MATHEMATICS ... bwhere a and b are whole numbers and b (Ï 0. ... In some books, you can find the expression equivalent fractions for this statement, but the.
LLG Paris-Abu Dhabi Advanced Math and Science Pilot Class
MATHEMATICS Gr11 Lesson 1.
Rationals numbers
De…nition:
a where a and b are whole numbers and b 6= 0: b Notation: We will write Q the set of rational numbers.
A fraction is a ratio
Examples: 1 ; 0, 2
7, 1245 are rational numbers
Whole numbers are rationals. We write Z
Q
Decimal numbers are also rationals. (ex: 2:35683 =
235683 ) 100000
There are numbers even ratios that are not rationals numbers (ex
p
2 ) 2
Equality of two fractions De…nition: a A fraction
b
equals another one
c if ad = bc d
In some books, you can …nd the expression equivalent fractions for this statement, but the equivalence is a reserved word for another use that we will see later.
Examples: 30
3 = 90 and 45
15 = 3
5
2 = 90 therefore
30 2 equals 45 3
Reduced form De…nition: a A fraction
b
is called reduced when a and b don’t have any common divisor.
Examples: 2 30 is reduced contrary to . 3 45 30 2 2 30 2 is reduced and = so is reduced form of 3 45 3 3 45
Remark:
Simpli…cation involves lowering terms - cancelling common factors or divisors on top and bottom.
Examples: 30 2 = 45 3
15 2 = 15 3
7 420 2 5 6 7 = = 360 5 18 4 6 Every time we can do it, we will write a fraction with its reduced form. This form is very useful to compare fractions, and operate on them.
Multiplication Because of the commutativity of the operations
a a k a ka = = and b b b b
k
Examples:
7 3
4 28 = 9 27
and
12 5
4 48 3 = = 9 45 3
, we have : c ac = d bd
16 16 = 15 15
Remark: As we can see in the previous example, the explained method for multiplying fractions is not optimized. We could use commutativity of multiplication and division to simplify more quickly the calculation as follow : 12 4 3 4 4 4 4 16 = = = 5 9 5 3 3 5 3 15
Inverse of a fraction
De…nition: Inverse of a fraction :
a b is :0 doesn’t have any inverse. b a b ab = =1 a ab
Assume that a 6= 0; and b 6= 0 the inverse of a b
Examples: The inverse of
5 3 is . 3 5
Division Divide by a number is equivalent to multiply by the inverse of this number, so it comes a b
=
a b
c a b a c =b d
Example:
3 4 7 5
=
3 4
1 a = c bc d ad = c bc
5 15 = 7 28
Addition
a c a+c The value of the sum of fractions with the same denominator is obvious : + = b b b a c So we can easily add any fractions and by converting this fractions to forms based on the b d same denominator bd before adding them. It comes a c ad bc ad + bc + = + = b d bd bd bd
As we can see in the previous example, the explained method for adding fractions is not optimize. 2 3 We could write the terms and with the common denominator 12 which is the low common 6 4 multiple of 6 and 4. (written lcm (6; 4)) In order to optimized the calculation, it’s often useful to simply the fractions before to do anything else.
Annex 1:
p
2 is not a rational number.
First, let’s prove a small lemma: Lemma: For all whole number a, if a2 is even, then a is even. Proof: Let consider a a whole number such as a2 is even. Assume that a is odd, number and 1. Now if a2 is even and a odd, then a = 2k + 1 with k a whole number. Then a2 = (2k + 1)2 = (2k + 1)
(2k + 1) = 2k
(2k + 1) = 4k 2 + 4k + 1 = 4 k 2 + k + 1
(2k + 1) + 1
and a is odd as sum of an odd a2 is odd too ! =)(= So our hypothesis was false: a cannot be odd.
p 2is rational. a and b two whole numbers with no common divisors Assume that p Then 2there are p a 2 verifying 2 = . So a = 2b and a = 2b : Therefore a2 is even and by application of the lemma, b a is even too. a can be written 2k, with k 2 Z and it follows (2k)2 = 2b2 so 4k 2 = 2b2 and by simpli…cation, 2k 2 = b2 which, by an analog p way prove that b is even. Now we have written 2 as a quotient of 2 even numbers and therefore the fraction can be reduced.=)(= p So our hypothesis was false and 2 cannot be a rational number.
Annex 2: Stability of the rational numbers set under the algebraic operations. Remark that if a 2 Q and b 2 Q then a + b 2 Q, a
a + b. ââ. BA = b a + b. ââ. AB. We just prove the following theorem: Theorem: If a + b = 0, there exists a unique point G satisfying a. ââ. AG + b. âââ. BG = ââ.
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