Why the Question Matters
An indifference map represents how a consumer ranks possible bundles. Each curve groups together bundles the consumer regards as equally good. If two curves cross, they share a bundle. That shared bundle links the rankings represented by both curves, so it is not possible for the curves to represent different levels of preference.
The key reasoning is about consistency, not the shape of a curve. Transitivity lets us follow a chain of indifferences: if one bundle is indifferent to a second, and the second to a third, then the first and third are indifferent too. A crossing therefore cannot join two genuinely different ranking levels.
Transitivity means that if \(A\) is weakly preferred to \(B\), and \(B\) is weakly preferred to \(C\), then \(A\) is weakly preferred to \(C\). It applies in both directions when the comparisons are indifferences. This gives the useful result that if \(A \sim B\) and \(B \sim C\), then \(A \sim C\).
The Core Argument
Suppose two curves cross at bundle \(I\). Because \(I\) lies on the first curve, every bundle on that curve is indifferent to \(I\). Because \(I\) also lies on the second curve, every bundle on the second curve is indifferent to \(I\). Transitivity then makes every bundle on either curve indifferent to every bundle on the other.
That conclusion is incompatible with treating the curves as different ranking levels. The intersection does not create two rankings for \(I\); it connects both sets to the same ranking. If the curves are complete indifference curves, they must represent the same indifference class. A sketch that labels them as distinct levels is inconsistent.
The implication follows directly from transitivity. From \(A \sim I\), we have \(A \succeq I\) and \(I \succeq A\). From \(I \sim B\), we have \(I \succeq B\) and \(B \succeq I\). Transitivity gives \(A \succeq B\) from \(A \succeq I \succeq B\), and \(B \succeq A\) from \(B \succeq I \succeq A\). Both directions together mean \(A \sim B\).
Call the crossing bundle \(I\). It belongs to both alleged curves.
Every bundle \(A\) on the first curve satisfies \(A \sim I\); every bundle \(B\) on the second satisfies \(B \sim I\).
The two indifferences imply \(A \sim B\). Thus the curves cannot represent different rankings.
Worked Examples
Worked Example: Checking Two Proposed Curves Algebraically
A preference map proposes two curves for bundles \((x,y)\) of two goods: \(x+y=8\) and \(x+2y=10\). Treat both as indifference curves for the same consumer. Find their intersection, then use a second pair of bundles to test whether the map can be consistent with strict monotonicity, under which more of one good and the same amount of the other is strictly preferred.
Step 1: Find the crossing. Subtract the first equation from the second:
Therefore \(y=2\). Substituting into \(x+y=8\) gives \(x+2=8\), so \(x=6\). The curves intersect at \(I=(6,2)\).
Step 2: Choose bundles on each curve with the same \(x\). Set \(x=5\). On the first curve, \(5+y=8\), so \(y=3\) and \(A=(5,3)\). On the second curve, \(5+2y=10\), so \(2y=5\), \(y=2.5\), and \(B=(5,2.5)\).
Step 3: Compare the rankings. Since \(A\) and \(I\) lie on the first proposed indifference curve, \(A \sim I\). Since \(B\) and \(I\) lie on the second, \(B \sim I\). Transitivity implies \(A \sim B\). But \(A\) has the same amount of good \(x\) as \(B\) and \(0.5\) more units of good \(y\). Strict monotonicity implies \(A \succ B\), a contradiction.
Check: Substitution verifies \(5+3=8\) and \(5+2(2.5)=10\). Both selected bundles are on the proposed curves, and their strict ranking conflicts with the indifference chain. The method is to solve the intersection, then compare points on each curve. A common mistake is to stop after calculating the intersection; the shared bundle must also be used to test the rankings.
Worked Example: Using a Shared Quantity to Expose a Conflict
A map shows the curves \(x+y=10\) and \(2x+y=14\), with the first labelled as a higher ranking level than the second. Find their intersection. Then compare points on the curves that contain the same quantity of good \(y\), and determine whether the labels are consistent if preferences are strictly monotonic.
Step 1: Solve for the intersection. Subtract the first equation from the second:
Thus \(x=4\). Substituting into \(x+y=10\) gives \(y=6\), so \(I=(4,6)\).
Step 2: Select one point from each curve at \(y=5\). On \(x+y=10\), \(x+5=10\), hence \(x=5\) and \(A=(5,5)\). On \(2x+y=14\), \(2x+5=14\), hence \(2x=9\), \(x=4.5\), and \(B=(4.5,5)\).
Step 3: Trace indifference and compare. The crossing gives \(A \sim I\) and \(B \sim I\), so transitivity requires \(A \sim B\). Yet \(A\) and \(B\) contain the same amount of good \(y\), while \(A\) contains \(0.5\) more units of good \(x\). Strict monotonicity requires \(A \succ B\). Therefore the curves cannot both represent the distinct ranking levels claimed by the map.
Check: The intersection satisfies \(4+6=10\) and \(2(4)+6=14\). The comparison points satisfy \(5+5=10\) and \(2(4.5)+5=14\). A common mistake is to compare points with different amounts of both goods and infer a ranking without justification. Holding \(y\) fixed makes the monotonicity comparison clear.
Worked Example: Separating Transitivity From a Map’s Ranking Labels
A diagram shows \(C_1: x+y=8\) and \(C_2: 2x+3y=20\), and labels a point \(A=(2,6)\) on \(C_1\) as strictly preferred to a point \(B=(7,2)\) on \(C_2\). Check whether the curves intersect. Then use transitivity to assess the labels, without using a geometric claim about which point has more goods.
Step 1: Verify the marked points. For \(A\), \(2+6=8\), so \(A\) lies on \(C_1\). For \(B\), \(2(7)+3(2)=14+6=20\), so \(B\) lies on \(C_2\).
Step 2: Find the intersection. From \(x+y=8\), \(x=8-y\). Substitute into \(2x+3y=20\):
Then \(x=8-4=4\), so \(I=(4,4)\). This point satisfies both equations: \(4+4=8\) and \(2(4)+3(4)=20\).
Step 3: Test the ranking claim. If \(C_1\) and \(C_2\) are indifference curves, \(A \sim I\) and \(B \sim I\). By transitivity, \(A \sim B\). The map’s separate claim \(A \succ B\) contradicts that result: strict preference cannot hold between bundles that are indifferent.
Check: Both marked bundles satisfy their stated curve equations, and the calculated intersection satisfies both equations. This example uses transitivity directly; it does not rely on comparing the quantities of the goods in \(A\) and \(B\). A common mistake is to treat the curve labels as evidence of ranking even when the shared point makes those labels inconsistent.
What the Geometric Argument Does—and Does Not—Require
The central result comes from transitivity and the definition of an indifference curve. Once two curves share \(I\), any bundle on either curve is indifferent to \(I\), and therefore the bundles on both curves are indifferent to one another. If a diagram also declares that the curves represent different ranking levels, the declaration conflicts with that implication.
Strict monotonicity provides a second way to spot the problem in a conventional crossing sketch. Near a crossing, one can often select a point on each curve with the same quantity of one good, but different quantities of the other. If one point has more of that other good, strict monotonicity ranks it strictly above the other point. The intersection and transitivity, however, imply that they are indifferent. This is a contradiction. This check uses strict monotonicity; it is not needed for the basic shared-bundle argument.
Be precise about the assumptions. Transitivity and indifference within each curve imply that the two curves cannot be distinct ranking classes if they intersect. They do not, by themselves, make every possible pair of plotted paths a contradiction. For example, two sketches could both depict portions of the same indifference class. In that case they do not represent different levels, even if their paths meet or cross.
A further common error is to argue that crossing alone proves one curve is preferred to the other. A crossing gives a shared bundle, not an automatic ranking between all other bundles. Establish a contradiction by tracing indifference through the shared bundle, and, if using a geometric comparison, state the preference assumption that justifies the strict ranking.
Check Your Understanding
For each proposed map, check the relevant equations or ranking statements before deciding whether the curves can represent distinct indifference levels.
- The curves \(x+y=12\) and \(x+2y=16\) are proposed as distinct indifference curves. Find their intersection and state the transitivity implication for bundles on the two curves. (Answer: They intersect at \((8,4)\); every bundle on either curve must be indifferent to every bundle on the other, so they cannot represent distinct levels.)
- Two curves share bundle \(I\). Bundle \(A\) lies on the first, bundle \(B\) lies on the second, and the map states \(A \sim I\) and \(B \sim I\). What relation must hold between \(A\) and \(B\), and can the map also state \(A \succ B\)? (Answer: \(A \sim B\); no, that conflicts with \(A \succ B\).)
- The proposed curves are \(x+2y=14\) and \(2x+y=13\). Find their intersection. At \(y=4\), find the point on each curve and, assuming strict monotonicity, explain the conflict. (Answer: The intersection is \((4,5)\); at \(y=4\) the points are \((6,4)\) and \((4.5,4)\), so the first is strictly preferred under strict monotonicity although transitivity implies indifference.)
- A map draws two crossing paths, but its author says they are two portions of the same complete indifference class, not different ranking levels. Does the crossing alone contradict transitivity? Explain. (Answer: No; transitivity makes the bundles on both paths indifferent, which is consistent if the paths depict the same indifference class.)
- Two complete curves intersect at \(I\). The map says a point \(P\) on the first curve is indifferent to \(I\), a point \(Q\) on the second curve is indifferent to \(I\), and \(P\) is strictly preferred to \(Q\). Identify the minimum logical conflict, without assuming monotonicity. (Answer: Transitivity gives \(P \sim Q\), contradicting \(P \succ Q\).)