2003 AMC 10A Problem 6

P1-10~1 minPrint

Absolute Value & Inequalities Functions Use the answer choices (mod checks, size, form) to eliminate or select Test small/specific values or special cases to find or verify the answer

Define $x \heartsuit y$ to be $|x-y|$ for all real numbers $x$ and $y$ . Which of the following statements is not true?

Show solution without answering

Key insight

x heart 0 is the absolute value of x, which is never negative. So for a negative x, like -3, it cannot equal x.

Solution

The symbol $x \heartsuit y$ just means "the distance between $x$ and $y$ ", which is $|x - y|$ . Translate each choice into ordinary absolute values and test it.

(A) says $|x - y| = |y - x|$ . The distance from $x$ to $y$ is the same as from $y$ to $x$ . True.
(B) says $2|x - y| = |2x - 2y|$ . The right side is $|2(x - y)|$ , and doubling a number doubles its size. For example $x = 1, y = 4$ : $2 \cdot 3 = 6$ and $|2 - 8| = 6$ . True.
(C) says $|x - 0| = x$ , that is, $|x| = x$ . Try a negative number, $x = -3$ : $|-3| = 3$ , but $x = -3$ . They are not equal. False.
(D) says $|x - x| = 0$ . That is $|0| = 0$ . True.
(E) says $|x - y| > 0$ when $x \ne y$ . Two different numbers are a positive distance apart. True.

Only (C) fails.

The answer is $\boxed{\textbf{(C)}\ x \heartsuit 0 = x\ \text{for all}\ x}$ .

Why this works

A statement that claims to hold "for all $x$ " is knocked down by a single example where it fails. With absolute values, the first thing to try is a negative number, since that is where $|x|$ and $x$ differ. Rewriting a made-up symbol in familiar terms turns each choice into a fact you already know.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Translate each statement using $x \heartsuit y = |x - y|$ .

(A) $|x - y| = |y - x|$ : true, since a number and its negative have the same absolute value.
(B) $2|x - y| = |2x - 2y|$ : true, because $|2t| = 2|t|$ .
(C) $x \heartsuit 0 = |x|$ . Taking $x = -3$ gives $|-3| = 3 \neq -3$ . False.
(D) $|x - x| = 0$ : true.
(E) $|x - y| > 0$ when $x \neq y$ : true, since only $0$ has absolute value $0$ .

The statement that fails is $\boxed{\textbf{(C)}\ x \heartsuit 0 = x\ \text{for all}\ x}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 16: Functions
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)
Introduction to Algebra, Chapter 17: Graphing Functions (§17.2)