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
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
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
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
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
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