# Let S be the set of all real numbers. Show that the relation R = {(a, b) : a2 + b2 = 1} is symmetric but neither reflexive nor transitive.

Given that, a, b S, R = {(a, b) : a2 + b2 = 1 }

Now,

R is Reflexive if (a,a) R a S

For any a S, we have

a2+a2 = 2 a2 ≠ 1

(a,a) R

Thus, R is not reflexive.

R is Symmetric if (a,b) R (b,a) R a,b S

(a,b) R

a2 + b2 = 1

b2 + a2 = 1

(b,a) R

Thus, R is symmetric .

R is Transitive if (a,b) R and (b,c) R (a,c) R a,b,c S

Let (a,b) R and (b,c) R a, b,c S

a2 + b2 = 1 and b2 + c2 = 1

a2+ c2+2b2 = 2

a2+ c2= 2-2b2 ≠ 1

(a, c) R

Thus, R is not transitive.

Thus, R is symmetric but neither reflexive nor transitive.

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