Given the following two conditions:

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


$x\prec y$ 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 $x+a\succsim y+a$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – VARulle
    Mar 21, 2020 at 15:34
  • $\begingroup$ @VARulle I am truly sorry that I made a mistake in my partial "proof". We must show that $x\sim y$ implies $x+a\succsim y+a$ $\endgroup$
    – High GPA
    Mar 22, 2020 at 1:02

1 Answer 1


You cannot prove this. It is wrong.

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.


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