🍫 **Steps to Write a Set in Set-builder Form**

To convert the given set in set—builder form, we use the following steps

Step I. Describe the elements of the set by using a symbol *x* or any other symbol *y*, *z* etc.

Step II. Write the symbol colon ‘:’.

Step III. After the sign of colon, write the characteristic property possessed by the elements of the set.

Step IV. Enclose the whole description within braces i.e., { }.

**Representation of a Statement in Both Forms**

Statement | Roster form | Set-builder form |

The set of all natural numbers between 10 and 14. | {11, 12, 13} | {∈ℕ, 10<x:x x<14} |

The set of all schools in Ottawa beginning with letter A. | {Abraar School, Académie Réveil, Ashbury College, Astolot Educational Centre, …} | { is name of school Delhi beginning with letter A}x:x |

The set of all distinct letters used in the word ‘Friend’. | {F,r,i,e,n,d} | { is the distinct letters used in the word ‘Friend’}x:x |

Example 1:**Worked out a Problem**

Write the following set, *A*={14,21,28,35,42,…,98} in set—builder form.

Step I. | Describe the elements of the set by using a symbol.Let x represent the elements of given set. |

Step II. | Write the symbol colon.Write the symbol ‘:’ after x. |

Step III. | Find the characteristic property possessed by the elements of the set.Given numbers are all natural numbers greater than 7 which are multiples of 7 and less than 100. |

Step IV. | Enclose the whole description within braces.Thus, A={x:x is a set of natural numbers greater than 7 which are multiples of 7 and less than 100}. |

Example 2:**Using Set-Builder Notation**

a) Write set *B*={1,2,3,4,5} in set-builder notation.

b) Write. in words, how you would read set *B* in set-builder notation.

Solution:

a) Because set *B* consists of the natural numbers less than 6. we write

*B*={

*x*|

*x*∈ℕ and

*x*<6}

Another acceptable answer is

*B*={

*x*|

*x*∈ℕ and

*x*≤5}.

b) Set

*B*is the set of all elements

*x*such that

*x*is a natural number and

*x*is less than 6.

Example 3:**Roster Form to Set-Builder Notation**

a) Write set *C*={North America, South America, Europe, Asia, Australia. Africa. Antarctica} in set-builder notation.

b) Write in words how you would read set *C* in set-builder notation.

Solution:

a) *C*={*x*|*x* is a continent}.

b) Set *C* is the set of all elements *x* such that *x* is a continent.

Ex4. Write the set *D*={1,4,9,16,25,…}in set—builder form.

✍ Solution:We may write the set *D* as *D*={*x*:*x* is the square of a natural number} Alternatively, we can write *D*={*x*:*x*=*n*^{2}, where *n*∈ℕ}

Ex5. Write the set {½,⅔,¾,⅘,⅚,^{6}⁄_{7}} in set—builder form.

✍ Solution:We see that each member in the given set has the numerator one less than the denominator. Also, the numerator begin from 1 and do not exceed 6. Hence, in the set—builder form the given set is

{*x*:*x* =^{n}/_{(n+1)}, where *n* is a natural number and 1≤*n*≤6}

Ex6. Write set *E*={3,6,9,12,15} in set-builder form.

Sol. *E*={*x*:*x* is a natural number multiple of 3 and *x*<18}.
Ex7. Write set *F*={1,4,9,…,100} in set—builder form.

Sol. *F*={*x*:*x* =*n*^{2}, *n*∈ℕ and *n*<11}
Ex8. Write set *G*={½,⅖,^{3}⁄_{10},^{4}⁄_{17},^{5}⁄_{26},^{6}⁄_{37},^{7}⁄_{50}} in set-builder form.

Sol. *G*={*x*:*x* =n/(*n*^{2}+1), *n*∈ℕ and *n*≤7}

💪 How to Insert the appropriate Symbol either an Element or not an element to a Set?

Ex9. Here are some further illustrations of set-builder notation.

1. {*n*:*n* is a prime number}={2,3,5,7,11,13,17,…}

2. {*n*∈ℕ:*n* is prime}={2,3,5,7,11,13,17,…}

3. {*n*^{2}:*n*∈ℤ}={0,1,4,9,16,25,…}

4. {*x*∈ℝ:*x*^{2}-2=0}={√2,- √2}

5. {*x*∈ℤ:*x*^{2}-2=0}=Ø

6. {*x*∈ℤ:|*x*|<4}={-3,-2,-1,0,1,2,3}
7. {2*x*∶*x*∈ℤ,|*x*|<4}={-6,-4,-2,0,2,4,6}
8. {*x*∈ℤ:|2*x*|<4}={-1,0,1}

These last three examples highlight a conflict of notation that we must always be alert to. The expression |*X*| means **absolute value** if *X* is a number and **cardinality** if *X* is a set. The distinction should always be clear from context. Consider {*x*∈ℤ:|*x*|<4} in Ex9 (6) above. Here *x*∈ℤ, so *x* is a number (not a set), and thus the bars in |*x*| must mean absolute value, not cardinality. On the other hand, suppose *A*={{1,2},{3,4,5,6},{7}} and *B*={*X*∈A:|*X*|<3}. The elements of *A* are sets (not numbers), so the |*X*| in the expression for *B* must mean cardinality. Therefore *B*={{1,2},{7}}.