<p>Let me pick up the discussion in the comments.
Consider any diagonal through the origin. Suppose it intersects some IC more than once. Pick two of these points and call them A and B. Because they are on a straight increasing line, one of the bundles, say B, has more of both good 1 and good 2. By monotonicity, bundle B should be preferred over A, which contradicts the statement that they are on the same IC.</p>

<p>If that is what you mean, your intuition is correct. However, there seems to be a jump in your argument: By saying
"if one indifference curve intersects the diagonal line n times (say), then (n-1) points belong to different indifference curves" you seem to be already assuming that the statement to be shown is correct and the n intersections could not exist in the first place.</p>

CC BY-SA 4.0
2024-03-31T05:35:54.613Z
Thanks for this very helpful answer. One question though: this argument proves that the line so constructed intersects each IC at most once. How do we argue that it intersects each IC *at least* once?
EE18
CC BY-SA 4.0
2019-04-24T18:11:40.953Z
2019-04-24T18:11:40.953Z
Bayesian
<p>Question from Intermediate Microeconomics by Hal Varian:<br>
"We claimed in the text that if preferences were monotonic, then a diagonal line through the origin would intersect each indifference curve exactly once. Can you prove this rigorously?"<br>
What I've done:<br>
Preferences in the indifference curve can be represented using utility function and all the bundles that are indifferent are assigned a single value. And each value of the utility function also represents a fixed point on the diagonal line drawn through the origin. So, if one indifference curve intersects the diagonal line n times (say), then (n-1) points belong to different indifference curves and thus it would mean that the indifference curve in question has intersected another indifference curve, which is fundamentally not possible.<br>
Does it prove the statement satisfactorily? </p>

CC BY-SA 4.0
2019-04-24T11:59:05.617Z
Consider any diagonal through the origin. Suppose it intersects some IC more than once. Pick two of these points and call them A and B. Because they are on a straight increasing line, one of the bundles, say B, has more of both good 1 and good 2. By monotonicity, bundle B should be preferred over A, which contradicts the statement that they are on the same IC.
Bayesian
CC BY-SA 4.0
2019-04-24T12:55:56.123Z
Thanks. That's what I wanted to mean as well. ICs cannot intersect each other because that would mean that the consumer will prefer one bundle over the rest of bundles in the same IC. Which is not possible.
Anubhab Giri
CC BY-SA 4.0
2019-04-24T16:13:26.237Z
I'm not sure I quite follow your logic in this sentence: "So, if one indifference curve intersects the diagonal line n times (say), then (n-1) points belong to different indifference curves". At any rate, @Bayesian's approach is most straightforward.
Herr K.
CC BY-SA 4.0
2019-04-24T17:37:39.220Z
@Bayesian This should be an answer, not a comment. Now the answer will remain open although it has been completely answered by your comment.
Alecos Papadopoulos
CC BY-SA 4.0
2019-04-24T18:13:16.030Z
@AlecosPapadopoulos Alright. I thought the question was about the specific argument in the question which I could not follow clearly.
Bayesian
<microeconomics><utility><preferences>
Proof of monotonocity of preferences
Anubhab Giri
# Proof of monotonocity of preferences

(Asked by: username_0 on 2019-04-24)

Question from Intermediate Microeconomics by Hal Varian:  
"We claimed in the text that if preferences were monotonic, then a diagonal line through the origin would intersect each indifference curve exactly once. Can you prove this rigorously?"  
What I've done:  
Preferences in the indifference curve can be represented using utility function and all the bundles that are indifferent are assigned a single value. And each value of the utility function also represents a fixed point on the diagonal line drawn through the origin. So, if one indifference curve intersects the diagonal line n times (say), then (n-1) points belong to different indifference curves and thus it would mean that the indifference curve in question has intersected another indifference curve, which is fundamentally not possible.  
Does it prove the statement satisfactorily?

**Tags:** microeconomics, utility, preferences
> **Comments:**
> **username_1**: Consider any diagonal through the origin. Suppose it intersects some IC more than once. Pick two of these points and call them A and B. Because they are on a straight increasing line, one of the bundles, say B, has more of both good 1 and good 2. By monotonicity, bundle B should be preferred over A, which contradicts the statement that they are on the same IC.  
> **username_0**: Thanks. That's what I wanted to mean as well. ICs cannot intersect each other because that would mean that the consumer will prefer one bundle over the rest of bundles in the same IC. Which is not possible.  
> **username_2**: I'm not sure I quite follow your logic in this sentence: "So, if one indifference curve intersects the diagonal line n times (say), then (n-1) points belong to different indifference curves". At any rate, @Bayesian's approach is most straightforward.  
> **username_3**: @Bayesian This should be an answer, not a comment. Now the answer will remain open although it has been completely answered by your comment.  
> **username_1**: @AlecosPapadopoulos Alright. I thought the question was about the specific argument in the question which I could not follow clearly.  

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## 1 answers

### Answer by username_1 on 2019-04-24. Score: 7 | Accepted answer

Let me pick up the discussion in the comments.
Consider any diagonal through the origin. Suppose it intersects some IC more than once. Pick two of these points and call them A and B. Because they are on a straight increasing line, one of the bundles, say B, has more of both good 1 and good 2. By monotonicity, bundle B should be preferred over A, which contradicts the statement that they are on the same IC.

If that is what you mean, your intuition is correct. However, there seems to be a jump in your argument: By saying
"if one indifference curve intersects the diagonal line n times (say), then (n-1) points belong to different indifference curves" you seem to be already assuming that the statement to be shown is correct and the n intersections could not exist in the first place.

> **Comments:**
> **username_4**: Thanks for this very helpful answer. One question though: this argument proves that the line so constructed intersects each IC at most once. How do we argue that it intersects each IC *at least* once?  

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