Rational preferences/individual decision-making theory

Question

I am taking advanced micro course this semester. In one of the problems we need to determine whether the preference relation is rational (i.e. complete and transitive). Since we have not really discussed the set of two-dimensional real vectors $X=R^2$ (non-negative), I am wondering whether these preference relations are indeed rational, and if yes, how can it be shown

(i) The relation ≽ defined by $(x_1,x_2)$ ≽ $(y_1,y_2)$ if and only if $x_1≥y_1$ and $x_2≥y_2$
(ii) The relation ≽ defined by $(x_1,x_2)$ ≽ $(y_1,y_2)$ if and only if $x_1≥y_1$
(iii) The relation ≽ defined by $(x_1,x_2)$ ≽ $(y_1,y_2)$ if and only if $\min\{x_1,x_2\} ≥ \min\{y_1,y_2\}$
(iv) The relation ≽ defined by $(x_1,x_2)$ ≽ $(y_1,y_2)$ if and only if $x_1>y_1$ or $x_1=y_1$ and $x_2≥y_2$

Answer 1

(i) Is not complete. For instance, (10,5) is not $\succeq$ (9,6), because $10>9$, but $5<6$. However, (9,6) is also not $\succeq$ (10,5) for the same reason. Hence, there exists a pair of bundles $A,B$ such that neither $A\succeq B$ nor $B \succeq A$. Thus, it is not rational.

(ii) and (iii) are both rational. You can either see this by showing transitivity and completeness directly or you exploit a central result in microeconomics: A preference $\succeq$ can be represented by a utility function $u$ only if it is rational. For (ii), you can define $u(x_1,x_2)=x_1$ such that $x_2$ doesn't matter for comparisons as the preference states. For (iii), you can define Leontief preferences, $u(x_1,x_2)=\min(x_1,x_2)$.

(iv) is a funny case, because it shows why the result we exploited above is just "only if" and not "if and only if." Lexicographic preferences are rational, but they are not representable by a standard utility function.

For any two bundles either the first dimension differs, or the first one is equal and the second one differs or both are equal. So for any two bundles we have either $\succeq$ or $\preceq$ or both, i.e., the preferences are complete. They are also transitive because $X\succeq Y$ implies that either $x_1>y_1$ or $x_1=y_1$ and $x_2\geq y_2$ and $Y\succeq Z$ implies that either $y_1>z_1$ or $y_1=z_1$ and $y_2\geq z_2$. Therefore, either $x_1>z_1$ or $x_1=z_1$ and $x_2\geq z_2$, implying $X\succeq Z$.