# If all indifference curves are parallel lines, then preference has linear representation

Given a continuous preference relation $\succeq$ over $X=\mathbb{R}^2_{+}$ where all sets: $$I_x\equiv\{y\in X:y\sim x\}$$ are lines on $X,\forall x\in X$, and are parallel to $I_y,\forall y\notin I_x$.

How can I show that $\succeq$ has a linear representation?

It seems intuitive that if every indifference curve is a line, then the utility itself must be a line, but I'm not sure how to go back to it. Would someone be willing to help me?

Thanks! Any helpful tips are appreciated! :D

The indifference curves are constructed by viewing the utility function as an equation (for a fixed utility index value per curve). So from

$$U = U(x_1,x_2)$$

where the left side is just a symbol, we move to

$$\bar U = U(x_1,x_2)$$ where now the left side is a specific number.

Take the total differential on both sides to obtain

$$0 = U_1dx_1 + U_2dx_2 \implies \frac {dx_2}{dx_1} = -\frac {U_1}{U_2}$$

Any straight line in the two-dimensional plane has a constant slope so also

$$\frac {dx_2}{dx_1} =c$$

Same for $y$-bundle. Etc.

• Thanks for your answer Alecos! I'm still trying to understand it perfectly before accepting it. I'm not sure what would happen if $U$ wasn't differentiable. Thanks for helping! :D – Guilherme Salomé Sep 2 '15 at 0:50
• @GuilhermeSalomé Good point. Think it over. – Alecos Papadopoulos Sep 2 '15 at 2:02