Setting: We have two choices of goods $(x_1,y_1)$ and $(x_2,y_2)$ from the set of choices $[-1,1]^2$. Moreover, we have the following preference relation $$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff |x_1|\geq|x_2|\>\>\text{or}\>\> |y_1|\geq|y_2|$$

Question: We have to check if there exists a utility function reprensation of this preference relation.

My attempt: So from what I have learned, we know that a preference relation admits a utility function representation if it is rational (reflexive, complete, transitive) and continuous. I have found that this preference relation is not transitive, but this does not mean that there does not exist a utility function representation, because the aforementioned statement is not an if and only if statement.

Moreover, I thought we could try to derive a contradiction from the fact that if there exists a utility function $u$ representation of the preference relation, then we have $$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff u(x_1,y_1)\geq u(x_2,y_2)$$ I tried to use the fact that the relation is not transitive to derive a contradiction by using the statement above, but was unsuccessful.

Sadly, these are the two main theorems/propositions that I've learned to solve these problems.

Any help is appreciated!


The existence of a utility function $u$ implies transitivity. Let $A$, $B$ and $C$ be objects (pairs, in your example) for which $ARB$ and $BRC$. Then $$ u(A) \ge u(B) \text{ and } u(B) \ge u(C) $$ so $$ u(A) \ge u(C) $$ so $ARC$.

Since you have an instance that contradicts transitivity there can be no utility function.

  • $\begingroup$ That's actually exactly what I came up with also, but I wasn't sure if it was a complete proof. Thanks alot! $\endgroup$ – Charlie Shuffler Jul 2 at 13:54

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