<p>Terms such as <span class="math-container">$a$</span> or <span class="math-container">$ta+(1-t)b$</span> are consumption bundles, not numbers, and <span class="math-container">$\succ$</span> is not the same as <span class="math-container">$&gt;$</span>, so you cannot simply &quot;subtract&quot; <span class="math-container">$ta$</span> from both sides of an expression or &quot;divide&quot; both sides by <span class="math-container">$1-t$</span> as though this were an inequality in the real numbers. To work in the real numbers you have to use utilities, e.g. write <span class="math-container">$u(ta+(1-t)b)&lt;u(a)$</span>. Then it should become clear why you can't go on as you did.</p>

CC BY-SA 4.0
2022-01-17T08:48:23.180Z
This seems to depend more on $\prec$ being different from $<$? Consumption bundles usually have an algebraic representation (a vector) that would admit such manipulations, i.e. if $$ta +(1-t)b << a$$ then $$b << a.$$ So $a$ and $b$ being bundles is not really the problem.
Giskard
CC BY-SA 4.0
2022-01-17T12:13:04.223Z
@Giskard, I think this is in some sense implied by $a$ and $b$ not being numbers. The relation $<$ is only defined on numbers, after all. (The relation $<<$ is defined on vectors, but this can't be the interpretation the questioner had in mind, since then even the first line of his "proof" would be clearly wrong.)
VARulle
CC BY-SA 4.0
2022-01-17T13:34:13.743Z
I could also define $\prec$ on *numbers* in such a way that the above calculation is invalid, so I still think the crucial point is that $\prec$ does not allow these manipulations the way $<$ does.
Giskard
CC BY-SA 4.0
2022-01-17T15:08:04.863Z
@Giskard, can we say it is both? If I prefer 3 apples to 5 bananas, then it is wrong to write "$3 > 5$". But neither "$3\succ 5$" nor "3 apples $>$ 5 bananas" is well-defined, so this translation is wrong because (i) bundles are not numbers AND (ii) $\succ$ is not $>$.
VARulle
CC BY-SA 4.0
2022-01-17T08:41:33.133Z
2022-01-18T11:16:29.313Z
2022-01-18T11:16:29.313Z
VARulle
<p>The answer in short is that adding the same vector to both the sides need not preserve the preference relation because the added utilities to both sides may be different even when the same vector is added. The additional utility also depends on the quantities of all the goods already owned. For example, suppose having 10 bananas is indifferent to having 10 oranges. We add the same vector of 5 bananas and 0 oranges in both the cases. But it is obvious that the 5 bananas would add much more utility in the first case (as the preferences are concave in your example) than in the second case.</p>
<p>More mathematically, let's say the number of commodities is <span class="math-container">$n$</span>, then <span class="math-container">$a, b$</span> are both <span class="math-container">$n$</span> dimensional vectors, <span class="math-container">$t$</span> is a positive real number less than 1.</p>
<p>The initial LHS is: <span class="math-container">$ta + (1-t)b$</span> (<span class="math-container">$:=c$</span>) which, say, gives you the utility, <span class="math-container">$u(ta + (1-t)b)$</span></p>
<p>The initial RHS is: <span class="math-container">$a$</span> which gives you the utility <span class="math-container">$u(a)$</span></p>
<p><span class="math-container">$ta + (1-t)b \prec a$</span> simply means <span class="math-container">$u(ta + (1-t)b) &lt; u(a)$</span></p>
<p>Then you added <span class="math-container">$-ta$</span> to both the sides.</p>
<p>The new LHS is: <span class="math-container">$(1-t)b$</span>, which gives you the utility <span class="math-container">$u((1-t)b)$</span></p>
<p>The increase in utility in the LHS is: <span class="math-container">$u((1-t)b) - u(ta + (1-t)b)$</span></p>
<p>The new RHS is: <span class="math-container">$(1-t)a$</span> which gives you the utility <span class="math-container">$u((1-t)a)$</span></p>
<p>The increase in utility in the RHS is: <span class="math-container">$u((1-t)a) - u(a)$</span></p>
<p>Adding <span class="math-container">$-ta$</span> to both sides would mean adding equal utility to both sides iff <span class="math-container">$u((1-t)b) - u(ta + (1-t)b) = u((1-t)a) - u(a)$</span>.</p>
<p>This is a condition only some and not all utility functions would satisfy. Adding unequal amounts of utility may lead tobreversal of the preference relation.</p>
<p>For example, let <span class="math-container">$u(x,y) = y/(2- x^2), a=(1,1), b=(0,2), t=0.5$</span>.</p>
<p>The initial relations hold, that is, <span class="math-container">$a$</span> ~ <span class="math-container">$b$</span> and <span class="math-container">$c \prec a$</span>. But adding <span class="math-container">$-ta$</span> to both sides leads to adding unequal amounts of utility, LHS's utility increases by <span class="math-container">$-5/14$</span> while the RHS's utility increases by <span class="math-container">$-5/7$</span>, and now the LHS is preferred.</p>

CC BY-SA 4.0
2022-01-20T10:43:17.220Z
The example is somewhat suboptimal, with infinite and negative utilities for $x\ge\sqrt{2}$.
VARulle
CC BY-SA 4.0
2022-01-19T15:00:02.127Z
2022-01-19T15:07:22.583Z
2022-01-19T15:07:22.583Z
Ishan Kashyap Hazarika
<p>I'm going through various preference relation proofs using the various axioms (eg convexity), trying to get more comfortable with the mathematical tools before my class starts in about six weeks' time.</p>
<p>Can someone please explain to me why the highlighted step is invalid? Ie, basic algebra rearranging seems to lead to an obviously incorrect conclusion. Is the algebraic operation assuming that the preference relation maps linearly, or something like that? Many thanks.</p>
<p><a href="https://i.sstatic.net/Ghe8u.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Ghe8u.png" alt="Concave preferences" /></a></p>

<microeconomics>
Microeconomics - preference relations proof - Why is this logic invalid?
cashman
# Microeconomics - preference relations proof - Why is this logic invalid?

(Asked by: username_0 on 2022-01-17)

I'm going through various preference relation proofs using the various axioms (eg convexity), trying to get more comfortable with the mathematical tools before my class starts in about six weeks' time.

Can someone please explain to me why the highlighted step is invalid? Ie, basic algebra rearranging seems to lead to an obviously incorrect conclusion. Is the algebraic operation assuming that the preference relation maps linearly, or something like that? Many thanks.

[![Concave preferences](https://i.sstatic.net/Ghe8u.png)](https://i.sstatic.net/Ghe8u.png)

**Tags:** microeconomics

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## 2 answers

### Answer by username_1 on 2022-01-17. Score: 8

Terms such as $a$ or $ta+(1-t)b$ are consumption bundles, not numbers, and $\succ$ is not the same as $>$, so you cannot simply "subtract" $ta$ from both sides of an expression or "divide" both sides by $1-t$ as though this were an inequality in the real numbers. To work in the real numbers you have to use utilities, e.g. write $u(ta+(1-t)b). Then it should become clear why you can't go on as you did.

> **Comments:**
> **username_2**: This seems to depend more on $\prec$ being different from $<$? Consumption bundles usually have an algebraic representation (a vector) that would admit such manipulations, i.e. if $$ta +(1-t)b << a$$ then $$b << a.$$ So $a$ and $b$ being bundles is not really the problem.  
> **username_1**: @Giskard, I think this is in some sense implied by $a$ and $b$ not being numbers. The relation $<$ is only defined on numbers, after all. (The relation $<<$ is defined on vectors, but this can't be the interpretation the questioner had in mind, since then even the first line of his "proof" would be clearly wrong.)  
> **username_2**: I could also define $\prec$ on *numbers* in such a way that the above calculation is invalid, so I still think the crucial point is that $\prec$ does not allow these manipulations the way $<$ does.  
> **username_1**: @Giskard, can we say it is both? If I prefer 3 apples to 5 bananas, then it is wrong to write "$3 > 5$". But neither "$3\succ 5$" nor "3 apples $>$ 5 bananas" is well-defined, so this translation is wrong because (i) bundles are not numbers AND (ii) $\succ$ is not $>$.  

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### Answer by username_3 on 2022-01-19. Score: 4

The answer in short is that adding the same vector to both the sides need not preserve the preference relation because the added utilities to both sides may be different even when the same vector is added. The additional utility also depends on the quantities of all the goods already owned. For example, suppose having 10 bananas is indifferent to having 10 oranges. We add the same vector of 5 bananas and 0 oranges in both the cases. But it is obvious that the 5 bananas would add much more utility in the first case (as the preferences are concave in your example) than in the second case.

More mathematically, let's say the number of commodities is $n$, then $a, b$ are both $n$ dimensional vectors, $t$ is a positive real number less than 1.

The initial LHS is: $ta + (1-t)b$ ($:=c$) which, say, gives you the utility, $u(ta + (1-t)b)$

The initial RHS is: $a$ which gives you the utility $u(a)$

$ta + (1-t)b \prec a$ simply means $u(ta + (1-t)b) < u(a)$

Then you added $-ta$ to both the sides.

The new LHS is: $(1-t)b$, which gives you the utility $u((1-t)b)$

The increase in utility in the LHS is: $u((1-t)b) - u(ta + (1-t)b)$

The new RHS is: $(1-t)a$ which gives you the utility $u((1-t)a)$

The increase in utility in the RHS is: $u((1-t)a) - u(a)$

Adding $-ta$ to both sides would mean adding equal utility to both sides iff $u((1-t)b) - u(ta + (1-t)b) = u((1-t)a) - u(a)$.

This is a condition only some and not all utility functions would satisfy. Adding unequal amounts of utility may lead tobreversal of the preference relation.

For example, let $u(x,y) = y/(2- x^2), a=(1,1), b=(0,2), t=0.5$.

The initial relations hold, that is, $a$ ~ $b$ and $c \prec a$. But adding $-ta$ to both sides leads to adding unequal amounts of utility, LHS's utility increases by $-5/14$ while the RHS's utility increases by $-5/7$, and now the LHS is preferred.

> **Comments:**
> **username_1**: The example is somewhat suboptimal, with infinite and negative utilities for $x\ge\sqrt{2}$.  

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