TRUE OR FALSE — NUMBER SETS
Set 1: Classifying numbers as rational or irrational
PREAMBLE
The given number is a rational number.
1. $-5$
ANSWER: True
2. $0.4545...$ (recurring)
ANSWER: True
3. $\sqrt7$
ANSWER: False
SOLUTION
$-5$ is an integer, and every integer is rational since it can be written as itself over 1. A recurring decimal like $0.4545...$ always converts back into an exact fraction (here $5/11$), so it is rational too. $\sqrt7$ is irrational because 7 is not a perfect square, so its decimal expansion is non-terminating and non-recurring, and it cannot be written as a fraction of two integers.
Set 2: Membership of zero in number sets
PREAMBLE
The number 0 is a member of the given set.
1. Natural Numbers
ANSWER: False
2. Whole Numbers
ANSWER: True
3. Integers
ANSWER: True
SOLUTION
Natural numbers begin at 1 and exclude 0 entirely. Whole numbers are formed by adding 0 to the natural numbers, so 0 is included there. Integers extend the whole numbers to include negative values, keeping every whole number, including 0, as a member.
Set 3: Closure of whole numbers under an operation
PREAMBLE
The set of whole numbers is closed under the given operation.
1. Addition
ANSWER: True
2. Subtraction
ANSWER: False
3. Multiplication
ANSWER: True
SOLUTION
Adding two whole numbers always produces another whole number, so whole numbers are closed under addition. Subtracting a larger whole number from a smaller one can give a negative result, such as $3-5=-2$, which is not a whole number, so closure fails under subtraction. Multiplying two whole numbers always produces another whole number, so closure holds under multiplication.
Set 4: Identifying the associative property
PREAMBLE
1. $3\times4=4\times3$ demonstrates the associative property
ANSWER: False
2. $3+(4+5)=(3+4)+5$ demonstrates the associative property
ANSWER: True
3. $(2\times3)\times4=2\times(3\times4)$ demonstrates the associative property
ANSWER: True
SOLUTION
$3\times4=4\times3$ swaps the order of the two factors, which is the commutative property, not the associative property. $3+(4+5)=(3+4)+5$ and $(2\times3)\times4=2\times(3\times4)$ both keep the numbers in the same order but change how they are grouped, which is exactly what the associative property describes.
Set 5: Finding the multiplicative inverse of 5
PREAMBLE
The given value is the multiplicative inverse of 5.
1. $1/5$
ANSWER: True
2. $0.2$
ANSWER: True
3. $-1/5$
ANSWER: False
SOLUTION
The multiplicative inverse of 5 is the value that multiplies with 5 to give 1, which is $1/5$; since $1/5=0.2$, both representations are correct. $-1/5$ is not the multiplicative inverse, since $5\times(-1/5)=-1$, not 1 — it is also easy to confuse with the additive inverse of 5, which is $-5$, not $-1/5$.
Set 6: Testing subset relationships between A = {1, 2} and B = {1, 2, 3}
PREAMBLE
1. A ⊆ B
ANSWER: True
2. B ⊆ A
ANSWER: False
3. A ⊂ B
ANSWER: True
SOLUTION
Every element of A = {1, 2} is also in B = {1, 2, 3}, so A ⊆ B is true. B contains the element 3, which is not in A, so B ⊆ A is false. Since A ⊆ B and A is not equal to B, A is also a proper subset of B, so A ⊂ B is true.
Set 7: Using the two-set counting formula for a class of 40 students, 25 studying Mathematics, 18 studying Physics, and 10 studying both
PREAMBLE
1. The number of students who study Mathematics or Physics (or both) is 33
ANSWER: True
2. The number of students who study only Mathematics is 15
ANSWER: True
3. The number of students who study neither subject is 5
ANSWER: False
SOLUTION
$n(M\cup P)=n(M)+n(P)-n(M\cap P)=25+18-10=33$, so the first statement is true. The number studying only Mathematics is $n(M)-n(M\cap P)=25-10=15$, so the second statement is true. The number studying neither subject is $40-n(M\cup P)=40-33=7$, not 5, so the third statement is false.
Set 8: Applying De Morgan's laws with U = {1,2,3,4,5,6}, A = {1,2,3} and B = {3,4,5}
PREAMBLE
1. $(A\cup B)'=\{6\}$
ANSWER: True
2. $(A\cap B)'=\{1,2,4,5,6\}$
ANSWER: True
3. $A'\cup B'=\{6\}$
ANSWER: False
SOLUTION
$A\cup B=\{1,2,3,4,5\}$, so its complement in U is $\{6\}$, making the first statement true. $A\cap B=\{3\}$, so its complement is $\{1,2,4,5,6\}$, making the second statement true. By De Morgan's laws, $A'\cup B'$ actually equals $(A\cap B)'$, not $(A\cup B)'$; directly, $A'=\{4,5,6\}$ and $B'=\{1,2,6\}$, so $A'\cup B'=\{1,2,4,5,6\}$, not $\{6\}$, making the third statement false.