<p>As you can tell from the comments, there are multiple versions of monotonicity in preference theory, but let us assume the (in my opinion) most intuitive one for now, and say</p>
<p>A preference <span class="math-container">$\succeq$</span> represented by utility function <span class="math-container">$u(\cdot)$</span> is <strong>strictly monotonic</strong> if whenever <span class="math-container">$x &gt; y$</span>, <span class="math-container">$u(x) &gt; u(y)$</span></p>
<p>Let us consider the most degenerate satiated preference over <span class="math-container">$\mathbb{R}$</span>,
<span class="math-container">$$
 u(x) = 5 \quad \text{ for any }  x 
$$</span>
(why 5? because why not).</p>
<p>Consider <span class="math-container">$348050348509485$</span> and <span class="math-container">$0$</span>.</p>
<p>It is quite clear that <span class="math-container">$348050348509485 &gt; 0$</span>.</p>
<p>However, <span class="math-container">$u(348050348509485) = 5 = u(0)$</span>.</p>
<p>This preference fails the strict monotoncitiy assumption.</p>

CC BY-SA 4.0
2021-10-22T21:56:04.133Z
2021-10-22T21:56:04.133Z
Walrasian Auctioneer
<p>Satiation means there is a consumption bundle that is at least as good as any other consumption bundle. Monotonicity means that larger bundles are always better. Since there is always a larger bundle, no bundle can be at least as good as any other consumption bundle when preferences are monotone. So monotone preferences are never satiated.</p>

CC BY-SA 4.0
2021-10-23T08:19:12.120Z
2021-10-23T08:19:12.120Z
Michael Greinecker
<p>Here's one way to think about it:</p>
<p>Assume your preferences are differentiable, U = f(x).
If your preferences are monotonically increasing, meaning:</p>
<ul>
<li><span class="math-container">$U_1 = f(x)$</span></li>
<li><span class="math-container">$U_2 = f(x+e)$</span></li>
<li><span class="math-container">$U_2 &gt; U_1; e &gt; 0$</span></li>
</ul>
<p>Then the first derivative is always positive, since a derivative is:
<span class="math-container">$\lim\limits_{e \to 0} \frac{f(x+e)-f(x)}{e}$</span></p>
<p>If a first derivative is positive, the agent prefers more.
Since the this is true everywhere, the agent always prefers more, they are <em>insatiable</em>.</p>

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2021-10-22T16:30:13.007Z
2021-10-23T18:25:00.733Z
2021-10-23T18:25:00.733Z
RegressForward
<p>From looking at the graphical illustrations, I understand that satiated preferences violate the monotonicity assumption. I was wondering if there is algebra which can be used to show this?</p>

CC BY-SA 4.0
2021-10-21T23:42:10.440Z
There are several monotonicity assumptions. Which one do you have in mind?
Michael Greinecker
CC BY-SA 4.0
2021-10-22T05:34:52.553Z
Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
Community
CC BY-SA 4.0
2021-10-22T20:47:40.353Z
@MichaelGreinecker I am not sure sorry I have just learnt about this topic and know that we assume more is always preferred to less. But this isn't the case for satiated preferences and I'm wondering if there is a numerical example that can show this?
Sofia
<microeconomics><preferences>
How to explain satiated preferences with algebra?
Sofia
# How to explain satiated preferences with algebra?

(Asked by: username_0 on 2021-10-21)

From looking at the graphical illustrations, I understand that satiated preferences violate the monotonicity assumption. I was wondering if there is algebra which can be used to show this?

**Tags:** microeconomics, preferences
> **Comments:**
> **username_1**: There are several monotonicity assumptions. Which one do you have in mind?  
> **username_2**: Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.  
> **username_0**: @MichaelGreinecker I am not sure sorry I have just learnt about this topic and know that we assume more is always preferred to less. But this isn't the case for satiated preferences and I'm wondering if there is a numerical example that can show this?  

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## 3 answers

### Answer by username_3 on 2021-10-22. Score: 3

As you can tell from the comments, there are multiple versions of monotonicity in preference theory, but let us assume the (in my opinion) most intuitive one for now, and say

A preference $\succeq$ represented by utility function $u(\cdot)$ is **strictly monotonic** if whenever $x > y$, $u(x) > u(y)$

Let us consider the most degenerate satiated preference over $\mathbb{R}$,
$$
u(x) = 5 \quad \text{ for any } x
$$
(why 5? because why not).

Consider $348050348509485$ and $0$.

It is quite clear that $348050348509485 > 0$.

However, $u(348050348509485) = 5 = u(0)$.

This preference fails the strict monotoncitiy assumption.


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### Answer by username_1 on 2021-10-23. Score: 2

Satiation means there is a consumption bundle that is at least as good as any other consumption bundle. Monotonicity means that larger bundles are always better. Since there is always a larger bundle, no bundle can be at least as good as any other consumption bundle when preferences are monotone. So monotone preferences are never satiated.


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### Answer by username_4 on 2021-10-22. Score: 2

Here's one way to think about it:

Assume your preferences are differentiable, U = f(x).
If your preferences are monotonically increasing, meaning:

* $U\_1 = f(x)$
* $U\_2 = f(x+e)$
* $U\_2 > U\_1; e > 0$

Then the first derivative is always positive, since a derivative is:
$\lim\limits\_{e \to 0} \frac{f(x+e)-f(x)}{e}$

If a first derivative is positive, the agent prefers more.
Since the this is true everywhere, the agent always prefers more, they are *insatiable*.


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