Question

Show that the relation '≥' on the set R of all real numbers is reflexive and transitive but not symmetric.

Answer 1

Let R be the set such that R = {(a, b) : a, b$\in R;a\ge b$}

Reflexivity:
$$\begin{gathered} \mathrm{Let}a\mathit{}\mathrm{be}\mathrm{an}\mathrm{arbitrary}\mathrm{element}\mathrm{of}R. \\ \Rightarrow a\in R \\ \Rightarrow a=a \\ \Rightarrow a\ge a\mathrm{is}\mathrm{true}\mathrm{for}a=a \\ \Rightarrow \left(a,a\right)\in R \\ \mathrm{Hence},R\mathrm{is}\mathrm{reflexive}. \end{gathered}$$

Symmetry:
$$\begin{gathered} \mathrm{Let}\left(a,b\right)\in R \\ \Rightarrow a\ge b\mathrm{is}\mathrm{same}\mathrm{as}b\le a,\mathrm{but}\mathrm{not}b\ge a \\ \mathrm{Thus},\left(b,a\right)\notin R \\ \mathrm{Hence},R\mathrm{is}\mathrm{not}\mathrm{symmetric}. \end{gathered}$$

Transitivity:
$$\begin{gathered} \mathrm{Let}\left(a,b\right)\mathrm{and}\left(b,c\right)\in R \\ \Rightarrow a\ge b\mathrm{and}b\ge c \\ \Rightarrow a\ge b\ge c \\ \Rightarrow a\ge c \\ \Rightarrow \left(a,c\right)\in R \\ \mathrm{Hence},R\mathrm{is}\mathrm{transitive}. \end{gathered}$$