Mat246h1-S – Lec0201/9201 – Lec0101

MAT246H1-S – LEC0201/9201 Concepts in Abstract Mathematics

January 26th, 2021

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

Definition: absolute value of an integer 푛 if 푛 ∈ ℕ For 푛 ∈ ℤ, we define the absolute value of 푛 by |푛| ≔ . { −푛 if 푛 ∈ (−ℕ)

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 2 / 6 Proof. 1 If 푛 ∈ ℕ then |푛| = 푛 ∈ ℕ. If 푛 ∈ (−ℕ) then 푛 = −푚 for some 푚 ∈ ℕ and |푛| = −푛 = −(−푚) = 푚 ∈ ℕ. 2 First case: 푛 ∈ ℕ. Then 푛 ≤ 푛 = |푛|. Second case: 푛 ∈ (−ℕ). Then 푛 ≤ 0 ≤ |푛|. 3 Note that |0| = 0 and that if 푛 ≠ 0 then |푛| ≠ 0. 4 You have to study separately the four cases depending on the signs of 푎 and 푏. 5 If 푏 < 0 then |푎| ≤ 푏 and −푏 ≤ 푎 ≤ 푏 are both false. So we may assume that 푏 ∈ ℕ. Then First case: 푎 ∈ ℕ. Then |푎| ≤ 푏 ⇔ 푎 ≤ 푏 ⇔ −푏 ≤ 푎 ≤ 푏. Second case: 푎 ∈ (−ℕ). Then |푎| ≤ 푏 ⇔ −푎 ≤ 푏 ⇔ −푏 ≤ 푎 ⇔ −푏 ≤ 푎 ≤ 푏. ■

Absolute value – 2 Proposition

1 ∀푛 ∈ ℤ, |푛| ∈ ℕ 2 ∀푛 ∈ ℤ, 푛 ≤ |푛| 3 ∀푛 ∈ ℤ, |푛| = 0 ⇔ 푛 = 0 4 ∀푎, 푏 ∈ ℤ, |푎푏| = |푎||푏| 5 ∀푎, 푏 ∈ ℤ, |푎| ≤ 푏 ⇔ −푏 ≤ 푎 ≤ 푏

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 3 / 6 Absolute value – 2 Proposition

Proof. 1 If 푛 ∈ ℕ then |푛| = 푛 ∈ ℕ. If 푛 ∈ (−ℕ) then 푛 = −푚 for some 푚 ∈ ℕ and |푛| = −푛 = −(−푚) = 푚 ∈ ℕ. 2 First case: 푛 ∈ ℕ. Then 푛 ≤ 푛 = |푛|. Second case: 푛 ∈ (−ℕ). Then 푛 ≤ 0 ≤ |푛|. 3 Note that |0| = 0 and that if 푛 ≠ 0 then |푛| ≠ 0. 4 You have to study separately the four cases depending on the signs of 푎 and 푏. 5 If 푏 < 0 then |푎| ≤ 푏 and −푏 ≤ 푎 ≤ 푏 are both false. So we may assume that 푏 ∈ ℕ. Then First case: 푎 ∈ ℕ. Then |푎| ≤ 푏 ⇔ 푎 ≤ 푏 ⇔ −푏 ≤ 푎 ≤ 푏. Second case: 푎 ∈ (−ℕ). Then |푎| ≤ 푏 ⇔ −푎 ≤ 푏 ⇔ −푏 ≤ 푎 ⇔ −푏 ≤ 푎 ≤ 푏. ■ Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 3 / 6 Euclidean division – 1 Theorem: Euclidean division Given 푎 ∈ ℤ and 푏 ∈ ℤ ⧵ {0}, there exists a unique couple (푞, 푟) ∈ ℤ2 such that

푎 = 푏푞 + 푟 { 0 ≤ 푟 < |푏|

The integers 푞 and 푟 are respectively the quotient and the remainder of the division of 푎 by 푏.

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 4 / 6 Euclidean division – 1 Theorem: Euclidean division Given 푎 ∈ ℤ and 푏 ∈ ℤ ⧵ {0}, there exists a unique couple (푞, 푟) ∈ ℤ2 such that

푎 = 푏푞 + 푟 { 0 ≤ 푟 < |푏|

The integers 푞 and 푟 are respectively the quotient and the remainder of the division of 푎 by 푏.

Existence: First case: 푏 > 0. We set 퐸 = {푝 ∈ ℤ ∶ 푏푝 ≤ 푎}. • 퐸 ≠ ∅, indeed if 0 ≤ 푎 then 0 ∈ 퐸, otherwise 푎 ∈ 퐸. • |푎| is an upper bound of 퐸 (check it). Thus 퐸 is a non-empty subset of ℤ which is bounded from above. Hence it admits a greatest element, i.e. there exists 푞 ∈ 퐸 such that ∀푝 ∈ 퐸, 푝 ≤ 푞. We set 푟 = 푎 − 푏푞. Since 푞 ∈ 퐸, 푟 = 푎 − 푏푞 ≥ 0. And 푞 + 1 ∉ 퐸 since 푞 + 1 > 푞 whereas 푞 is the greatest element of 퐸. Therefore 푏(푞 + 1) > 푎, so 푟 = 푎 − 푏푞 < 푏 = |푏|. We wrote 푎 = 푏푞 + 푟 with 0 ≤ 푟 < |푏| as expected.

푎 = 푏푞 + 푟 { 0 ≤ 푟 < |푏|

The integers 푞 and 푟 are respectively the quotient and the remainder of the division of 푎 by 푏.

Existence: Second case: assume that 푏 < 0. Then we apply the first case to 푎 and −푏 > 0: there exists (푞, 푟) ∈ ℤ2 such that 푎 = −푏푞 + 푟 = 푏(−푞) + 푟 with 0 ≤ 푟 < −푏 = |푏|.

푎 = 푏푞 + 푟 { 0 ≤ 푟 < |푏|

The integers 푞 and 푟 are respectively the quotient and the remainder of the division of 푎 by 푏.

Uniqueness: Let (푞, 푟) and (푞′, 푟′) be two suitable couples. Then 푟′ − 푟 = (푎 − 푏푞′) − (푎 − 푏푞) = 푏(푞 − 푞′). Besides

0 ≤ 푟 < |푏| −|푏| < −푟 ≤ 0 ⟹ ⟹ −|푏| < 푟′ − 푟 < |푏| { 0 ≤ 푟′ < |푏| { 0 ≤ 푟′ < |푏|

Thus −|푏| < 푏(푞 − 푞′) < |푏|, from which we get |푏||푞 − 푞′| = |푏(푞 − 푞′)| < |푏|. Since |푏| > 0, we obtain 0 ≤ |푞 − 푞′| < 1. Therefore |푞 − 푞′| = 0, which implies that 푞 − 푞′ = 0, i.e. 푞 = 푞′. Finally, 푟′ = 푏 − 푎푞′ = 푏 − 푎푞 = 푟. ■

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 4 / 6 Euclidean division – 2 Examples • Division of 22 by 5: 22 = 5 × 4 + 2 The quotient is 푞 = 4 and the remainder is 푟 = 2.

• Division of −22 by 5: −22 = 5 × (−5) + 3 The quotient is 푞 = −5 and the remainder is 푟 = 3.

• Division of 22 by −5: 22 = (−5) × (−4) + 2 The quotient is 푞 = −4 and the remainder is 푟 = 2.

• Division of −22 by −5: −22 = (−5) × 5 + 3 The quotient is 푞 = 5 and the remainder is 푟 = 3.

Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Jan 26, 2021 5 / 6 Proof. Let 푛 ∈ ℤ. By Euclidean division by 2, there exist 푘, 푟 ∈ ℤ such that 푛 = 2푘 + 푟 and 0 ≤ 푟 < 2. Hence either 푟 = 0 or 푟 = 1. And these cases are exclusive by the uniqueness of the Euclidean division. ■

Euclidean division – 3

Proposition: parity of an integer Given 푛 ∈ ℤ, exactly one of the followings occurs: • either 푛 = 2푘 for some 푘 ∈ ℤ (then we say that 푛 is even), • or 푛 = 2푘 + 1 for some 푘 ∈ ℤ (then we say that 푛 is odd).

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

Proof. Let 푛 ∈ ℤ. By Euclidean division by 2, there exist 푘, 푟 ∈ ℤ such that 푛 = 2푘 + 푟 and 0 ≤ 푟 < 2. Hence either 푟 = 0 or 푟 = 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