     1. Chapter 2 Class 11 Relations and Functions
2. Serial order wise
3. Examples

Transcript

Example 19, Let R be a relation from Q to Q defined by R = {(a, b): a, b Q and a b Z}. Show that (i) (a, a) R for all a Q Given R = {(a, b): a, b Q and a b Z} Hence we can say that (a, b) is in relation R if a, b Q i.e. both a & b are in set Q a b Z i.e. difference of a & b is an integer We need to prove both these conditions for (a,a) a, a Q , i.e. a is in set Q Also, a a = 0 & 0 is an integer, a a Z Since both conditions are satisfied, (a, a) R Example 19, Let R be a relation from Q to Q defined by R = {(a,b): a,b Q and a b Z}. Show that (ii) (a,b) R implies that (b, a) R Given R = {(a, b): a, b Q and a b Z} Given (a, b) R , i.e. (a, b) is in relation R . So, the following conditions are true a, b Q i.e. both a & b are in set Q a b Z i.e. difference of a & b is an integer We need to prove both these conditions for (b,a) 1. a, b Q,hence b,a Q 2. Given (a b) is an integer, So, negative of integer is also an integer, i.e. (a b) is also integer a + b is integer, b a is integer b a Z Since both conditions are satisfied, (b, a) R Example 19, Let R be a relation from Q to Q defined by R = {(a,b): a,b Q and a b Z}. Show that (iii) (a, b) R and (b, c) R implies that (a, c) R Given R = {(a, b): a, b Q and a b Z} We need to prove both these conditions for (a, c) 1. Given a, b & b, c Q, hence a, c Q 2. (a b) is an integer, So, negative of integer is also an integer, i.e. (a-b) is also integer -a+b is integer, b-a is integer b a Z Since both conditions are satisfied, (b, a) R 2. Given (a b) & (b c) is an integer, Sum of integers is also an integer So, (a b) + (b c) is also integer a b + b c is integer, a c is integer a c Z Since both conditions are satisfied, (a, c) R

Examples 