{1}{2}\)

- Ask whether the ellipsis gives a unique continuation. - Imagine two different tails that begin after the displayed digits. - Classification depends on the infinite decimal behavior.

1. A finite beginning of a decimal does not determine its behavior forever. 2. The digits could eventually terminate or repeat, producing a rational number. 3. The digits could continue without any eventual repeating block, producing an irrational number. 4. A rule for all later digits is needed.

No. The displayed prefix is not enough. A complete rule describing whether the decimal eventually terminates, repeats, or remains nonrepeating is needed.

- Remember that an equation involving a square can have two real solutions. - Simplify the square root of the fraction. - Classify every possible solution, not only the positive one.

1. Taking both square-root solutions gives \(x=\pm\sqrt{\frac{81}{16}}\). 2. \(\sqrt{\frac{81}{16}}=\frac{9}{4}\). 3. Thus, \(x=\frac{9}{4}\) or \(x=-\frac{9}{4}\). 4. Both possible values are rational, so \(x\) must be rational.

Yes. The possible values are \(x=\frac{9}{4}\) and \(x=-\frac{9}{4}\), and both are rational.

- Convert each mixed number to an improper fraction first. - Reduce the fraction before deciding whether its square root is rational. - The square root of a reduced fraction is rational only when both the numerator and denominator are perfect squares.

1. Convert the mixed number: \(3\frac{1}{16}=\frac{49}{16}\). Then \(\sqrt{\frac{49}{16}}=\frac{7}{4}\), which is rational. 2. Convert the mixed number: \(2\frac{14}{25}=\frac{64}{25}\). Then \(\sqrt{\frac{64}{25}}=\frac{8}{5}\), which is rational. 3. Convert the mixed number: \(1\frac{4}{5}=\frac{9}{5}\). The fraction is in lowest terms, and its denominator is not a perfect square, so \(\sqrt{\frac{9}{5}}\) is irrational. 4. Convert the mixed number: \(7\frac{1}{9}=\frac{64}{9}\). Then \(\sqrt{\frac{64}{9}}=\frac{8}{3}\), which is rational.

a) Rational: \(\frac{7}{4}\) b) Rational: \(\frac{8}{5}\) c) Irrational: \(\sqrt{\frac{9}{5}}\) d) Rational: \(\frac{8}{3}\)

- Write each terminating decimal as a reduced fraction. - When is the square root of a reduced fraction rational? - Check whether both the numerator and denominator are perfect squares.

1. \(A = \sqrt{0.64} = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0.8\). 2. \(D = \sqrt{0.0064} = \sqrt{\frac{4}{625}} = \frac{2}{25} = 0.08\). 3. For \(B\), \(6.4 = \frac{32}{5}\). In lowest terms, the numerator and denominator are not both perfect squares, so \(\sqrt{\frac{32}{5}}\) is irrational. 4. For \(C\), \(0.064 = \frac{8}{125}\). In lowest terms, the numerator and denominator are not both perfect squares, so \(\sqrt{\frac{8}{125}}\) is irrational.

a) \(A = 0.8\) and \(D = 0.08\) b) \(B = \sqrt{\frac{32}{5}}\) and \(C = \sqrt{\frac{8}{125}}\) are irrational because the reduced numerator and denominator are not both perfect squares.

- For part a), choose a simple one-decimal-place number for \(\sqrt{z}\), then square it. - Convert \(0.1\) to a reduced fraction. - Test whether both parts of the reduced fraction are perfect squares.

1. Choose, for example, \(\sqrt{z}=0.5\). Then \(z=0.5^2=0.25\), and \(0.01<0.25<1\). 2. Write \(0.1=\frac{1}{10}\). This fraction is in lowest terms. The numerator \(1\) is a perfect square, but the denominator \(10\) is not. 3. The square root of a reduced fraction is rational only when both its numerator and denominator are perfect squares. Therefore, \(\sqrt{0.1}=\sqrt{\frac{1}{10}}\) is irrational.

a) One example is \(z=0.25\), because \(\sqrt{0.25}=0.5\). b) \(0.1=\frac{1}{10}\); \(1\) is a perfect square, but \(10\) is not. c) \(\sqrt{0.1}\) is irrational.

- Recall the definition of a rational number. - Look for counterexamples involving perfect squares. - Think about how repeating decimals can be written as fractions. - Test products of simple irrational square roots.

1. Statement a) is true. Every integer \(z\) can be written as \(\frac{z}{1}\). 2. Statement b) is false. For example, \(\sqrt{9} = 3\), which is rational. 3. Statement c) is true. Every repeating decimal can be converted to a fraction of integers; for example, \(0.\overline{3} = \frac{1}{3}\). 4. Statement d) is false. For example, \(\sqrt{2} \cdot \sqrt{2} = 2\), which is rational even though both factors are irrational.

a) True b) False; for example, \(\sqrt{9} = 3\). c) True d) False; for example, \(\sqrt{2} \cdot \sqrt{2} = 2\).

- Square decimal values to compare them with \(7\). - What property separates rational and irrational numbers? - Think about the decimal expansion of an irrational number.

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.

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.

- Check whether each decimal terminates or repeats. - A radical symbol does not automatically make a number irrational. - Simplify any square root whose numerator and denominator are perfect squares.

1. \(-\sqrt{49}=-7\) and \(0.125=\frac{1}{8}\), so both are rational and already placed correctly. 2. \(\sqrt{12}\) is irrational because \(12\) is not a perfect square, so it moves to the irrational bin. 3. \(0.\overline{18}\) is rational because its decimal expansion repeats, so it moves to the rational bin. 4. \(\pi-3\) is irrational because subtracting a rational number from \(\pi\) cannot make it rational, so it is already placed correctly. 5. \(\sqrt{\frac{64}{121}}=\frac{8}{11}\), so it moves to the rational bin. 6. Exactly \(3\) cards move.

Rational: \(-\sqrt{49}\), \(0.125\), \(0.\overline{18}\), \(\sqrt{\frac{64}{121}}\) Irrational: \(\sqrt{12}\), \(\pi-3\) Exactly \(3\) cards were moved.

- Separate “nonterminating” from “nonrepeating.” - Look only at what happens forever after any initial digits. - Ask whether one fixed block can repeat from some point onward.

1. A rational decimal may be nonterminating if it eventually repeats. 2. In \(a\), the digit \(5\) repeats forever after the first two decimal places, so \(a\) is rational. 3. In \(b\), the gaps of zeros keep changing, so no fixed block eventually repeats. 4. Therefore, \(b\) is irrational.

Kai’s statement is false because a nonterminating decimal can still be eventually repeating. The number \(a\) is rational, and the number \(b\) is irrational.

- List the perfect squares near the given interval. - Decide when either radicand could equal one of those squares. - Check both endpoints of each consecutive pair.

1. For a whole-number radicand, the square root is rational exactly when the radicand is a perfect square. 2. The only perfect square from \(30\) through \(41\) is \(36\). 3. If \(n+1=36\), then \(n=35\). 4. If \(n=36\), then \(n=36\). 5. Both values make exactly one of the two square roots rational.

\(n=35\) or \(n=36\)

- Use the area of a square to identify its side length. - Relate the diagonal to two equal perpendicular sides. - Classify the exact simplified values, not their appearances before simplification.

1. The side length is \(s=\sqrt{50}\,\text{cm}\), which is irrational because \(50\) is not a perfect square. 2. The diagonal satisfies \(d^2=s^2+s^2=50+50=100\). 3. Thus, \(d=\sqrt{100}=10\,\text{cm}\), which is rational. 4. Different expressions involving irrational numbers can simplify to rational values.

a) The side length \(\sqrt{50}\,\text{cm}\) is irrational. b) The diagonal is \(10\,\text{cm}\), which is rational. c) Squaring and combining the side lengths produces the perfect square \(100\), so the diagonal simplifies to a rational number.

- Start from the definition of a rational number. - Examine the prime factors of the denominator in lowest terms. - Recall the two possible decimal behaviors of rational numbers.

1. \(\frac{17}{24}\) is a ratio of two integers with a nonzero denominator, so it is rational. 2. The fraction is already in lowest terms, and \(24=2^3\cdot3\). 3. Because the reduced denominator has a prime factor other than \(2\) or \(5\), the decimal does not terminate. 4. Every rational decimal that does not terminate eventually repeats.

\(\frac{17}{24}\) is rational, and its decimal expansion is nonterminating but eventually repeating.

- Keep \(\pi\) in the exact expression. - Identify the type of each multiplier. - Consider what division by a nonzero rational factor would imply.

1. The circumference is \(C=14\pi\,\text{ft}\). 2. Since \(14\) is a nonzero rational number and \(\pi\) is irrational, \(14\pi\) is irrational. 3. Three laps have length \(3\cdot14\pi=42\pi\,\text{ft}\), which is also irrational. 4. Multiplication by a nonzero rational number cannot turn an irrational number into a rational number.

a) The circumference \(14\pi\,\text{ft}\) is irrational. b) The distance \(42\pi\,\text{ft}\) is irrational. c) Both multipliers are nonzero rational numbers, so the products remain irrational.

- Isolate the unknown. - Classify the exact expression that results. - Compare that classification with the condition placed on the unknown.

1. Solving for \(q\) gives \(q=7-\sqrt{11}\). 2. The number \(7\) is rational and \(\sqrt{11}\) is irrational. 3. Their difference is irrational. 4. Therefore, the required value of \(q\) is not rational, so no rational solution exists.

No rational number \(q\) satisfies the equation.

- Look for an irrational number created by a familiar inverse operation. - Choose its radicand so the number does not simplify to an integer. - Check each condition separately after choosing the example.

1. One valid choice is \(x=\sqrt{7}\). 2. Since \(7\) is not a perfect square, \(\sqrt{7}\) is irrational. 3. \(x^2=(\sqrt{7})^2=7\), which is rational. 4. \(x+2=\sqrt{7}+2\) is irrational because adding a rational number to an irrational number cannot make it rational.

One example is \(x=\sqrt{7}\). Then \(x^2=7\) is rational, while \(x+2=\sqrt{7}+2\) is irrational.

- List the two-digit perfect squares. - Keep only those with the required ones digit. - Relate a two-decimal radicand to a one-decimal square root.

1. A two-decimal radicand has a rational terminating square root when its two-digit numerator over \(100\) is a perfect square fraction. 2. Among the numbers ending in \(6\) from \(06\) through \(96\), the perfect squares are \(16\) and \(36\). 3. Thus, \(0.16=(0.4)^2\) and \(0.36=(0.6)^2\). 4. The valid digits are \(1\) and \(3\).

The box can be \(1\) or \(3\): \(\sqrt{0.16}=0.4\) and \(\sqrt{0.36}=0.6\).

- Test a familiar irrational number related to squaring. - Separate the classification of a number from the classification of its square. - One valid counterexample is enough to disprove an “always” statement.

1. Choose \(x=\sqrt{2}\). 2. The number \(x\) is irrational. 3. Its square is \(x^2=(\sqrt{2})^2=2\), which is rational. 4. Therefore, the statement is false.

The statement is false. For example, \(x=\sqrt{2}\) is irrational, but \(x^2=2\) is rational.

- Relate the width, height, and diagonal of the rectangle. - Classify the exact radical before considering any rounded value. - Distinguish a measurement approximation from the mathematical length.

1. The diagonal satisfies \(d^2=3^2+5^2=9+25=34\). 2. Thus, \(d=\sqrt{34}\,\text{m}\). 3. Since \(34\) is not a perfect square, \(\sqrt{34}\) is irrational. 4. The decimal \(5.83\) is a rational approximation, but it does not change the classification of the exact length.

a) \(\sqrt{34}\,\text{m}\) b) Irrational c) No. \(5.83\,\text{m}\) is only a rational approximation to the irrational exact length.

- Track how each block length changes. - A decimal can follow a rule without being periodic. - Add the numbers digit by digit and look for a pattern. - Recall how to convert a purely repeating decimal to a fraction.

1. For \(x\), the decimal is built from blocks containing \(n\) consecutive \(1\)s followed by one \(0\), for \(n = 1, 2, 3, \ldots\). 2. For \(y\), the decimal is built from blocks containing \(n\) consecutive \(3\)s followed by one \(4\), for \(n = 1, 2, 3, \ldots\). 3. The block lengths keep increasing, so neither decimal can become periodic. Both decimals are nonterminating and nonrepeating, so both numbers are irrational. 4. At every decimal place, the aligned digits are either \(1+3\) or \(0+4\). Thus, \(x+y = 0.4444\ldots = 0.\overline{4}\). 5. Since \(0.\overline{4} = \frac{4}{9}\), \(x+y = \frac{4}{9}\).

a) In \(x\), a block of \(n\) ones is followed by a zero. In \(y\), a block of \(n\) threes is followed by a four. In both patterns, \(n\) increases by \(1\) each time. b) Both numbers are irrational because their decimal expansions are nonterminating and nonrepeating. c) \(x+y = 0.\overline{4} = \frac{4}{9}\)

- What kinds of decimal expansions represent rational numbers? - How can you guarantee that a fixed block repeats forever? - How can a decimal follow a rule but never become periodic?

1. A rational example is formed by repeating the block \(45\): \(0.454545454545\ldots\). The next ten digits are \(4545454545\). The decimal is rational because it is periodic. 2. An irrational example is \(0.451010010001\ldots\), where successive \(1\)s are separated by one zero, then two zeros, then three zeros, and so on. The next ten digits after the given \(45\) are \(1010010001\). 3. Because the gaps between the \(1\)s keep increasing, no fixed digit block repeats forever. The decimal is nonterminating and nonrepeating, so it is irrational.

a) Example: \(0.454545454545\ldots\). The next ten digits are \(4545454545\), and the repeating pattern makes the number rational. b) Example: \(0.451010010001\ldots\). The next ten digits are \(1010010001\). The increasing gaps prevent periodic repetition, so the number is irrational.

- Add decimal places to create a value between the endpoints. - A fixed repeating block produces a rational number. - To create an irrational decimal, use a nonterminating pattern that never becomes periodic.

1. A terminating example is \(0.455\). 2. A purely repeating example is \(0.\overline{455} = \frac{455}{999}\), which is approximately \(0.455455\ldots\) and lies between \(0.45\) and \(0.46\). 3. For \(\frac{n}{200}\), solve \(0.45 < \frac{n}{200} < 0.46\). Multiplying by \(200\) gives \(90 < n < 92\), so \(n = 91\). Thus, \(\frac{91}{200} = 0.455\). 4. An irrational example is \(0.4501001000100001\ldots\), where the number of zeros between successive \(1\)s increases. This decimal is nonterminating and nonrepeating.

a) \(0.455\) b) \(0.\overline{455}\) c) \(\frac{91}{200}\) d) \(0.4501001000100001\ldots\), with an increasing number of zeros between successive \(1\)s

- 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?

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 \(828\).

a) \(s = 2\), so the sum is rational. b) The first three digits after the decimal point are \(828\).

- Begin by temporarily assuming the student’s possibility is true. - Isolate the square-root term using an operation that preserves rationality. - Compare the result with what is known about the radicand.

1. Assume \(4+\sqrt{6}\) is rational. 2. Subtracting the rational number \(4\) would make \(\sqrt{6}\) rational. 3. Since \(6\) is not a perfect square, \(\sqrt{6}\) is irrational. 4. The assumption creates a contradiction, so \(4+\sqrt{6}\) is irrational.

\(4+\sqrt{6}\) is irrational.

- Evaluate the product with the exceptional factor first. - For a nonzero factor, imagine dividing the product by that factor. - Ask what would follow if one of the nonzero products were rational.

1. \(0\cdot x=0\), so part a) is rational. 2. A nonzero rational multiple of an irrational number cannot be rational; otherwise division by the nonzero rational factor would make \(x\) rational. 3. Therefore, \(7x\) and \(-\frac{3}{5}x\) are irrational.

a) Rational b) Irrational c) Irrational The zero factor is the exceptional case because it makes the product \(0\).

- Think about reversing addition or subtraction. - For the product, test an exceptional rational value. - Then consider what division would imply when the rational factor is nonzero.

1. \(a+b\) is always irrational; otherwise subtracting rational \(a\) would make \(b\) rational. 2. \(a-b\) is always irrational for the same reason. 3. The product \(ab\) needs more information. If \(a=0\), the product is rational; if \(a\ne0\), the product is irrational.

a) Guaranteed irrational b) Guaranteed irrational c) More information is needed; the result depends on whether \(a=0\).

- Translate the rational-root condition into a condition on the radicand. - List relevant perfect squares within the possible range of \(6k\). - Check that each resulting \(k\) is not itself a perfect square.

1. The condition that \(\sqrt{6k}\) is rational requires \(6k\) to be a perfect square. 2. The perfect squares from \(6\) through \(180\) that are divisible by \(6\) are \(36\) and \(144\). 3. From \(6k=36\), \(k=6\), and \(\sqrt{6}\) is irrational. 4. From \(6k=144\), \(k=24\), and \(\sqrt{24}\) is irrational. 5. Therefore, both values satisfy the conditions.

\(k=6\) and \(k=24\)

- Solve the product equation for one factor. - Test the possibility that the unknown factor is rational. - Use division to see what that would force about the given irrational factor.

1. Since \(xy=5\), neither \(x\) nor \(y\) is zero, and \(y=\frac{5}{x}\). 2. If \(y\) were rational, then it would be a nonzero rational number. 3. It would follow that \(x=\frac{5}{y}\) is rational. 4. That contradicts the given fact that \(x\) is irrational, so \(y\) must be irrational.

\(y\) must be irrational.

- For a rational difference, begin with two equal irrational numbers. - For an irrational difference, compare an irrational number with a nonzero rational multiple of itself. - A single contrasting pair is enough to show the outcome is not fixed.

1. The pair \(\sqrt{7}\) and \(\sqrt{7}\) has rational difference \(0\). 2. The numbers \(\sqrt{3}\) and \(2\sqrt{3}\) are both irrational because a nonzero rational multiple of an irrational number is irrational. 3. Their difference is \(\sqrt{3}-2\sqrt{3}=-\sqrt{3}\), which is irrational. 4. Therefore, the difference of two irrational numbers can be rational or irrational, so the claim is false.

Example with rational difference: \(\sqrt{7}-\sqrt{7}=0\). Example with irrational difference: \(\sqrt{3}-2\sqrt{3}=-\sqrt{3}\). The claim is false.

- Rational decimals terminate or eventually repeat. - Examine the appended numbers \(10\), \(100\), \(1000\), and so on. - Can a fixed repeating block produce arbitrarily long strings of zeros?

1. A real number is rational exactly when its decimal expansion terminates or eventually repeats. 2. The decimal for \(a\) does not terminate because natural numbers continue to be appended forever. 3. It also cannot eventually repeat. For every natural number \(m\), the appended number \(10^m\) contains \(m\) consecutive zeros after its leading \(1\). Thus, the decimal contains arbitrarily long strings of zeros. 4. An eventually repeating decimal with a fixed period cannot contain arbitrarily long strings of zeros unless the repeating period consists only of zeros, which would make the decimal terminate. Neither case applies here. 5. Therefore, the decimal is nonterminating and nonrepeating, so \(a\) is irrational. A predictable rule alone does not make a number rational.

The claim is false. The decimal neither terminates nor eventually repeats. The blocks \(10^m\) create arbitrarily long strings of zeros, which no fixed nonzero repeating period can produce. Therefore, \(a\) is irrational.

- Reduce the radicand and examine the possible denominator. - Recall what the denominator of a squared fraction looks like in lowest terms. - Once the radicand is a whole number, generate the relevant perfect squares.

1. Reduce \(\frac{n}{2}\) to lowest terms. If \(n\) is odd, its denominator is \(2\). 2. The square of a rational number written in lowest terms has a perfect-square denominator, so a reduced denominator of \(2\) is impossible. Therefore, \(n\) must be even and \(\frac{n}{2}\) must be a whole number. 3. A whole number has a rational square root only when it is a perfect square, so write \(\frac{n}{2}=m^2\) for a whole number \(m\). Then \(n=2m^2\). 4. For \(m=1,2,3,4,5\), the values of \(n\) are \(2,8,18,32,50\). The next value exceeds \(50\).

\(n\in\{2,8,18,32,50\}\)