<p>Any preference relation represented by an utility function is transtive.</p>

<p>Suppose <span class="math-container">$x \succsim y$</span> and  <span class="math-container">$y \succsim z$</span> and <span class="math-container">$\succsim$</span> is represented by <span class="math-container">$U$</span>. Then <span class="math-container">$U(x) \geq U(y)  $</span> and <span class="math-container">$U(y) \geq U(z)$</span>, so <span class="math-container">$U(x) \geq U(z)$</span>  and then <span class="math-container">$x \succsim z$</span>.</p>

CC BY-SA 4.0
2022-12-26T04:44:10.913Z
This is just a consequence of the transitivity in the real numbers to which the utility function maps, right?
Dave
CC BY-SA 4.0
2018-10-10T01:54:18.033Z
2018-10-10T01:54:18.033Z
Nícolas Pinto
<p>That this utility function represents monotonic preferences, I think it's clear. Both goods have positive and constant marginal utilities. What I think is less clear is if this preference relation is transitive. How can I assess it?</p>

CC BY-SA 4.0
2018-10-04T21:06:22.390Z
How about applying the definitions of "utility function represents preferences" and of "transitive preferences"?
Giskard
CC BY-SA 4.0
2018-10-04T21:41:35.820Z
A preference relation is transitive if $A \succ B, B \succ C \implies A \succ C$. So if there's $p,q, r \in \mathbb{R}^2$ and $U(p) > U(q), U(q) > U(r) \implies U(p) > U(r)$ then $U()$ represents a transitive preference relation. Hmm... Yeah, positive and constant derivatives should do the trick to guarantee it. Thanks, man.
Pedro Cavalcante
CC BY-SA 4.0
2018-10-05T01:56:56.667Z
$U(x,y) = (x+y)^2$ is a monotonic transformation of $U^{'}(x,y) = x+y$.(the transformation is $f(z) = z^2$)
superhulk
CC BY-SA 4.0
2018-10-05T04:21:13.650Z
@PedroCavalcanteOliveira Why do you need anything about derivaties? Why is
$$
U(A) > U(B) \text{ and } U(B) > U(C) \Rightarrow U(A) > U(C)
$$
not enough?
Giskard
<microeconomics><preferences>
Does $U(x,y) = x^2 + y^2 + 2xy$ represent transitive, monotonic preferences?
Pedro Cavalcante
# Does $U(x,y) = x^2 + y^2 + 2xy$ represent transitive, monotonic preferences?

(Asked by: username_0 on 2018-10-04)

That this utility function represents monotonic preferences, I think it's clear. Both goods have positive and constant marginal utilities. What I think is less clear is if this preference relation is transitive. How can I assess it?

**Tags:** microeconomics, preferences

> **Comments:**
> **username_1**: How about applying the definitions of "utility function represents preferences" and of "transitive preferences"?  
> **username_0**: A preference relation is transitive if $A \succ B, B \succ C \implies A \succ C$. So if there's $p,q, r \in \mathbb{R}^2$ and $U(p) > U(q), U(q) > U(r) \implies U(p) > U(r)$ then $U()$ represents a transitive preference relation. Hmm... Yeah, positive and constant derivatives should do the trick to guarantee it. Thanks, man.  
> **username_2**: $U(x,y) = (x+y)^2$ is a monotonic transformation of $U^{'}(x,y) = x+y$.(the transformation is $f(z) = z^2$)  
> **username_1**: @PedroCavalcanteOliveira Why do you need anything about derivaties? Why is $$ U(A) > U(B) \text{ and } U(B) > U(C) \Rightarrow U(A) > U(C) $$ not enough?  

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## 1 answers

### Answer by username_3 on 2018-10-10. Score: 1

Any preference relation represented by an utility function is transtive.

Suppose $x \succsim y$ and $y \succsim z$ and $\succsim$ is represented by $U$. Then $U(x) \geq U(y) $ and $U(y) \geq U(z)$, so $U(x) \geq U(z)$ and then $x \succsim z$.

> **Comments:**
> **username_4**: This is just a consequence of the transitivity in the real numbers to which the utility function maps, right?  

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