<p>You have to show that something does not hold universally true. To show this, you just have to show that there is at least one exception- a counterexample. For this counterexample, you can make any assumption that does not contradict the assumptions of your problem.</p>

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2021-10-16T23:00:14.943Z
In general can you assume that more is better, as in more quantity brings you more utility?
Rumi
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2021-10-16T23:55:55.483Z
Yes, and you can even find one explicit utility function and one explicit increasing function $f$ that gives you a counterexample.
Michael Greinecker
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2021-10-16T17:42:29.473Z
2021-10-16T17:42:29.473Z
Michael Greinecker
<p>I am required to show that if: <span class="math-container">$f:R^L \rightarrow R$</span> is a strictly monotonic function and <span class="math-container">$u:R^L \rightarrow R$</span> is a utility function representing a preference relation <span class="math-container">$\succsim$</span>, then the function <span class="math-container">$v:R^L \rightarrow R$</span> defined by <span class="math-container">$v(x) = u(f(x))$</span> does not always represent the preference <span class="math-container">$\succsim$</span>.</p>
<p>For such a proposition, can I assume that if <span class="math-container">$x&gt;y$</span>, then <span class="math-container">$u(x) &gt; u(y)$</span>. The context is that I am trying to prove this by contradiction and my contradiction hinges on <span class="math-container">$x &gt;y \implies u(x) &gt; u(y)$</span>, where x,y are elements of <span class="math-container">$R^L$</span>.</p>

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2021-10-17T04:55:19.750Z
If the domain of $u$ is $R^L$, and the codomain of $f$ is $R$, then $u(f(x))$ doesn't make sense. Also, if $x$ and $y$ are in $R^L$, then $x>y$ doesn't make sense unless you've defined an ordering on $R^L$.
Acccumulation
<microeconomics><proof>
Can I assume utility functions strictly increasing?
Kinno
# Can I assume utility functions strictly increasing?

(Asked by: username_0 on 2021-10-16)

I am required to show that if: $f:R^L \rightarrow R$ is a strictly monotonic function and $u:R^L \rightarrow R$ is a utility function representing a preference relation $\succsim$, then the function $v:R^L \rightarrow R$ defined by $v(x) = u(f(x))$ does not always represent the preference $\succsim$.

For such a proposition, can I assume that if $x>y$, then $u(x) > u(y)$. The context is that I am trying to prove this by contradiction and my contradiction hinges on $x >y \implies u(x) > u(y)$, where x,y are elements of $R^L$.

**Tags:** microeconomics, proof

> **Comments:**
> **username_1**: If the domain of $u$ is $R^L$, and the codomain of $f$ is $R$, then $u(f(x))$ doesn't make sense. Also, if $x$ and $y$ are in $R^L$, then $x>y$ doesn't make sense unless you've defined an ordering on $R^L$.  

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## 1 answers

### Answer by username_2 on 2021-10-16. Score: 4 | Accepted answer

You have to show that something does not hold universally true. To show this, you just have to show that there is at least one exception- a counterexample. For this counterexample, you can make any assumption that does not contradict the assumptions of your problem.

> **Comments:**
> **username_3**: In general can you assume that more is better, as in more quantity brings you more utility?  
> **username_2**: Yes, and you can even find one explicit utility function and one explicit increasing function $f$ that gives you a counterexample.  

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