Binary Operations

24 soru

Soru 21Soru →

A binary operation Δ\Delta defined on the set of real numbers R\mathbb{R} is given by aΔb=a+b−ab4a \Delta b = a + b - \frac{ab}{4}. What is the inverse of 22 under this operation?

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Cevap: −4-4

Cevap

The inverse of 22 under the binary operation is −4-4.
To find the inverse of 22, we first determine the identity element ee using aΔe=aa \Delta e = a, which yields e=0e = 0. Setting 2Δx=02 \Delta x = 0 gives 2+x−2x4=02 + x - \frac{2x}{4} = 0, which simplifies to 2+x2=02 + \frac{x}{2} = 0, resulting in x=−4x = -4.

Adım Adım Çözüm

1
Determine the identity element ee of the operation.
e=0e = 0
By definition, aΔe=aa \Delta e = a. Thus, a+e−ae4=a  ⟹  e(1−a4)=0a + e - \frac{ae}{4} = a \implies e\left(1 - \frac{a}{4}\right) = 0, which gives e=0e = 0 for all real aa.
2
Set up the inverse equation for the element 22.
2+x−2x4=02 + x - \frac{2x}{4} = 0
Let xx be the inverse of 22. By definition, 2Δx=e2 \Delta x = e, where e=0e = 0.
3
Simplify and solve for xx.
x=−4x = -4
Simplify 2+x−x2=0  ⟹  2+x2=0  ⟹  x2=−2  ⟹  x=−42 + x - \frac{x}{2} = 0 \implies 2 + \frac{x}{2} = 0 \implies \frac{x}{2} = -2 \implies x = -4.

Anahtar Kavram

Inverse Element in Binary Operations
Soru 22Soru →

A binary operation ∗\ast is defined on the set of real numbers R\mathbb{R} by a∗b=a+b−4a \ast b = a + b - 4. If x−1x^{-1} represents the inverse of an element xx under the operation ∗\ast, what is the value of xx such that x∗3=10−1x \ast 3 = 10^{-1}?

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Cevap: −1-1

Cevap

The value of xx is −1-1.
First, the identity element ee is determined using a∗e=aa \ast e = a, which gives a+e−4=a  ⟹  e=4a + e - 4 = a \implies e = 4. Next, the inverse 10−110^{-1} is calculated from 10∗10−1=410 \ast 10^{-1} = 4, yielding 10+10−1−4=4  ⟹  10−1=−210 + 10^{-1} - 4 = 4 \implies 10^{-1} = -2. Finally, setting x∗3=−2x \ast 3 = -2 gives x+3−4=−2  ⟹  x−1=−2  ⟹  x=−1x + 3 - 4 = -2 \implies x - 1 = -2 \implies x = -1.

Adım Adım Çözüm

1
Find the identity element ee under the operation ∗\ast.
e=4e = 4
By definition of an identity element, a∗e=a  ⟹  a+e−4=a  ⟹  e=4a \ast e = a \implies a + e - 4 = a \implies e = 4.
2
Determine the inverse of 1010, denoted 10−110^{-1}, under ∗\ast.
10−1=−210^{-1} = -2
By definition of an inverse element, 10∗10−1=e  ⟹  10+10−1−4=4  ⟹  6+10−1=4  ⟹  10−1=−210 \ast 10^{-1} = e \implies 10 + 10^{-1} - 4 = 4 \implies 6 + 10^{-1} = 4 \implies 10^{-1} = -2.
3
Substitute 10−1=−210^{-1} = -2 into the given equation x∗3=10−1x \ast 3 = 10^{-1} and solve for xx.
x=−1x = -1
Using the definition of the operation, x∗3=x+3−4=x−1x \ast 3 = x + 3 - 4 = x - 1. Setting x−1=−2x - 1 = -2 yields x=−1x = -1.

Anahtar Kavram

Identity and Inverse Elements in Binary Operations
Soru 23Soru →

A binary operation ∘\circ defined on the set of real numbers R\mathbb{R} is given by a∘b=3a+2b−1a \circ b = 3a + 2b - 1. What is the value of xx for which (2∘x)∘3=26(2 \circ x) \circ 3 = 26?

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Cevap: 11

Cevap

The value of xx for which (2∘x)∘3=26(2 \circ x) \circ 3 = 26 is 11.
Applying the binary operation rule a∘b=3a+2b−1a \circ b = 3a + 2b - 1 to the inner expression gives 2∘x=2x+52 \circ x = 2x + 5. Then applying the rule to (2x+5)∘3(2x + 5) \circ 3 gives 3(2x+5)+2(3)−1=6x+203(2x + 5) + 2(3) - 1 = 6x + 20. Equating 6x+20=266x + 20 = 26 yields 6x=66x = 6, giving x=1x = 1.

Adım Adım Çözüm

1
Evaluate the inner binary operation expression 2∘x2 \circ x using the rule a∘b=3a+2b−1a \circ b = 3a + 2b - 1.
2∘x=3(2)+2(x)−1=6+2x−1=2x+52 \circ x = 3(2) + 2(x) - 1 = 6 + 2x - 1 = 2x + 5
The definition of the binary operation replaces aa with 22 and bb with xx.
2
Substitute the result (2x+5)(2x + 5) into the outer binary operation expression (2∘x)∘3(2 \circ x) \circ 3.
(2x+5)∘3=3(2x+5)+2(3)−1=6x+15+6−1=6x+20(2x + 5) \circ 3 = 3(2x + 5) + 2(3) - 1 = 6x + 15 + 6 - 1 = 6x + 20
Apply the rule a∘b=3a+2b−1a \circ b = 3a + 2b - 1 where a=2x+5a = 2x + 5 and b=3b = 3.
3
Equate the simplified expression to 2626 and solve for xx.
6x+20=26  ⟹  6x=6  ⟹  x=16x + 20 = 26 \implies 6x = 6 \implies x = 1
Setting the calculated expression equal to the given value allows isolation of xx.

Anahtar Kavram

Evaluation of non-commutative composite binary operations
Soru 24Soru →

A binary operation ⋆\star defined on the set of real numbers R∖{12}\mathbb{R} \setminus \left\{\frac{1}{2}\right\} is given by a⋆b=a+b−2aba \star b = a + b - 2ab. If x−1x^{-1} represents the inverse of an element xx under ⋆\star, what is the value of xx for which x−1=3x^{-1} = 3?

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Cevap: 35\frac{3}{5}

Cevap

The value of xx is 35\frac{3}{5}.
First find the identity element ee using a⋆e=aa \star e = a, which gives a+e−2ae=a  ⟹  e(1−2a)=0  ⟹  e=0a + e - 2ae = a \implies e(1 - 2a) = 0 \implies e = 0. Next, use the inverse definition x⋆x−1=e  ⟹  x+x−1−2xx−1=0x \star x^{-1} = e \implies x + x^{-1} - 2x x^{-1} = 0. Substituting x−1=3x^{-1} = 3 yields x+3−2x(3)=0  ⟹  −5x+3=0  ⟹  x=35x + 3 - 2x(3) = 0 \implies -5x + 3 = 0 \implies x = \frac{3}{5}.

Adım Adım Çözüm

1
Find the identity element ee under the operation ⋆\star.
e=0e = 0
By definition, a⋆e=a  ⟹  a+e−2ae=a  ⟹  e(1−2a)=0  ⟹  e=0a \star e = a \implies a + e - 2ae = a \implies e(1 - 2a) = 0 \implies e = 0 for all a≠12a \neq \frac{1}{2}.
2
Set up the inverse equation for element xx.
x⋆x−1=0  ⟹  x+x−1−2xx−1=0x \star x^{-1} = 0 \implies x + x^{-1} - 2x x^{-1} = 0
An element operating on its inverse yields the identity element e=0e = 0.
3
Substitute x−1=3x^{-1} = 3 into the inverse equation and solve for xx.
x+3−6x=0  ⟹  −5x+3=0  ⟹  5x=3  ⟹  x=35x + 3 - 6x = 0 \implies -5x + 3 = 0 \implies 5x = 3 \implies x = \frac{3}{5}
Substitute the given inverse value and perform simple linear algebraic rearrangement.

Anahtar Kavram

Identity and Inverse Elements in Binary Operations
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Binary Operations Alıştırma Soruları — JAMB UTME — Sayfa 2 | Examkin