<p>You cannot prove this. It is wrong.</p>

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

CC BY-SA 4.0
2020-03-23T16:11:48.380Z
2020-03-24T10:06:02.463Z
2020-03-24T10:06:02.463Z
VARulle
<p>Given the following two conditions:</p>

<p><span class="math-container">$x\succ y$</span> implies <span class="math-container">$x+a\succsim y+a$</span>,</p>

<p>And, </p>

<p><span class="math-container">$x\prec y$</span> implies  <span class="math-container">$x+a\precsim y+a$</span></p>

<p>We want to prove that <span class="math-container">$\succsim$</span> is a linear preference.</p>

<hr>

<p>One of the definition of linear preference is that:
<span class="math-container">$x\succsim y \Leftrightarrow x+a\succsim y+a$</span> </p>

<p>So I am trying to do this:</p>

<p>Since <span class="math-container">$x\succsim y$</span> means that <span class="math-container">$x\succ y$</span> or <span class="math-container">$x\sim y$</span></p>

<p>We already know that <span class="math-container">$x\succ y$</span> implies <span class="math-container">$x+a\succsim y+a$</span>,</p>

<p>all things left is to prove that <span class="math-container">$x\sim y$</span> also implies that <span class="math-container">$x+a\succsim y+a$</span>.</p>

CC BY-SA 4.0
2020-03-21T15:34:03.787Z
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
CC BY-SA 4.0
2020-03-22T01:02:28.097Z
@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
<macroeconomics><preferences><theory>
Prove that a preference is linear
High GPA
# Prove that a preference is linear

(Asked by: username_0 on 2020-03-21)

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 $x+a\succsim y+a$.

**Tags:** macroeconomics, preferences, theory

> **Comments:**
> **username_1**: 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.  
> **username_0**: @username_1 I am truly sorry that I made a mistake in my partial "proof". We must show that $x\sim y$ implies $x+a\succsim y+a$  

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## 1 answers

### Answer by username_1 on 2020-03-23. Score: 2 | Accepted answer

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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