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Question: Let \[N\] denote the set of all the natural numbers. Define two binary relations on \(N\) as \({R...

Let NN denote the set of all the natural numbers. Define two binary relations on NN as
R1=(x,y)∈N×N:2x+y=10{R_1} = \\{ (x,y) \in N \times N:2x + y = 10\\} and R2=(x,y)∈N×N:x+2y=10{R_2} = \\{ (x,y) \in N \times N:x + 2y = 10\\}
Then,
A. both R1,R2{R_1},{R_2} are transitive relations.
B. both R1,R2{R_1},{R_2} are symmetric relations.
C. Range of R2{R_2} is 1,2,3,4\\{ 1,2,3,4\\}
D. Range of R1{R_1} is 2,4,8\\{ 2,4,8\\}

Explanation

Solution

In this question, we are given two binary relations on NN which are R1,R2{R_1},{R_2}. Now firstly we find all the (x,y)∈N×N(x,y) \in N \times N satisfying R1{R_1} and then for R2{R_2}. And then, we will solve according to the given options.

Complete step-by-step answer:
In the above question, we are given that NN is the set of all the natural numbers andR1,R2{R_1},{R_2} are the two binary operations onNN.
R1=(x,y)∈N×N:2x+y=10{R_1} = \\{ (x,y) \in N \times N:2x + y = 10\\} −−−−−(1) - - - - - (1)
R2=(x,y)∈N×N:x+2y=10{R_2} = \\{ (x,y) \in N \times N:x + 2y = 10\\} −−−−(2) - - - - (2)
Now firstly we find all the (x,y)∈N×N(x,y) \in N \times N satisfying R1{R_1} and then for R2{R_2}.
We will consider the relation R1{R_1}.
We know that
R1=(x,y)∈N×N:2x+y=10{R_1} = \\{ (x,y) \in N \times N:2x + y = 10\\}
2x+y=102x + y = 10, (x,y)∈N(x,y) \in N
x=10−y2x = \dfrac{{10 - y}}{2}
So as x∈N⇒10−y2∈Nx \in N \Rightarrow \dfrac{{10 - y}}{2} \in N
Now if 10−y2∈N\dfrac{{10 - y}}{2} \in N, y∈Ny \in N
Then 10−y,y∈N10 - y,y \in N belongs to even number.
So we can say that the possible values ofy∈Ny \in N for x∈Nx \in N will be
y=2,4,6,8y = \\{ 2,4,6,8\\}
So we can say the range of theR1=2,4,6,8{R_1} = \\{ 2,4,6,8\\}
Now considering the relation R2{R_2} on NN
R2=(x,y)∈N×N:x+2y=10{R_2} = \\{ (x,y) \in N \times N:x + 2y = 10\\}
x+2y=10x + 2y = 10, (x,y)∈N(x,y) \in N
x=10−2yx = 10 - 2y
Now for x∈Nx \in N
10−2y∈N10 - 2y \in N
Now if 10−2y∈N10 - 2y \in N is the natural number, then
y∈1,2,3,4y \in \\{ 1,2,3,4\\}
So we can say that the range of R2=1,2,3,4{R_2} = \\{ 1,2,3,4\\}

Now from options given, we can say that option C is correct.

Note: In the above question, when we considered the relation R1{R_1}, we have taken xx in the terms of yy. We can also find yy in the term of xx.
2x+y=102x + y = 10
y=10−2xy = 10 - 2x
From this also, we can find the possible values of x,yx,y. Then we will use the value of xx to find the range of R1{R_1} which is y.y.