mthsc 412 section 1.1 the division algorithm
TRANSCRIPT
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MTHSC 412 Section 1.1 – The DivisionAlgorithm
Kevin James
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
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Theorem (Well-Ordering Principle)
Every nonempty set S of nonnegative integers has a least element.That is, there is m ∈ S such that x ∈ S ⇒ m ≤ x.
Note
The well ordering principle is equivalent to the principle ofmathematical induction.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 3: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/3.jpg)
Theorem (Well-Ordering Principle)
Every nonempty set S of nonnegative integers has a least element.That is, there is m ∈ S such that x ∈ S ⇒ m ≤ x.
Note
The well ordering principle is equivalent to the principle ofmathematical induction.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 4: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/4.jpg)
Theorem (The Division Algorithm)
Suppose that a, b ∈ Z with b > 0. Then there exist uniqueq, r ∈ Z such that
1 a = bq + r , and
2 0 ≤ r < b.
Example
1 Given a = 14 and b = 3, we can write 14 = 3 ∗ 4 + 2. So,q = 4 and r = 2.
2 Given a = −14 and b = 3, we can write −14 = 3 ∗ (−5) + 1.So, q = −5 and r = 1.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 5: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/5.jpg)
Theorem (The Division Algorithm)
Suppose that a, b ∈ Z with b > 0. Then there exist uniqueq, r ∈ Z such that
1 a = bq + r , and
2 0 ≤ r < b.
Example
1 Given a = 14 and b = 3, we can write
14 = 3 ∗ 4 + 2. So,q = 4 and r = 2.
2 Given a = −14 and b = 3, we can write −14 = 3 ∗ (−5) + 1.So, q = −5 and r = 1.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 6: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/6.jpg)
Theorem (The Division Algorithm)
Suppose that a, b ∈ Z with b > 0. Then there exist uniqueq, r ∈ Z such that
1 a = bq + r , and
2 0 ≤ r < b.
Example
1 Given a = 14 and b = 3, we can write 14 = 3 ∗ 4 + 2. So,q = 4 and r = 2.
2 Given a = −14 and b = 3, we can write −14 = 3 ∗ (−5) + 1.So, q = −5 and r = 1.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 7: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/7.jpg)
Theorem (The Division Algorithm)
Suppose that a, b ∈ Z with b > 0. Then there exist uniqueq, r ∈ Z such that
1 a = bq + r , and
2 0 ≤ r < b.
Example
1 Given a = 14 and b = 3, we can write 14 = 3 ∗ 4 + 2. So,q = 4 and r = 2.
2 Given a = −14 and b = 3, we can write
−14 = 3 ∗ (−5) + 1.So, q = −5 and r = 1.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 8: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/8.jpg)
Theorem (The Division Algorithm)
Suppose that a, b ∈ Z with b > 0. Then there exist uniqueq, r ∈ Z such that
1 a = bq + r , and
2 0 ≤ r < b.
Example
1 Given a = 14 and b = 3, we can write 14 = 3 ∗ 4 + 2. So,q = 4 and r = 2.
2 Given a = −14 and b = 3, we can write −14 = 3 ∗ (−5) + 1.So, q = −5 and r = 1.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 9: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/9.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.
Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 10: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/10.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.
Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 11: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/11.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.
Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 12: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/12.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0.
Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
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Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .
Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
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Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0.
Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
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Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1
⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 16: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/16.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a|
≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 17: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/17.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a
⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 18: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/18.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.
Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 19: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/19.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .
So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
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Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.
By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 21: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/21.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.
Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 22: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/22.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .
Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 23: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/23.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .
Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 24: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/24.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note that
r − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 25: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/25.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b =
= a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 26: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/26.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).
Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
![Page 27: MTHSC 412 Section 1.1 The Division Algorithm](https://reader033.vdocument.in/reader033/viewer/2022042403/625d87e53dbf6336b742ace9/html5/thumbnails/27.jpg)
Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows that
r − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
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Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.
Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
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Proof.
Existence: Let a, b ∈ Z with b > 0.Let S = {a− bq | q ∈ Z; a− bq ≥ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then a− b(−1) = b > 0. Thus, b ∈ S .Now, assume a 6= 0. Recall thatb ≥ 1⇒ |a|b ≥ |a| ≥ −a⇒ a+ b|a| ≥ 0.Thus, a− b(−|a|) ∈ S .So, S 6= ∅.By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r ≥ 0 and we have for some q ∈ Z, a−bq = r ⇒ a = bq+ r .Also, note thatr − b = (a− bq)− b = = a− b(q + 1).Since r − b < r and r is the least element of S , it follows thatr − b < 0⇒ r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm
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Corollary
Let a, c ∈ Z with c 6= 0. Then there exist unique q, r ∈ Z such that
a = cq + r and 0 ≤ r < |c |.
Proof.
Exercise.
Kevin James MTHSC 412 Section 1.1 – The Division Algorithm