Question

Show that the relation R defined by (a, b) R(c, d) if and only if a+d=b+c on set N*N is an equivance relation. ​

Answer 1

Not sure but maybe I am right just check




Let (a,b)=x, (b,c)=y and (c,d)=z; Now to prove that R on A is equivalence. We'll have prove that it is also reflexive(xRx),

symmetric[(xRy)=(yRx)] and transitive[(xRy) and (yRz) implies (xRz)]. And the condition is: (a,b)R(c,d) implies a+d=b+c

xRx=(a,b)R(a,b) implies a+b=b+a which is true;
therefore it is reflexive


xRy=(a,b)R(b,c) implies a+c=b+b….1 & yRx=(b,c)R(a,b) implies b+b=c+a…2 here 1 and 2 are equal therefore R is symmetric.

xRy=a+c=b+b…1 & yRz=(b,c)R(c,d) implies b+d=2c…..2 implies that xRz=(a,b)R(c,d) implies a+d=b+c……3 Now Adding 1 and 2 we get: a+c+b+d=2b+2c implies a+d=b+c which is to be proved for transitivity

Answer 2

Given:

Relation R defined by (a, b) R(c, d) by a+d=b+c on set N*N.

To Prove:

Relation R is an equivalence relation.

Solution:

For a relation R to be an equivalence relation, it should be reflexive, symmetric and transitive.

1. Reflexive :

• An object x , is reflexive, if xRx.
• Here x = (a,b)
• (a,b)R(a,b) => a + b = b + a
• R is reflexive.

    2. Symmetric :

• Consider 2 object x and y.
• R is symmetric if xRy = yRx
• x = (a,b)
• y = (c,d)
• xRy = a + d = b + c
• yRx = c + b = a + d
• Since both are same, R is symmetric.

    3. Transitive:

• Consider 3 objects x, y and z.
• if xRy, yRz then if xRz, R is transitive.
• x = (a,b)
• y = (c,d)
• z = (k,l ) (say)
• xRy = > a + d = c + b -(1)
• yRz = > c + l = k + d -(2)
• (1) + (2) = > a + c + d + l = c + b + k + d
• a + l = k + b - (3)
• xRz = a + l = k + b
• This is same as equation 3.
• Therefore R is transitive.

Hence R is reflexive, symmetric and transitive. R is an equivalence relation.