<p>By contradiction, assume $z \succsim x$. However, we already know that $y \succsim z$, which is a contradiction (after a couple more steps). Can you now see this?</p>

CC BY-SA 3.0
2016-09-29T12:02:49.143Z
Hi @Jay, thank you, yes I see how it is proved this way; I was wondering if we could prove it quicker another way which Kitsune Cavalry answered.
Sunhwa
CC BY-SA 3.0
2016-09-29T06:32:18.680Z
2016-10-03T03:46:38.300Z
2016-10-03T03:46:38.300Z
Jay
<blockquote>
  <p>Prove if $\succsim$ is rational then: if $x \succ y \succsim z$, then $x \succ z$</p>
</blockquote>

<p>By definition of $\succ$, </p>

<p>\begin{equation} \tag{1}
x \succ y \iff x \succsim y, \; \neg \; y \succsim x
\end{equation}</p>

<p>Where $\neg$ is the negation symbol.</p>

<p>So we want to show both</p>

<p>$$ x \succsim z \; and \; \neg z \succsim x $$</p>

<p>We are given </p>

<p>\begin{equation} \tag{2}
y \succsim z
\end{equation}</p>

<p>By $(1)$ and $(2)$ and transitivity of $\succsim$ (since $\succsim$ is rational), </p>

<p>\begin{equation} \tag{3}
x \succsim y \succsim z \Rightarrow x \succsim z
\end{equation}</p>

<p>We also need $\neg \; z \succsim x$</p>

<p>Suppose the contrary that $z \succsim x$, but then</p>

<p>$$ z \succsim x \succ y \succsim z $$</p>

<p>Such that </p>

<p>\begin{equation} \tag{4}
z \succ z
\end{equation}</p>

<p>is a contradiction  (as $\succ$ is irreflexive) so it must be that </p>

<p>\begin{equation} \tag{5}
\neg \; z \succsim x
\end{equation}</p>

<p>Applying $(3)$ and $(5)$ we have the desired property</p>

<p>$$ x \succsim z, \; \neg \; z \succsim x \Rightarrow x \succ z $$</p>

<p>My question is if I am allowed to arrive at the statement $(4)$ from the line above it?</p>

CC BY-SA 3.0
2016-09-28T22:12:01.840Z
No. Getting to $(4)$ the way you do requires the sort of "transitivity" on $\succ$ that you are trying to show in the first place.
Kitsune Cavalry
<microeconomics>
Prove if $\succsim$ is rational then: if $x \succ y \succsim z$, then $x \succ z$
Sunhwa
# Prove if $\succsim$ is rational then: if $x \succ y \succsim z$, then $x \succ z$

(Asked by: username_0 on 2016-09-28)

> Prove if $\succsim$ is rational then: if $x \succ y \succsim z$, then $x \succ z$

By definition of $\succ$,

\begin{equation} \tag{1}
x \succ y \iff x \succsim y, \; \neg \; y \succsim x
\end{equation}

Where $\neg$ is the negation symbol.

So we want to show both

$$ x \succsim z \; and \; \neg z \succsim x $$

We are given

\begin{equation} \tag{2}
y \succsim z
\end{equation}

By $(1)$ and $(2)$ and transitivity of $\succsim$ (since $\succsim$ is rational),

\begin{equation} \tag{3}
x \succsim y \succsim z \Rightarrow x \succsim z
\end{equation}

We also need $\neg \; z \succsim x$

Suppose the contrary that $z \succsim x$, but then

$$ z \succsim x \succ y \succsim z $$

Such that

\begin{equation} \tag{4}
z \succ z
\end{equation}

is a contradiction (as $\succ$ is irreflexive) so it must be that

\begin{equation} \tag{5}
\neg \; z \succsim x
\end{equation}

Applying $(3)$ and $(5)$ we have the desired property

$$ x \succsim z, \; \neg \; z \succsim x \Rightarrow x \succ z $$

My question is if I am allowed to arrive at the statement $(4)$ from the line above it?

**Tags:** microeconomics
> **Comments:**
> **username_1**: No. Getting to $(4)$ the way you do requires the sort of "transitivity" on $\succ$ that you are trying to show in the first place.  

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## 1 answers

### Answer by username_2 on 2016-09-29. Score: 2

By contradiction, assume $z \succsim x$. However, we already know that $y \succsim z$, which is a contradiction (after a couple more steps). Can you now see this?

> **Comments:**
> **username_0**: Hi @Jay, thank you, yes I see how it is proved this way; I was wondering if we could prove it quicker another way which Kitsune Cavalry answered.  

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