mthsc 412 section 1.1 the division algorithm
TRANSCRIPT
MTHSC 412 Section 1.1 β The DivisionAlgorithm
Kevin James
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Theorem (Well-Ordering Principle)
Every nonempty set S of nonnegative integers has a least element.That is, there is m β S such that x β S β m β€ x.
Note
The well ordering principle is equivalent to the principle ofmathematical induction.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Theorem (Well-Ordering Principle)
Every nonempty set S of nonnegative integers has a least element.That is, there is m β S such that x β S β m β€ x.
Note
The well ordering principle is equivalent to the principle ofmathematical induction.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Theorem (The Division Algorithm)
Suppose that a, b β Z with b > 0. Then there exist uniqueq, r β Z such that
1 a = bq + r , and
2 0 β€ r < b.
Example
1 Given a = 14 and b = 3, we can write 14 = 3 β 4 + 2. So,q = 4 and r = 2.
2 Given a = β14 and b = 3, we can write β14 = 3 β (β5) + 1.So, q = β5 and r = 1.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Theorem (The Division Algorithm)
Suppose that a, b β Z with b > 0. Then there exist uniqueq, r β Z such that
1 a = bq + r , and
2 0 β€ r < b.
Example
1 Given a = 14 and b = 3, we can write
14 = 3 β 4 + 2. So,q = 4 and r = 2.
2 Given a = β14 and b = 3, we can write β14 = 3 β (β5) + 1.So, q = β5 and r = 1.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Theorem (The Division Algorithm)
Suppose that a, b β Z with b > 0. Then there exist uniqueq, r β Z such that
1 a = bq + r , and
2 0 β€ r < b.
Example
1 Given a = 14 and b = 3, we can write 14 = 3 β 4 + 2. So,q = 4 and r = 2.
2 Given a = β14 and b = 3, we can write β14 = 3 β (β5) + 1.So, q = β5 and r = 1.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Theorem (The Division Algorithm)
Suppose that a, b β Z with b > 0. Then there exist uniqueq, r β Z such that
1 a = bq + r , and
2 0 β€ r < b.
Example
1 Given a = 14 and b = 3, we can write 14 = 3 β 4 + 2. So,q = 4 and r = 2.
2 Given a = β14 and b = 3, we can write
β14 = 3 β (β5) + 1.So, q = β5 and r = 1.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Theorem (The Division Algorithm)
Suppose that a, b β Z with b > 0. Then there exist uniqueq, r β Z such that
1 a = bq + r , and
2 0 β€ r < b.
Example
1 Given a = 14 and b = 3, we can write 14 = 3 β 4 + 2. So,q = 4 and r = 2.
2 Given a = β14 and b = 3, we can write β14 = 3 β (β5) + 1.So, q = β5 and r = 1.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.
Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.
Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.
Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0.
Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .
Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0.
Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1
β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a|
β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βa
β a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.
Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .
So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .
By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.
Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .
Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .
Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note that
r β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b =
= aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).
Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows that
r β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.
Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Proof.
Existence: Let a, b β Z with b > 0.Let S = {aβ bq | q β Z; aβ bq β₯ 0}.Note that S is a subset of the nonnegative integers.Now, note that if a = 0 then aβ b(β1) = b > 0. Thus, b β S .Now, assume a 6= 0. Recall thatb β₯ 1β |a|b β₯ |a| β₯ βaβ a+ b|a| β₯ 0.Thus, aβ b(β|a|) β S .So, S 6= β .By the Well Ordering Principle, S has a smallest element.Let r be the smallest element of S .Then r β₯ 0 and we have for some q β Z, aβbq = r β a = bq+ r .Also, note thatr β b = (aβ bq)β b = = aβ b(q + 1).Since r β b < r and r is the least element of S , it follows thatr β b < 0β r < b.Uniqueness: Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm
Corollary
Let a, c β Z with c 6= 0. Then there exist unique q, r β Z such that
a = cq + r and 0 β€ r < |c |.
Proof.
Exercise.
Kevin James MTHSC 412 Section 1.1 β The Division Algorithm