Real Numbers

The roster form lists all elements in a set within curly brackets and separates them with commas. We are asked to list all integers greater than -3 and less than 5. These numbers are to the right side of -3 and to the left side of 5 in a number line.

Now that we identified the set, let's list its elements. -2, -1, 0, 1, 2, 3, 4 We can now write the roster form of set A. A={ -2, -1, 0, 1, 2, 3, 4 } Although we can write the elements of a set in any order, writing them in ascending order is a good practice. We just need to be careful to write each once.

We now need to find the whole numbers that are less than 10. Recall that whole numbers start from 0 and increase by one unit indefinitely. The whole numbers less than 10 are to the right side of 0 included and to the left side of 10 in a number line.

Now that we identified the numbers in the set, let's write them down. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Finally, let's enclose them within curly brackets and add its name. Remember that the order of the elements can be disregarded. B={ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }

Consider the given set. C={ 2, 3, 5, 7, 11, 13, 17, 19} This set contains odd and even numbers starting from 2 and continuing up to 19. Yet, we need a more specific property to describe these numbers. After going deeper, we can see that these numbers are the first prime numbers up to 19 because the only factors of each are 1 and the numbers themselves.

C contains the first prime numbers less than or equal to 19

We can now use this statement and a variable to represent the elements of set C. In this case, let's use the variable x. We can then say that x is a prime number such that x is less than or equal to 19. Let's write the set-builder notation of C. C={ x | x is a prime number, x ≤ 19 }

Let's look at the given set. D={ ... , -4, -2, 2, 4,  ... } The numbers in this set are integers. These numbers are also even numbers and the set does not contain 0. This means that D contains even integers different than 0.

D contains even integers different than 0.

Let's now use this statement and the variable x to write the set-builder notation. In this case, x is an even integer different than 0. D={ x | x is an even integer , x ≠ 0 }

We want to identify the common elements between sets A and B. We are given set A in set-builder notation. Let's first write set A into its listing form. A={x | xis an even whole number less than9} ⇕ A={0, 2, 4, 6, 8} A common element is an element that is in both sets. In other words, it is an element that belongs to A and B at the same time. Let's identify these common elements. A&={0, 2, 4, 6, 8} B&={ 4, 5, 6, 7, 8} The common elements are 4, 6, and 8. { 4, 6, 8}

This time we will write all the elements that belong to sets A and B. Recall that each element in a set is unique, so if any element exists in both sets, we will write them once. A&={0, 2, 4, 6, 8} B&={ 4, 5, 6, 7, 8} Great! Let's now write all elements that are in both sets. {0, 2, 4, 5, 6, 7, 8}

We are asked to list the elements of set A. Let's only consider the elements inside the red circle representing set A.

Let's keep in mind that the elements in the overlapping section of A and B are also part of A. We can now list the elements of A. A={-12,7,-8,-5,15,-23,54} Recall that the order is not important while listing the elements of a set.

This time we will list only the elements that are common in the sets A and B. Let's now examine only the overlapping region.

We can see from the diagram that only two elements belong to both sets. {54, -23 }

Finally, we will find the total number of elements in the universal set U. We will list all the elements that we see inside the bigger rectangle.

Note that we should also consider the elements that do not belong to sets A and B but are still inside set U. U={& -12,7,-8, -5,15,54,-23,-13,10, & 67, 98,9,-21,-43,-2,25,0,6,17} There are a total of nineteen elements in the universal set.

At first, we will list the elements of the given set B. Recall that the elements of a set should be written only once which means we will disregard the duplicates of the letters E and T for this case. L-E-T-T-E-R ⇓ B={ L,E,T,R } Set B has four elements. Recall that all elements in a subset are also elements of the original set. Now, we will list all the subsets starting with the ones having zero element, followed by one, two, three, and four elements at most since we have maximum of four elements for this case.

Subsets with Zero-Element

The only subset with zero elements of any set is the empty set. We can write this using the empty set notation. ∅ or { }

Subsets with One-Element

There are four elements in our original set. We can use each of these to form the single element subsets. {L}, {E}, {T}, {R}

Subsets with Two-Elements

Next, we will write the subsets including only two-elements. { L, E }, { L, T }, { L, R }, {E, T }, {E, R }, {T, R}

Subsets with Three-Elements

After that, we will write all the three-element subsets. {L, E, T }, {L, E, R }, {L, T, R }, {E, T, R }

Subsets with Four-Elements

The original set has four elements, so we can only form one subset having four elements: the original set itself. { L, E, T, R }

Summary

Let's see all these subsets in a table.

Now, we will consider the given options and see whether they match with the obtained subsets. l { } & ✓ {E,T} & ✓ {T,R } & ✓ {E,B} & * {E,E,T} & * {L,T,R} & ✓ {L,E,T,R} & ✓ {L,E,T,T,E,R} & * Notice that the given set B does not include the letter B. Therefore, it does not include the subset {E,B}. Also, since the elements in the sets {E,E,T} and {L,E,T,T,E,R} are not written only once, they do not even represent a set itself.

Let's consider the table we created in Part A of the subsets of B.

Let's add all the number of subsets. 1+4+6+4+1=16 Set B has a total of 16 subsets. This applies to any set with four elements. A shortcut for calculating the number of subsets of a set with n elements is with the formula 2^n. In this case, n is 4. 2^4=16

We will determine which of the given numbers are irrational. Let's recall the definition of irrational numbers.

An irrational number is a number that is not rational. So, an irrational number cannot be written as ab where a and b are integers and b≠0.

According to this definition, - 109 is a rational number because it is a fraction. -10/9= -10/9 → Rational Notice that 0 is also a rational number since it can be written as a fraction by choosing a nonzero integer for the denominator. For instance, 03=0. 0= 0/3 → Rational Next, let's recall the fact that the decimal form of an irrational number neither terminates nor repeats. In other words, if a decimal number terminates or repeats then it is rational. This means that both 5.70 and 0.13 are rational. 5.70 → Rational 0.13 → Rational We will now have a look at π. It is also one of the famous irrational number because its expanded form never terminates. π =3.14159265359... → Irrational Note that the cube root of any integer that is not a perfect cube is irrational. This information helps us examine sqrt(71). Notice that 71 is not a perfect cube which makes sqrt(71) an irrational number. sqrt(71) → Irrational Last, we will examine -sqrt(16). Recall that the square root of any whole number is rational if the radicand is a perfect square. Since 16 is a perfect square, we can calculate this root.

This means that -sqrt(16) is rational. Let's see the types of the given numbers as rational or irrational. - 109 &→ Rational 0 &→ Rational 5.70 &→ Rational 0.13 &→ Rational π &→ Irrational sqrt(71) &→ Irrational -sqrt(16) &→ Rational