# Prove that a preference is linear

Given the following two conditions:

$$x\succ y$$ implies $$x+a\succsim y+a$$,

And,

$$x\prec y$$ implies $$x+a\precsim y+a$$

We want to prove that $$\succsim$$ is a linear preference.

One of the definition of linear preference is that: $$x\succsim y \Leftrightarrow x+a\succsim y+a$$

So I am trying to do this:

Since $$x\succsim y$$ means that $$x\succ y$$ or $$x\sim y$$

We already know that $$x\succ y$$ implies $$x+a\succsim y+a$$,

all things left is to prove that $$x\sim y$$ also implies that $$x+a\succsim y+a$$.

• The two conditions you provide are really only one condition (by exchanging $x$ and $y$ in the first you get the second). And what is "left to prove" according to your partial answer is actually true by definition, so that's not what is left to prove. Mar 21 '20 at 15:34
• @VARulle I am truly sorry that I made a mistake in my partial "proof". We must show that $x\sim y$ implies $x+a\succsim y+a$ Mar 22 '20 at 1:02

Define $$u(x)=\min{\{x,0\}}$$. Let $$\succsim$$ be the preference relation represented by $$u$$. This preference relation satisfies $$x\succ y \Longrightarrow x+a\succsim y+a$$ for all $$x,y,a\in\mathbb R$$. But let $$x=0$$, $$y=1$$, and $$a=-1$$. Then $$x\sim y$$, but $$y+a=0\succ -1=x+a$$, thus $$x+a\not\succsim y+a$$ and $$\succsim$$ is not linear.