Let R be a relation on N × ℕ defined by (a, b) R (c, d) if and only if a d(b-c)=b c(a-d). Then R is

Solution

Maths sir · 4 steps

Divide both sides by abcd and the messy condition becomes a clean statement about reciprocals.

1. Rewrite the condition. All four numbers are natural, so none is zero and you can divide ad(b-c)=bc(a-d) by abcd. This gives 1/c-1/b=1/d-1/a. Write g(x,y)=1/x-1/y for each pair.

   (a,b) R (c,d) ⇔ g(a,b) = - g(c,d)

2. Reflexive and symmetric. For (a,b) R (a,b) you need g(a,b)=-g(a,b), so g(a,b)=0, which means a=b. Then (1,2) is not related to itself, so R is not reflexive. Swapping the pairs turns g(a,b)=-g(c,d) into g(c,d)=-g(a,b), the same equation, so R is symmetric.
3. Transitive. From g(a,b)=-g(c,d) and g(c,d)=-g(e,f), the two minus signs cancel and give g(a,b)=+g(e,f), not the -g(e,f) you need. Example: (1,2) R (2,1) and (2,1) R (1,2), since both sides are 0, but (1,2) R (1,2) fails, since 2 ≠ -2. So R is not transitive.
4. So. R is symmetric but neither reflexive nor transitive, which is option (d).

The idea to keep. When a relation looks messy, rearrange it into the form f(first pair) = ± f(second pair). A minus sign means symmetric, but not reflexive and not transitive.

Answer from the NTA final answer key. Solution by PrepWiser.