# 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. – VARulle Mar 21 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 $x+a\succsim y+a$ – High GPA Mar 22 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.