Relation and Functions

Q1. As R is a relation defined on natural numbers, we have to provide some natural x such that y is also natural.

Natural x such that y is also natural would be even, due to the division by 2. They are -: 2, 4, 6. If we put x = 8, we get y = 0, which is not natural.

Q2. To prove that it is an equivalence relation,

Reflexivity
(a, b) R (a, b) => a + b = a + b. This relation holds (pretty obvious) and R is reflexive.

Symmetry
(a, b) R (b, a) => a + b = b + a. This relation holds due to commutative law of addition. Hence R is symmetric.

Transitivity
We have (a, b) R (c, d) => a + b = c + d.
And (c, d) R (e, f) => c + d = e + f
Then we should have (a, b) R (e, f) as a valid relation.
(a, b) R (e, f) => a + b = e + f must be proved.

Now as a + b = c + d and c + d = e + f, a + b = e + f.
Hence (a, b) R (e, f) holds. Hence R is transitive.
So R is an equivalence relation.

Q4. [x] > k implies [x] ≥ k + 1 where k is an integer.
Let x be 3.6 and k be 1.
Obviously [3.6] > 1.
This also implies [3.6] ≥ 2 which is true as 3 > 2.
Try k = 2.
[3.6] > 2
=> [3.6] ≥ 3 which is true as 3 = 3.

Now let's try negative integers.
Let x = -2.2 and k be -4.
[-2.2] > -4
=> [-2.2] ≥ -5 which is true.
And so on..

about the first part y=(8-x)/2................ so the function is from real to natural................... so 8-x/2 should be an integer... so domain is
(2,4,6). am i right...........................................

Thanks Pritish and Souradeep [1]