Algebra Quiz: Rational And Irrational Number Operations

What this quiz covers

This quiz focuses on Rational And Irrational Number Operations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Assume $r\times i$ is rational. Then $i=\frac{r\times i}{r}$ would be rational (rational divided by nonzero rational), contradicting that $i$ is irrational. Therefore $r\times i$ is irrational. (correct answer)
2. The claim is false because $0\times i=0$ is rational, so $r\times i$ can be rational even when $r$ is nonzero.
3. The claim is true because multiplication always makes decimals longer, so the result cannot be a fraction.
4. The claim is true because $r\times i$ is irrational by definition whenever $i$ is irrational.

Question 2

1. Sometimes it's rational and sometimes it's irrational; for example $\sqrt{2}\cdot\sqrt{8}=4$ (rational) but $\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$ (irrational). (correct answer)
2. It always equals an irrational number; for example $\sqrt{2}\cdot\sqrt{2}=\sqrt{4}$ is irrational and $\sqrt{2}\cdot\sqrt{3}=\sqrt{5}$ is irrational.
3. Sometimes it's rational and sometimes it's irrational; for example $\sqrt{2}\cdot\sqrt{2}=\sqrt{4}$ (irrational) but $\sqrt{2}\cdot\sqrt{3}=\sqrt{5}$ (rational).
4. It always equals a rational number; for example $\sqrt{2}\cdot\sqrt{8}=4$ and $\sqrt{3}\cdot\sqrt{12}=6$.

Question 3

1. It is always true because the sum of two non-repeating decimals cannot repeat.
2. It is always false because two irrationals always add to a rational number.
3. It is not always true because some pairs of irrationals add to a rational, for example $\sqrt{2}+(-\sqrt{2})=0$ (rational), even though other pairs like $\sqrt{2}+\sqrt{3}$ are irrational. (correct answer)
4. It is not always true because $\sqrt{2}+\sqrt{2}=\sqrt{4}=4$, which is rational.

Question 4

1. I only
2. II only
3. I and III only (correct answer)
4. I, II, and III

Question 5

1. Rational, because multiplying by $4$ removes the square root: $4\sqrt{5}=\sqrt{20}$, and all square roots are rational.
2. Irrational, because $\sqrt{5}$ is irrational and any time you add four equal sides the result is irrational only when the side is less than 3.
3. Irrational, because the perimeter is $4\sqrt{5}$ and $4$ is a nonzero rational; if $4\sqrt{5}$ were rational then dividing by $4$ would make $\sqrt{5}$ rational, a contradiction. (correct answer)
4. Rational, because perimeters are always whole numbers.

Question 6

1. The conclusion is wrong because $\sqrt{2} + \sqrt{8} = 3\sqrt{2}$ is actually rational
2. The rule stated is wrong; the sum of two irrational numbers can sometimes be rational, though $\sqrt{2} + \sqrt{8}$ happens to be irrational (correct answer)
3. Both the rule and conclusion are wrong; $\sqrt{2} + \sqrt{8}$ is rational and the sum of irrationals is always rational
4. Nothing is wrong; both the reasoning and conclusion are correct in this case

Question 7

1. $p \cdot q$ must be irrational because the product of two irrational numbers is always irrational
2. $p \cdot q$ must be rational because irrational times irrational cancels the irrationality
3. $p \cdot q$ could be either rational or irrational, depending on the specific values of $p$ and $q$ (correct answer)
4. $p \cdot q$ is undefined because you cannot multiply two irrational numbers together

Question 8

1. His simplification of the radicals is incorrect; $\sqrt{8} \neq 2\sqrt{2}$ and $\sqrt{18} \neq 3\sqrt{2}$
2. His arithmetic is wrong; $2\sqrt{2} + 3\sqrt{2} = 6\sqrt{2}$, not $5\sqrt{2}$
3. His conclusion about rationality is wrong; $5\sqrt{2}$ is irrational because it's a nonzero rational times an irrational (correct answer)
4. His reasoning is completely correct; $5\sqrt{2}$ is indeed a rational number

Question 9

1. It is rational because $\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{4} = 2$, and $2 + \frac{4}{7}$ is the sum of two rationals (correct answer)
2. It is irrational because it involves square roots in the original form
3. It cannot be determined without more information about the relationship between 12 and 3
4. It is irrational because $\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{4} = 2$, but adding any fraction makes it irrational

Question 10

1. Assume $5+\sqrt{3}$ is rational, say $q$. Then $\sqrt{3}=q-5$. Since $q$ and $5$ are rational and rationals are closed under subtraction, $q-5$ is rational, contradicting that $\sqrt{3}$ is irrational. Therefore $5+\sqrt{3}$ is irrational. (correct answer)
2. $5$ is rational and $\sqrt{3}$ is irrational, so their sum must be rational because the rational part "dominates."
3. $5+\sqrt{3}$ is irrational because $\sqrt{3}$ has a non-terminating decimal and adding $5$ does not change that decimal into a fraction.
4. Assume $5+\sqrt{3}$ is irrational. Then it cannot be written as $\frac{p}{q}$, so it is irrational.

Question 11

1. I only
2. I and II only
3. I and III only (correct answer)
4. I, II, and III

Question 12

1. It is sometimes rational because any two irrationals add to an integer.
2. It always equals irrational because irrational decimals never repeat, and adding two never makes them repeat.
3. It is sometimes rational and sometimes irrational; for example, $\sqrt{2}+(-\sqrt{2})=0$ (rational) but $\sqrt{2}+\sqrt{3}$ is irrational, so there is no single always-true rule. (correct answer)
4. It is always rational because two negatives cancel out in the decimals.

Question 13

1. The claim is false because $2\times \sqrt{2}=\sqrt{4}=4$, which is rational.
2. The claim is true because multiplying by a rational never changes irrational to rational, even when $r=0$.
3. The claim is true because $r\times i$ can be rewritten as $\frac{p}{q}\times i=\frac{p}{qi}$, a ratio of integers.
4. The claim is false because if $r=0$, then $r\times i=0$, which is rational. (correct answer)

Question 14

1. The sum $r+s$ is irrational unless $r=s$.
2. The sum $r+s$ is rational because all decimals are rational numbers.
3. The sum $r+s$ is rational because if $r=\frac{p}{q}$ and $s=\frac{m}{n}$, then $r+s=\frac{pn+mq}{qn}$, which is a ratio of integers with nonzero denominator. (correct answer)
4. The sum $r+s$ is irrational because adding fractions makes the decimal non-repeating.

Question 15

1. If $a=\frac{p}{q}$ and $b=\frac{r}{s}$, then $a+b=\frac{ps+qr}{qs}$, and $ps+qr$ and $qs$ are integers with $qs\ne 0$, so the sum is rational. (correct answer)
2. If $a$ and $b$ are rational, then their decimals terminate, so $a+b$ must terminate and be rational.
3. Since rational numbers are closed under multiplication, they must also be closed under addition, so $a+b$ is rational.
4. Because $\frac{p}{q}+\frac{r}{s}=\frac{p+r}{q+s}$, the sum of two rationals is rational.

Question 16

1. Irrational, because subtracting fractions creates non-repeating decimals.
2. Rational, because the sum of two rational numbers is rational: if $a=\frac{p}{q}$ and $b=\frac{r}{s}$, then $a+b=\frac{ps+qr}{qs}$ is a ratio of integers. (correct answer)
3. Irrational, because one of the numbers is negative, and irrational numbers are negative.
4. Rational, because $\frac{2}{3}+\left(-\frac{5}{6}\right)=\frac{2-5}{3-6}=\frac{-3}{-3}=1$.

Question 17

1. The claim is true because rational numbers are integers written as decimals.
2. The claim is false: rationals are closed under addition (so $a+b$ is rational), but a rational sum need not be an integer; for example $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$, which is rational but not an integer. (correct answer)
3. The claim is false because $a+b$ is always irrational when $a$ and $b$ are rational.
4. The claim is true because $a+b=\frac{ps+qr}{qs}$ is always a whole number.

Question 18

1. Assume $r+i$ is irrational. Then $i=(r+i)-r$ is irrational minus rational, which is irrational, so the assumption is consistent; therefore $r+i$ is irrational.
2. Assume $r+i$ is rational (call it $q$). Then $i=q-r$. Since $q$ and $r$ are rational and rationals are closed under subtraction, $q-r$ is rational. This contradicts that $i$ is irrational. Therefore $r+i$ is irrational. (correct answer)
3. Since $r$ is rational, it has a terminating or repeating decimal. Adding it to $i$ cannot change $i$, so $r+i$ must be irrational.
4. Assume $r+i$ is rational. Then $r=(r+i)-i$ is rational minus irrational, which is irrational, contradicting that $r$ is rational. Therefore $r+i$ is rational.

Question 19

1. Irrational, because adding two fractions usually makes a non-terminating decimal.
2. Rational, because $\frac{1}{2}$ and $\frac{1}{3}$ are rational and rationals are closed under addition: $\frac{p}{q}+\frac{r}{s}=\frac{ps+rq}{qs}$, a ratio of integers. (correct answer)
3. Irrational, because $\frac{1}{2}$ is rational but $\frac{1}{3}$ is irrational since it repeats.
4. Rational, because the denominator after adding will always be prime, which guarantees a rational number.

Question 20

1. I and III only (correct answer)
2. I, II, and III
3. I only
4. II only