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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 $m,n\in\mathbb{N}$.

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  • $\begingroup$ 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? $\endgroup$ – BB King Sep 11 '15 at 20:38
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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 $m \in \mathbb{N}, n=0$.

  1. 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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