Solution:

The relation is defined as `xRy` , iff

` 3 x + 4 y = 5`

If we take, `(x, y) = ( 1, 1/2)`

and `(2/3 , 3/4 )` then these pairs are

satisfied by the given relation.

`1 R 1/2 ` <=> `3 · 1 + 4 · 1/2 = 5`

and `2/3 R 3/4 ` <=> ` 2/3 . 3 + 4 . 3/4 = 5`

But `0 R 1 notin R` as

`0 R 1 ` <=> `0 xx 1 + 4 xx 1 = 4 != 5`

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Solution:

The number of ordered paris in the

equivalence class of `(3, 2)` is the number

of ordered pairs ` (a, b)` satisfying

`(a,b)` * `(3,2)` i.e. `2a = 3b => a/b = 3/2`

Clearly, such ordered pairs are

`(3, 2), (6, 4), (9, 6), (12, 8), (15, 10)` and

`(18, 12)`

`:.` Number of ordered pairs `= 6`

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Solution:

We have, `R_2 = {(x, y)} x` and `y` are integers and `x^2 + y^2= 64}`

Since, `64` is the sum of squares of `0` and `pm 8`.

When `x = 0`, then `y^2 = 64 => y = pm 8`

`x = 8`, then `y^2 = 64 - 8^2 => 64 - 64 = 0`

`x = - 8`, then `y^2 = 64 - ( -8)^2 = 64 - 64 = 0`

`:. R_2 = {(0, 8), (0,- 8), (8, 0), (- 8, 0)}`

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Solution:

`A = {-1,2,3}` and `B = {1,3}`

(i) `A xx B = {(-1, 1), (-1 , 3), (2, 1), (2, 3), (3, 1), (3, 3)}`

(ii) `B xx A = {(1, -1), (1, 2), (1, 3), (3, -1), (3, 2), (3, 3)}`

(iii) `B xx B = {(1, 1), (1, 3), (3, 1), (3, 3)}`

(iv) `A xx A = {(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)}`

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Solution:

It is obvious that the relation `R` is "the difference between `x` and `y` is `2"`.

In roster form, `R = {(5, 3), (6, 4), (7, 5)}`.

In set builder form, `R = {(x,y) : x - y = 2, x in P`,

`y in Q}`.

The domain of this relation is `{5, 6, 7}`.

The range of this relation is `{ 3, 4, 5}`.

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Solution:

`x ϵ R => x-x+sqrt5 = sqrt5` is irrational number


`therefore (x , x) ϵ R`

Hence `R ` is reflexive

`(sqrt5 , 1) ϵ R ` because `sqrt5-1 +sqrt5 = 2sqrt5-1`

which is an irrational number

`therefore (1 , sqrt5) ϵ R`

Hence R is not symmetric

we have `(sqrt5 ,1) , ( 1 ,2sqrt5) ∈ R` because

`sqrt5-1+sqrt5 = 2sqrt5-1` if `1-2sqrt5+sqrt5 = 1-sqrt5` are irrational
`therefore ( sqrt5 ,2sqrt5) ∈ R`

Hence R is not transitive

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Solution:

Let the given relation defined as
`R = {(a, b) | sin^2 a+ cos^2 b = 1}`

For reflexive, `sin^2 a+ cos^2 a= 1` `(because sin^2 theta+cos^2 theta = 1 ∀ theta ϵ R)`


`=> text()_aR_a => (a ,a) ϵ R`

`=> R` is reflexive

For symmetric `sin^2 a+cos^2 b = 1`

`=> 1-cos^2 a+1-sin^2 b = 1`


`=> sin^2 b +cos^2 a = 1`

`=> bRa`

Hence, R is symmetric.
For transitive

let `aRb , bR c`

`=> sin^2 a+cos^2 b = 1` ............(i)

`sin^2 b+ cos^2 c = 1 ` .......(ii)

On adding Eqs. (i) and (ii), we get

`sin^2 a+ (sin^2 b + cos^2 b)+ cos^2 c =2`
`=> sin^2 a+ cos^2 c + 1 =2`
`=> sin^2 a+ cos^2 c = 1`

Hence R is transitive also
Therefore, relation R is an equivalence relation.

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Solution:

If `A` and `B` are equivalence relations then
`A cap B` is an equivalence relation.

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Solution:

`|a-a| = 0 < 1 therefore aRa ∀ a ϵ R`


`therefore` R is reflexive

Again , `a R b => |a-b| le 1 => |b-a| le 1 => bRa`


`therefore` R is symmetric '

Further, `1 R 2` and `2R 3` but `1 R 3` `[because |1-3| = 2 > 1]`


`therefore` R is not transitive.

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Solution:

`(x, y) in R x cap R'`

`=> (x, y) in R` and `(x, y) in R'`

`=> (y, x) in R` and `(y, x) in R'`,


since `R` and `R'` are symmetric.

`=> (y, x) in R cap R'`

`=> R cap R'` is symmetric

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Solution:

We have, `R_3 = { (x, | x |) | x` is real number}

Clearly, domain of `R_3 = R`

Since, image of any real number under `R_3` is positive real number or zero.

`:. ` Range of `R_3 = R^(+) cup {0}` or `(0, oo)`

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