الأسبوع 6 · العمليات على الكسور الجبرية النسبية: الجمع والطرح والضرب والقسمة والمعكوسان وخواص العمليات · 3 دروس
3
Additive and multiplicative inverses (reciprocal) and the properties of the operations
Mathematics Prep 3 Term 1, Unit 2, Lesson 5
GoalWrite the additive inverse and the multiplicative inverse of a rational function, and use the properties of addition and multiplication.
New words
additive inverse · −f, the function that gives 0 when added to fmultiplicative inverse · 1 ÷ f, the function that gives 1 when multiplied by f (f ≠ 0)commutative · changing the order does not change the result: f + g = g + fidentity · 0 for addition and 1 for multiplication: f + 0 = f and f × 1 = f
Explanation
The additive inverse of p ÷ q is −p ÷ q: for example the additive inverse of (x − 2) ÷ (x + 2) is (2 − x) ÷ (x + 2), because (x − 2) + (2 − x) = 0. The multiplicative inverse (reciprocal) of p ÷ q is q ÷ p, but only when p ≠ 0. Its domain removes the zeros of both p and q, so the reciprocal of (x − 5) ÷ (x − 1) is (x − 1) ÷ (x − 5) with domain R − {1, 5}. Rational functions obey the same properties as numbers, on the domain where they are defined: addition and multiplication are commutative and associative, 0 is the identity for addition and 1 for multiplication, every function has an additive inverse, and every function that is not zero has a multiplicative inverse.
Property
Addition
Multiplication
commutative
f + g = g + f
f × g = g × f
associative
(f + g) + h = f + (g + h)
(f × g) × h = f × (g × h)
identity
f + 0 = f
f × 1 = f
inverse
f + (−f) = 0
f × (1 ÷ f) = 1 (f ≠ 0)
Worked examples
Find the additive inverse of f(x) = (x − 2) ÷ (x + 2) and check. The additive inverse is −f(x) = (2 − x) ÷ (x + 2). Check: (x − 2) ÷ (x + 2) + (2 − x) ÷ (x + 2) = ((x − 2) + (2 − x)) ÷ (x + 2) = 0 ÷ (x + 2) = 0 for every x ≠ −2.
Find the multiplicative inverse of f(x) = (x − 5) ÷ (x − 1) and its domain. The inverse is (x − 1) ÷ (x − 5). It is undefined where x − 5 = 0, and the original f is undefined where x − 1 = 0. Domain R − {1, 5}. Check: f(x) × (1 ÷ f(x)) = (x − 5) ÷ (x − 1) × (x − 1) ÷ (x − 5) = 1.
Let f(x) = (x − 4) ÷ (x − 1). Solve 1 ÷ f(x) = 3. The reciprocal is (x − 1) ÷ (x − 4), domain R − {1, 4}. Solve (x − 1) ÷ (x − 4) = 3: x − 1 = 3x − 12, so 2x = 11 and x = 11/2. It is in the domain. Check: (9/2) ÷ (3/2) = 3 ✓.
Class activity
Property detective. Pick f(x) = 1 ÷ x and g(x) = 1 ÷ (x + 1). Show with fractions that f + g = g + f and f × g = g × f. Which property did you test? Then find f + 0 and g × 1.
Practice
1
Write the additive inverse of (x + 1) ÷ (x − 3).
−(x + 1) ÷ (x − 3) = (−x − 1) ÷ (x − 3). The sum with the original is 0.
2
Find the reciprocal of f(x) = (x + 2) ÷ (x − 1) and its domain.
(x − 1) ÷ (x + 2). Domain: zeros of x + 2 and x − 1 removed: R − {−2, 1}.
3
Let f(x) = (x − 3) ÷ (x + 1). Solve 1 ÷ f(x) = 2.
(x + 1) ÷ (x − 3) = 2, so x + 1 = 2x − 6 and x = 7. Domain R − {−1, 3}. Check: 8 ÷ 4 = 2 ✓.
4
Check f + g = g + f for f = 1 ÷ x and g = 1 ÷ (x + 1).
Both orders give (x + 1 + x) ÷ (x(x + 1)) = (2x + 1) ÷ (x(x + 1)). Domain R − {0, −1}.
5
Name the property: (a) f × (1 ÷ f) = 1; (b) f + 0 = f; (c) (f × g) × h = f × (g × h).
