AMC 10 · 2003 · #6
Grade 6 algebraDefine x♡y to be ∣x−y∣ for all real numbers x and y. Which of the following statements is not true?
(A) x♡y=y♡x for all x and y
(B) 2(x♡y)=(2x)♡(2y) for all x and y
(C) x♡0=x for all x
(D) x♡x=0 for all x
(E) x♡y>0 if x=y
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A new operation $\heartsuit$ is defined by $x \heartsuit y = |x-y|$ for all real numbers $x$ and $y$. Four of the five statements (A)–(E) about this operation are always true; find the one statement that is not true.
Givens: $x \heartsuit y = |x-y|$ for all real numbers $x$ and $y$; (A) $x \heartsuit y = y \heartsuit x$ for all $x$ and $y$; (B) $2(x \heartsuit y) = (2x) \heartsuit (2y)$ for all $x$ and $y$; (C) $x \heartsuit 0 = x$ for all $x$; (D) $x \heartsuit x = 0$ for all $x$; (E) $x \heartsuit y > 0$ if $x \neq y$
Unknowns: Which one of the five statements (A)–(E) is not true
Understand
Restated: A new operation $\heartsuit$ is defined by $x \heartsuit y = |x-y|$ for all real numbers $x$ and $y$. Four of the five statements (A)–(E) about this operation are always true; find the one statement that is not true.
Givens: $x \heartsuit y = |x-y|$ for all real numbers $x$ and $y$; (A) $x \heartsuit y = y \heartsuit x$ for all $x$ and $y$; (B) $2(x \heartsuit y) = (2x) \heartsuit (2y)$ for all $x$ and $y$; (C) $x \heartsuit 0 = x$ for all $x$; (D) $x \heartsuit x = 0$ for all $x$; (E) $x \heartsuit y > 0$ if $x \neq y$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #6 Guess and Check, #14 Extreme Principle
The question asks which statement fails, so tool #3 (Eliminate Possibilities) fits exactly: test each of the five claims, throw out the ones that always hold, and the survivor is the answer. Because $x \heartsuit y = |x-y|$ is just the distance between $x$ and $y$, four of the claims are quick distance facts (symmetry, scaling, zero-distance, positive distance) and confirm as true. Tool #6 (Guess and Check) drives the disproof: a statement claiming something 'for all $x$' is broken by a single number that fails it. Tool #14 (Extreme Principle) tells us where to look for that number — absolute value only changes things for negatives, so probing a negative input is the case most likely to expose a false claim, and statement (C) breaks there.
Execute — Answer: C
6.NS.C.7 Step 1 Read the heart as distance
- The operation $x \heartsuit y = |x-y|$ is the distance between $x$ and $y$ on the number line.
- Distance is never negative.
- A statement that claims something holds 'for all' values is only true if it never fails, so to disprove one I just need a single example that breaks it.
💡 Seeing $|x-y|$ as a distance makes most of the claims obvious, and one bad example is enough to kill a 'for all' claim.
6.EE.B.5 Step 2 Confirm A, B, D, E are true
- (A) $x \heartsuit y = |x-y|$ and $y \heartsuit x = |y-x|$; since $|x-y| = |y-x|$ (distance is the same both directions), (A) holds.
- (B) $2(x \heartsuit y) = 2|x-y|$, while $(2x) \heartsuit (2y) = |2x-2y| = 2|x-y|$; the two sides match, so (B) is true — a tempting choice, but it checks out.
- (D) $x \heartsuit x = |x-x| = |0| = 0$, always true.
- (E) if $x \neq y$ then their distance is positive, so $|x-y| > 0$; true.
- All four are always true, so none of them is the answer.
💡 Checking each claim as a distance fact clears the four true statements and narrows the search to the one left over.
6.EE.A.2 Step 3 Break statement C with a negative
- Statement (C) claims $x \heartsuit 0 = x$ for all $x$.
- But $x \heartsuit 0 = |x-0| = |x|$, and $|x| = x$ only when $x \ge 0$.
- Test a negative value, where absolute value actually changes the sign: take $x = -3$.
- Then $(-3) \heartsuit 0 = |-3| = 3$, which is not equal to $-3$.
- So (C) fails, and since the correct formula is $x \heartsuit 0 = |x|$ (not $x$), statement (C) is the one that is not true.
- The answer is (C).
💡 Absolute value only bites on negatives, so a single negative input exposes the claim $x \heartsuit 0 = x$ as false.
6.NS.C.7 The operation $x \heartsuit y = |x-y|$ is the distance between $x$ and $y$ on th 6.EE.B.5 (A) $x \heartsuit y = |x-y|$ and $y \heartsuit x = |y-x|$; since $|x-y| = |y-x|$ 6.EE.A.2 Statement (C) claims $x \heartsuit 0 = x$ for all $x$. But $x \heartsuit 0 = |x- Review
Reasonableness: The correct identity is $x \heartsuit 0 = |x|$, which equals $x$ for non-negative $x$ but equals $-x$ for negative $x$. So statement (C) is true for half the number line and false for the other half — exactly the kind of 'almost right' claim that a 'for all' statement cannot afford. Every other statement is a genuine property of distance (symmetry, scaling, zero at equal points, positive when unequal), so (C) being the lone false one is consistent.
Alternative: Instead of hunting for a counterexample, rewrite each side in closed form. (C) simplifies to $|x|$ on the left and $x$ on the right; the equation $|x| = x$ is the definition of $x$ being non-negative, so it cannot hold 'for all $x$' — it already fails structurally, no specific number needed. The same closed-form check shows (A), (B), (D), (E) reduce to true identities, again pointing to (C).
CCSS standards used (min grade 6)
6.NS.C.7Understand ordering and absolute value of rational numbers (Reading $x \heartsuit y = |x-y|$ as a non-negative distance and knowing $|x| = x$ only for $x \ge 0$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Testing whether each 'for all' statement (A), (B), (D), (E) actually holds for every value.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating $x \heartsuit 0 = |x|$ at $x = -3$ to produce the counterexample that breaks statement (C).)
⭐ A statement that says something is true 'for all' numbers is wrong the moment one number breaks it, and with absolute value the number to try is a negative one.
⭐ A statement that says something is true 'for all' numbers is wrong the moment one number breaks it, and with absolute value the number to try is a negative one.
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