<p>Here is a proof by induction on $n$, for any $(n,m) \in \mathbb{N}^2$.</p>

<p>We will need the following preliminary result, which strengthens the assumption on $\succeq$:</p>

<p><strong>Lemma</strong>
\begin{equation*}
\forall x,y,z \in \mathbb{R}, x \succ y \Rightarrow x+z \succ y + z
\end{equation*}</p>

<p><strong>Proof</strong>
Suppose that $x+z \preceq y+z$. By your assumption on $\succeq$, we can add $-z$ to both sides of this inequality (since $-z \in X=\mathbb{R}$) and obtain
\begin{equation*}
x+z+(-z) \preceq y+z+(-z)
\end{equation*}
i.e. $x \preceq y$. This proves the lemma (by contrapositive).</p>

<p>Now, let us prove the main result.</p>

<ol>
<li>Basis ($n=0$)</li>
</ol>

<p>We need to prove that $mx \succeq 0$ for any $m \in \mathbb{N}$. This part can itself be proved by induction on $m$. It is trivial for $m=0$ and $m=1$. If the property holds for $m$, then 
\begin{equation*}
(m+1)x = mx +x \succeq mx + 0 \text{ since } x \succeq 0
\end{equation*}
Therefore $(m+1)x \succeq mx$, and since $mx \succeq 0$ by the induction hypothesis, by transitivity we obtain $(m+1)x \succeq 0$: the result holds for $m+1$. This proves the result for $m \in \mathbb{N}, n=0$.</p>

<ol start="2">
<li>Inductive step</li>
</ol>

<p>Let us fix $m,n \in \mathbb{N}^2$ and suppose that $\dfrac{m}{2^n}x \succeq 0$. Let us show that $\dfrac{m}{2^{n+1}}x \succeq 0$. To this end, we proceed by contradiction and we assume, otherwise, that $\dfrac{m}{2^{n+1}}x \prec 0$. Our lemma yields
\begin{equation*}
\dfrac{m}{2^{n+1}}x + \dfrac{m}{2^{n+1}}x \prec 0 + \dfrac{m}{2^{n+1}}x
\end{equation*}
and therefore (computing each side of the inequality)
\begin{equation*}
\dfrac{m}{2^{n}}x \prec \dfrac{m}{2^{n+1}}x
\end{equation*}
By assumption $\dfrac{m}{2^{n+1}}x \prec 0$, and we just showed that $\dfrac{m}{2^{n}}x \prec \dfrac{m}{2^{n+1}}x$. By transitivity we obtain $\dfrac{m}{2^{n}}x \prec 0$. This is a contradiction, since the result was supposed to hold for $n$.</p>

CC BY-SA 3.0
2015-09-12T11:01:12.680Z
2015-09-13T08:44:59.780Z
2015-09-13T08:44:59.780Z
Oliv
<p>Let $\succeq$ be a rational preference relation over a non-empty set $X\equiv\mathbb{R}$ and suppose it satisfies:
$$
x\succeq y \implies x+z\succeq y+z,\forall z\in X
$$
Then I want to show that if $x\succeq 0$ then, $\forall\lambda=\frac{m}{2^n}$, we have $\lambda x\succeq 0$.</p>

<p>How can I show that?</p>

<p>Thanks for helping! :D</p>

<p>P.S.: The book I'm using is not really clear on the math hypothesis, especially on what sets everything is defined and so on, so I'm assuming $m,n\in\mathbb{N}$.</p>

CC BY-SA 3.0
2015-09-11T20:38:02.270Z
Any extra information would be very welcome. Especially on what the relationship between n and m is? If m/2^n is > 1 or anything?
BB King
<expected-utility>
Finite's expected utility
Guilherme Salome
# Finite's expected utility

(Asked by: username_0 on 2015-09-11)

Let $\succeq$ be a rational preference relation over a non-empty set $X\equiv\mathbb{R}$ and suppose it satisfies:
$$
x\succeq y \implies x+z\succeq y+z,\forall z\in X
$$
Then I want to show that if $x\succeq 0$ then, $\forall\lambda=\frac{m}{2^n}$, we have $\lambda x\succeq 0$.

How can I show that?

Thanks for helping! :D

P.S.: The book I'm using is not really clear on the math hypothesis, especially on what sets everything is defined and so on, so I'm assuming $m,n\in\mathbb{N}$.

**Tags:** expected-utility
> **Comments:**
> **username_1**: Any extra information would be very welcome. Especially on what the relationship between n and m is? If m/2^n is > 1 or anything?  

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## 1 answers

### Answer by username_2 on 2015-09-12. Score: 3

Here is a proof by induction on $n$, for any $(n,m) \in \mathbb{N}^2$.

We will need the following preliminary result, which strengthens the assumption on $\succeq$:

**Lemma**
\begin{equation\*}
\forall x,y,z \in \mathbb{R}, x \succ y \Rightarrow x+z \succ y + z
\end{equation\*}

**Proof**
Suppose that $x+z \preceq y+z$. By your assumption on $\succeq$, we can add $-z$ to both sides of this inequality (since $-z \in X=\mathbb{R}$) and obtain
\begin{equation\*}
x+z+(-z) \preceq y+z+(-z)
\end{equation\*}
i.e. $x \preceq y$. This proves the lemma (by contrapositive).

Now, let us prove the main result.

1. Basis ($n=0$)

We need to prove that $mx \succeq 0$ for any $m \in \mathbb{N}$. This part can itself be proved by induction on $m$. It is trivial for $m=0$ and $m=1$. If the property holds for $m$, then
\begin{equation\*}
(m+1)x = mx +x \succeq mx + 0 \text{ since } x \succeq 0
\end{equation\*}
Therefore $(m+1)x \succeq mx$, and since $mx \succeq 0$ by the induction hypothesis, by transitivity we obtain $(m+1)x \succeq 0$: the result holds for $m+1$. This proves the result for $m \in \mathbb{N}, n=0$.

2. Inductive step

Let us fix $m,n \in \mathbb{N}^2$ and suppose that $\dfrac{m}{2^n}x \succeq 0$. Let us show that $\dfrac{m}{2^{n+1}}x \succeq 0$. To this end, we proceed by contradiction and we assume, otherwise, that $\dfrac{m}{2^{n+1}}x \prec 0$. Our lemma yields
\begin{equation\*}
\dfrac{m}{2^{n+1}}x + \dfrac{m}{2^{n+1}}x \prec 0 + \dfrac{m}{2^{n+1}}x
\end{equation\*}
and therefore (computing each side of the inequality)
\begin{equation\*}
\dfrac{m}{2^{n}}x \prec \dfrac{m}{2^{n+1}}x
\end{equation\*}
By assumption $\dfrac{m}{2^{n+1}}x \prec 0$, and we just showed that $\dfrac{m}{2^{n}}x \prec \dfrac{m}{2^{n+1}}x$. By transitivity we obtain $\dfrac{m}{2^{n}}x \prec 0$. This is a contradiction, since the result was supposed to hold for $n$.


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