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January26 ,2021 EUCLIDEANDIVISION ConceptsinAbstractMathematics

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MAT246H1-S – LEC0201/9201

Concepts in Abstract Mathematics

EUCLIDEAN DIVISION

January 26th, 2021

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 1 / 6

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Absolute value – 1

Definition: absolute value of an integer

For𝑛 ∈ ℤ, we define theabsolute value of𝑛by|𝑛| ≔ {

𝑛 if𝑛 ∈ ℕ

−𝑛 if𝑛 ∈ (−ℕ) .

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 2 / 6

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Absolute value – 2

Proposition

1 ∀𝑛 ∈ ℤ, |𝑛| ∈ ℕ

2 ∀𝑛 ∈ ℤ, 𝑛 ≤ |𝑛|

3 ∀𝑛 ∈ ℤ, |𝑛| = 0 ⇔ 𝑛 = 0

4 ∀𝑎, 𝑏 ∈ ℤ, |𝑎𝑏| = |𝑎||𝑏|

5 ∀𝑎, 𝑏 ∈ ℤ, |𝑎| ≤ 𝑏 ⇔ −𝑏 ≤ 𝑎 ≤ 𝑏

Proof.

1 If𝑛 ∈ ℕthen|𝑛| = 𝑛 ∈ ℕ.

If𝑛 ∈ (−ℕ)then𝑛 = −𝑚for some𝑚 ∈ ℕand|𝑛| = −𝑛 = −(−𝑚) = 𝑚 ∈ ℕ.

2 First case:𝑛 ∈ ℕ.Then𝑛 ≤ 𝑛 = |𝑛|.

Second case:𝑛 ∈ (−ℕ). Then𝑛 ≤ 0 ≤ |𝑛|.

3 Note that|0| = 0and that if𝑛 ≠ 0then|𝑛| ≠ 0.

4 You have to study separately the four cases depending on the signs of𝑎and𝑏.

5 If𝑏 < 0then|𝑎| ≤ 𝑏and−𝑏 ≤ 𝑎 ≤ 𝑏are both false. So we may assume that𝑏 ∈ ℕ. Then First case:𝑎 ∈ ℕ.Then|𝑎| ≤ 𝑏 ⇔ 𝑎 ≤ 𝑏 ⇔ −𝑏 ≤ 𝑎 ≤ 𝑏.

Second case:𝑎 ∈ (−ℕ). Then|𝑎| ≤ 𝑏 ⇔ −𝑎 ≤ 𝑏 ⇔ −𝑏 ≤ 𝑎 ⇔ −𝑏 ≤ 𝑎 ≤ 𝑏. ■

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 3 / 6

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Euclidean division – 1

Theorem: Euclidean division

Given𝑎 ∈ ℤand𝑏 ∈ ℤ ⧵ {0}, there exists a unique couple(𝑞, 𝑟) ∈ ℤ2such that

{

𝑎 = 𝑏𝑞 + 𝑟 0 ≤ 𝑟 < |𝑏|

The integers𝑞and𝑟are respectively thequotientand theremainderof the division of𝑎by𝑏.

The proof doesn’t appear in the handout: see either the slides or the lecture notes.

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 4 / 6

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Euclidean division – 2

Examples

• Division of22by5:

22 = 5 × 4 + 2 The quotient is𝑞 = 4and the remainder is𝑟 = 2.

• Division of−22by5:

−22 = 5 × (−5) + 3 The quotient is𝑞 = −5and the remainder is𝑟 = 3.

• Division of22by−5:

22 = (−5) × (−4) + 2 The quotient is𝑞 = −4and the remainder is𝑟 = 2.

• Division of−22by−5:

−22 = (−5) × 5 + 3 The quotient is𝑞 = 5and the remainder is𝑟 = 3.

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 5 / 6

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Euclidean division – 3

Proposition: parity of an integer

Given𝑛 ∈ ℤ, exactly one of the followings occurs:

• either𝑛 = 2𝑘for some𝑘 ∈ ℤ(then we say that𝑛is even),

• or𝑛 = 2𝑘 + 1for some𝑘 ∈ ℤ(then we say that𝑛is odd).

Proof.Let𝑛 ∈ ℤ.

By Euclidean division by2, there exist𝑘, 𝑟 ∈ ℤsuch that𝑛 = 2𝑘 + 𝑟and0 ≤ 𝑟 < 2.

Hence either𝑟 = 0or𝑟 = 1.

And these cases are exclusive by the uniqueness of the Euclidean division. ■

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 6 / 6

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