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Zorluk: OrtaFunctions and Custom Symbol Operations

For all real numbers xx and yy, the custom operation ⊙\odot is defined by x⊙y=x∣y∣−y∣x∣x \odot y = x|y| - y|x|. Which of the following statements must be true for all real numbers xx and yy? Select all such statements.

  1. x⊙y=0x \odot y = 0 whenever xx and yy have the same signCevap
  2. x⊙y=−(y⊙x)x \odot y = -(y \odot x)Cevap
  3. If x>0x > 0 and y<0y < 0, then x⊙y>0x \odot y > 0Cevap
  4. D
    x⊙(−x)=0x \odot (-x) = 0 for all non-zero real numbers xx
  5. E
    (x⊙y)⊙z=x⊙(y⊙z)(x \odot y) \odot z = x \odot (y \odot z)

Cevap

The correct statements are the statement asserting x⊙y=0x \odot y = 0 when xx and yy have the same sign, the statement asserting anti-commutativity x⊙y=−(y⊙x)x \odot y = -(y \odot x), and the statement asserting x⊙y>0x \odot y > 0 when x>0x > 0 and y<0y < 0.
The operation x⊙y=x∣y∣−y∣x∣x \odot y = x|y| - y|x| produces 0 whenever xx and yy share the same sign because terms evaluate to identical quantities. Swapping variables negates the expression, establishing anti-commutativity. When xx is positive and yy is negative, x⊙yx \odot y simplifies to −2xy-2xy, which is strictly greater than 0 since xy<0xy < 0.

Adım Adım Çözüm

1
Analyze the first statement regarding same-sign inputs
If x>0x > 0 and y>0y > 0, ∣x∣=x|x|=x and ∣y∣=y|y|=y, so x⊙y=xy−yx=0x \odot y = xy - yx = 0. If x<0x < 0 and y<0y < 0, ∣x∣=−x|x|=-x and ∣y∣=−y|y|=-y, so x⊙y=x(−y)−y(−x)=−xy+xy=0x \odot y = x(-y) - y(-x) = -xy + xy = 0. Thus, x⊙y=0x \odot y = 0 when xx and yy have the same sign.
Verifying the definition under both positive and negative cases of identical sign.
2
Analyze the second statement regarding operand order reversal
y⊙x=y∣x∣−x∣y∣=−(x∣y∣−y∣x∣)=−(x⊙y)y \odot x = y|x| - x|y| = -(x|y| - y|x|) = -(x \odot y), which holds universally for all real numbers.
Testing anti-commutativity by algebraic substitution into the custom operation.
3
Analyze the third statement for opposite signs (x>0x > 0 and y<0y < 0)
Since x>0x > 0, ∣x∣=x|x|=x. Since y<0y < 0, ∣y∣=−y|y|=-y. Substituting yields x(−y)−y(x)=−xy−xy=−2xyx(-y) - y(x) = -xy - xy = -2xy. Because x>0x > 0 and y<0y < 0, the product xyxy is negative, making −2xy-2xy strictly positive.
Determining the overall algebraic sign of the expression when variables have opposite signs.
4
Counter-test the remaining statements to verify incorrectness
For x⊙(−x)x \odot (-x) with x=1x=1: 1⊙(−1)=1∣−1∣−(−1)∣1∣=1−(−1)=2≠01 \odot (-1) = 1|-1| - (-1)|1| = 1 - (-1) = 2 \neq 0. For associativity with x=2,y=−1,z=−1x=2, y=-1, z=-1: (2⊙−1)⊙−1=4⊙−1=8(2 \odot -1) \odot -1 = 4 \odot -1 = 8, but 2⊙(−1⊙−1)=2⊙0=02 \odot (-1 \odot -1) = 2 \odot 0 = 0.
Demonstrating specific counterexamples for false generalizations.

Anahtar Kavram

Custom Binary Operations and Absolute Value Properties
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