UPSC NDA I and II 2020 Math Solved Paper

Concept:The logarithmic function with base $a = \frac{1}{2}$ (between $0$ and $1$) is strictly decreasing.Thus $\log_{1/2} x > \log_{1/2} y$ implies $x < y$.So the given relation reduces to $xRy \text{ iff } x < y$.Explanation:Given $a = \frac{1}{2}$, we have $0 < a < 1$.For such a base, the function $\log_a x$ is decreasing, meaning a larger input gives a smaller output.Therefore, $\log_a x > \log_a y$ directly implies $x < y$ (and vice versa).Now check the properties of the relation $xRy \text{ iff } x < y$ on $S = \{1,2,3,\dots\}$.Reflexive: A relation is reflexive if $xRx$ holds for all $x \in S$.Here $x < x$ is false — so not reflexive.Symmetric: If $x < y$, does $y < x$ follow? No — so not symmetric.Transitive: If $x < y$ and $y < z$, then $x < z$ always holds — so the relation is transitive.Hence only transitivity is satisfied.Answer:Transitive only.