Math 312, Lecture 2
Zinovy Reichstein
September 11, 2015
Math 312, Lecture 2
September 11, 2015
Mathematical induction
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains n then S contains n + 1.
Math 312, Lecture 2
September 11, 2015
Mathematical induction
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains n then S contains n + 1. Then S contains
every positive integer, i.e., S = N.
Math 312, Lecture 2
September 11, 2015
Mathematical induction
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains n then S contains n + 1. Then S contains
every positive integer, i.e., S = N.
In practice we use this to prove that some property Pn is satisfied by every
positive integer n as follows.
Math 312, Lecture 2
September 11, 2015
Mathematical induction
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains n then S contains n + 1. Then S contains
every positive integer, i.e., S = N.
In practice we use this to prove that some property Pn is satisfied by every
positive integer n as follows.
(i) First we prove that P1 is satisfied. This is called “the base case”.
Math 312, Lecture 2
September 11, 2015
Mathematical induction
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains n then S contains n + 1. Then S contains
every positive integer, i.e., S = N.
In practice we use this to prove that some property Pn is satisfied by every
positive integer n as follows.
(i) First we prove that P1 is satisfied. This is called “the base case”.
(ii) Then we prove that if Pn is satisfied, then Pn+1 is satisfied. This is
called “the induction step”.
Math 312, Lecture 2
September 11, 2015
Mathematical induction
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains n then S contains n + 1. Then S contains
every positive integer, i.e., S = N.
In practice we use this to prove that some property Pn is satisfied by every
positive integer n as follows.
(i) First we prove that P1 is satisfied. This is called “the base case”.
(ii) Then we prove that if Pn is satisfied, then Pn+1 is satisfied. This is
called “the induction step”.
If we can establish (i) and (ii), then property Pn will be true for every
integer n > 1. To see that this proof method is valid, denote the set of
integers n such that Pn is satisfied by S, and use the principle of
mathematical induction to show that S = N.
Math 312, Lecture 2
September 11, 2015
Mathematical induction examples/exercises
1. Show that 1 + 3 + 5 + · · · + (2n − 1) = n2 .
2. Show that 1 + q + q 2 + · · · + q n =
q n+1 − 1
for any real number q 6= 1.
q−1
3. Show that 2n > n2 for any n > 5.
4. Show that n lines in general position subdivide the plane into
n(n + 1)
+ 1 regions.
2
5. Show that n3 − n is divisible by 3 for any n > 1.
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn is valid for
every integer n > d, where d is not necessarily 1.
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn is valid for
every integer n > d, where d is not necessarily 1.
Indeed, we can use the substitution m = n − d + 1 and argue by induction
with respect to m instead of n. Here m > 1 corresponds to n > d.
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn is valid for
every integer n > d, where d is not necessarily 1.
Indeed, we can use the substitution m = n − d + 1 and argue by induction
with respect to m instead of n. Here m > 1 corresponds to n > d. In
practice we don’t do this, but simply use n = d as the base case.
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn is valid for
every integer n > d, where d is not necessarily 1.
Indeed, we can use the substitution m = n − d + 1 and argue by induction
with respect to m instead of n. Here m > 1 corresponds to n > d. In
practice we don’t do this, but simply use n = d as the base case.
For example, yesterday we proved the inequality 2n > n2 for every n > 5.
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn is valid for
every integer n > d, where d is not necessarily 1.
Indeed, we can use the substitution m = n − d + 1 and argue by induction
with respect to m instead of n. Here m > 1 corresponds to n > d. In
practice we don’t do this, but simply use n = d as the base case.
For example, yesterday we proved the inequality 2n > n2 for every n > 5.
Here the base case is n = 5, rather than n = 1.
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn is valid for
every integer n > d, where d is not necessarily 1.
Indeed, we can use the substitution m = n − d + 1 and argue by induction
with respect to m instead of n. Here m > 1 corresponds to n > d. In
practice we don’t do this, but simply use n = d as the base case.
For example, yesterday we proved the inequality 2n > n2 for every n > 5.
Here the base case is n = 5, rather than n = 1.
Why can use n = 5 as the base case here?
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn is valid for
every integer n > d, where d is not necessarily 1.
Indeed, we can use the substitution m = n − d + 1 and argue by induction
with respect to m instead of n. Here m > 1 corresponds to n > d. In
practice we don’t do this, but simply use n = d as the base case.
For example, yesterday we proved the inequality 2n > n2 for every n > 5.
Here the base case is n = 5, rather than n = 1.
Why can use n = 5 as the base case here? Because can make the
substitution m = n − 4. Then n = m + 4, and our inequality becomes
2m+4 > (m + 4)2 for every m > 1.
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn is valid for
every integer n > d, where d is not necessarily 1.
Indeed, we can use the substitution m = n − d + 1 and argue by induction
with respect to m instead of n. Here m > 1 corresponds to n > d. In
practice we don’t do this, but simply use n = d as the base case.
For example, yesterday we proved the inequality 2n > n2 for every n > 5.
Here the base case is n = 5, rather than n = 1.
Why can use n = 5 as the base case here? Because can make the
substitution m = n − 4. Then n = m + 4, and our inequality becomes
2m+4 > (m + 4)2 for every m > 1. The base case is now m = 1, and the
usual principle of mathematical induction applies.
Math 312, Lecture 2
September 11, 2015
Starting from n = d, instead of n = 1
Mathematical induction can be used to prove that property Pn is valid for
every integer n > d, where d is not necessarily 1.
Indeed, we can use the substitution m = n − d + 1 and argue by induction
with respect to m instead of n. Here m > 1 corresponds to n > d. In
practice we don’t do this, but simply use n = d as the base case.
For example, yesterday we proved the inequality 2n > n2 for every n > 5.
Here the base case is n = 5, rather than n = 1.
Why can use n = 5 as the base case here? Because can make the
substitution m = n − 4. Then n = m + 4, and our inequality becomes
2m+4 > (m + 4)2 for every m > 1. The base case is now m = 1, and the
usual principle of mathematical induction applies.
In practice we don’t make this substitution, but simply use n = 5 as the
base case.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction
The following variant on the principle of mathematical induction is often
convenient.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction
The following variant on the principle of mathematical induction is often
convenient.
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains every integer 6 n then S contains n + 1.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction
The following variant on the principle of mathematical induction is often
convenient.
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains every integer 6 n then S contains n + 1.
Then S contains every positive integer, i.e., S = N.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction
The following variant on the principle of mathematical induction is often
convenient.
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains every integer 6 n then S contains n + 1.
Then S contains every positive integer, i.e., S = N.
In practice we use this to prove that some property Pn is satisfied by every
positive integer n as follows.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction
The following variant on the principle of mathematical induction is often
convenient.
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains every integer 6 n then S contains n + 1.
Then S contains every positive integer, i.e., S = N.
In practice we use this to prove that some property Pn is satisfied by every
positive integer n as follows.
Base case: First we prove that P1 is satisfied.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction
The following variant on the principle of mathematical induction is often
convenient.
Suppose a set S of positive integers (i) contains 1, and (ii) has the
property that if S contains every integer 6 n then S contains n + 1.
Then S contains every positive integer, i.e., S = N.
In practice we use this to prove that some property Pn is satisfied by every
positive integer n as follows.
Base case: First we prove that P1 is satisfied.
Induction step: Then we prove that if P1 , . . . , Pn are all satisfied, then
Pn+1 is satisfied as well.
Math 312, Lecture 2
September 11, 2015
The well-ordering principle, the principle of mathematical induction
Math 312, Lecture 2
September 11, 2015
The well-ordering principle, the principle of mathematical induction and the
principle of strong mathematical induction
Math 312, Lecture 2
September 11, 2015
The well-ordering principle, the principle of mathematical induction and the
principle of strong mathematical induction are all equivalent to each other.
Math 312, Lecture 2
September 11, 2015
The well-ordering principle, the principle of mathematical induction and the
principle of strong mathematical induction are all equivalent to each other.
The textbook deduces the principles of mathematical induction
Math 312, Lecture 2
September 11, 2015
The well-ordering principle, the principle of mathematical induction and the
principle of strong mathematical induction are all equivalent to each other.
The textbook deduces the principles of mathematical induction (both
weak and strong) from the well ordering principle in Section 1.3.
Math 312, Lecture 2
September 11, 2015
The well-ordering principle, the principle of mathematical induction and the
principle of strong mathematical induction are all equivalent to each other.
The textbook deduces the principles of mathematical induction (both
weak and strong) from the well ordering principle in Section 1.3.
I will not reproduce these proofs in class.
Math 312, Lecture 2
September 11, 2015
The well-ordering principle, the principle of mathematical induction and the
principle of strong mathematical induction are all equivalent to each other.
The textbook deduces the principles of mathematical induction (both
weak and strong) from the well ordering principle in Section 1.3.
I will not reproduce these proofs in class. Our emphasis will be on applying
these principles to specific examples,
Math 312, Lecture 2
September 11, 2015
The well-ordering principle, the principle of mathematical induction and the
principle of strong mathematical induction are all equivalent to each other.
The textbook deduces the principles of mathematical induction (both
weak and strong) from the well ordering principle in Section 1.3.
I will not reproduce these proofs in class. Our emphasis will be on applying
these principles to specific examples, i.e., on practicing induction proofs.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 1
Exercise 1: Any integer amount of postage of 8 cents or more, can be paid
using only 3-cent and 5-cent stamps.
Solution: We denote the amount of postage to be paid by n, and use
(strong) induction.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 1
Exercise 1: Any integer amount of postage of 8 cents or more, can be paid
using only 3-cent and 5-cent stamps.
Solution: We denote the amount of postage to be paid by n, and use
(strong) induction.
Base case: n = 8. Here we can use one 5-cent stamp and one 3-cent
stamps.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 1
Exercise 1: Any integer amount of postage of 8 cents or more, can be paid
using only 3-cent and 5-cent stamps.
Solution: We denote the amount of postage to be paid by n, and use
(strong) induction.
Base case: n = 8. Here we can use one 5-cent stamp and one 3-cent
stamps. That is, 8 = 5 + 3.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 1
Exercise 1: Any integer amount of postage of 8 cents or more, can be paid
using only 3-cent and 5-cent stamps.
Solution: We denote the amount of postage to be paid by n, and use
(strong) induction.
Base case: n = 8. Here we can use one 5-cent stamp and one 3-cent
stamps. That is, 8 = 5 + 3.
For the subsequent argument it will also be helpful for us to verify that
n = 9 and n = 10 cents can be paid.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 1
Exercise 1: Any integer amount of postage of 8 cents or more, can be paid
using only 3-cent and 5-cent stamps.
Solution: We denote the amount of postage to be paid by n, and use
(strong) induction.
Base case: n = 8. Here we can use one 5-cent stamp and one 3-cent
stamps. That is, 8 = 5 + 3.
For the subsequent argument it will also be helpful for us to verify that
n = 9 and n = 10 cents can be paid. Indeed,
9 = 3 + 3 + 3 and 10 = 5 + 5.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 1
Exercise 1: Any integer amount of postage of 8 cents or more, can be paid
using only 3-cent and 5-cent stamps.
Solution: We denote the amount of postage to be paid by n, and use
(strong) induction.
Base case: n = 8. Here we can use one 5-cent stamp and one 3-cent
stamps. That is, 8 = 5 + 3.
For the subsequent argument it will also be helpful for us to verify that
n = 9 and n = 10 cents can be paid. Indeed,
9 = 3 + 3 + 3 and 10 = 5 + 5.
Induction step: Here we assume that n > 10, and that any amount
between 8 and n can be paid.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 1
Exercise 1: Any integer amount of postage of 8 cents or more, can be paid
using only 3-cent and 5-cent stamps.
Solution: We denote the amount of postage to be paid by n, and use
(strong) induction.
Base case: n = 8. Here we can use one 5-cent stamp and one 3-cent
stamps. That is, 8 = 5 + 3.
For the subsequent argument it will also be helpful for us to verify that
n = 9 and n = 10 cents can be paid. Indeed,
9 = 3 + 3 + 3 and 10 = 5 + 5.
Induction step: Here we assume that n > 10, and that any amount
between 8 and n can be paid. In particular, n − 2 cents can be paid,
because 8 6 n − 2 6 n. Now, to pay n + 1 cents, first pay n − 2 and then
add one more 3-cent stamp.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 2
Exercise 2. The nth Fibonacci number an is defined by the recursive
formula a0 = 0, a1 = 1, an+2 = an+1 + an for every n > 0. Show that
1
an = √ (αn − β n )
5
for any n > 1. Here
√
√
1+ 5
1− 5
α=
and β =
.
2
2
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 2
Exercise 2. The nth Fibonacci number an is defined by the recursive
formula a0 = 0, a1 = 1, an+2 = an+1 + an for every n > 0. Show that
1
an = √ (αn − β n )
5
for any n > 1. Here
√
√
1+ 5
1− 5
α=
and β =
.
2
2
Hints for the proof: Note that α and β are the roots of the quadratic
equation x 2 − x − 1 = 0.
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 2
Exercise 2. The nth Fibonacci number an is defined by the recursive
formula a0 = 0, a1 = 1, an+2 = an+1 + an for every n > 0. Show that
1
an = √ (αn − β n )
5
for any n > 1. Here
√
√
1+ 5
1− 5
α=
and β =
.
2
2
Hints for the proof: Note that α and β are the roots of the quadratic
equation x 2 − x − 1 = 0. That is, α2 = α + 1 and β 2 = β + 1
Math 312, Lecture 2
September 11, 2015
Strong mathematical induction, Exercise 2
Exercise 2. The nth Fibonacci number an is defined by the recursive
formula a0 = 0, a1 = 1, an+2 = an+1 + an for every n > 0. Show that
1
an = √ (αn − β n )
5
for any n > 1. Here
√
√
1+ 5
1− 5
α=
and β =
.
2
2
Hints for the proof: Note that α and β are the roots of the quadratic
equation x 2 − x − 1 = 0. That is, α2 = α + 1 and β 2 = β + 1 Multiplying
both sides of the firt equation by αn−1 we obtain αn+1 = αn + αn−1 .
Similarly, β n+1 = β n + β n−1 for every integer n.
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n.
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n. Here we assume that a 6= 0, but allow
b to be an arbitrary integer.
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n. Here we assume that a 6= 0, but allow
b to be an arbitrary integer.
The following expressions all mean the same thing:
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n. Here we assume that a 6= 0, but allow
b to be an arbitrary integer.
The following expressions all mean the same thing:
a divides b,
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n. Here we assume that a 6= 0, but allow
b to be an arbitrary integer.
The following expressions all mean the same thing:
a divides b,
a evenly divides b,
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n. Here we assume that a 6= 0, but allow
b to be an arbitrary integer.
The following expressions all mean the same thing:
a divides b,
a evenly divides b,
b is divisible by a,
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n. Here we assume that a 6= 0, but allow
b to be an arbitrary integer.
The following expressions all mean the same thing:
a divides b,
a evenly divides b,
b is divisible by a,
b is a multiple of a.
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n. Here we assume that a 6= 0, but allow
b to be an arbitrary integer.
The following expressions all mean the same thing:
a divides b,
a evenly divides b,
b is divisible by a,
b is a multiple of a.
We say that an integer n is a prime if n > 2 and n is not divisible by any
positive integer, other than 1 and n itself.
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n. Here we assume that a 6= 0, but allow
b to be an arbitrary integer.
The following expressions all mean the same thing:
a divides b,
a evenly divides b,
b is divisible by a,
b is a multiple of a.
We say that an integer n is a prime if n > 2 and n is not divisible by any
positive integer, other than 1 and n itself. Integers n > 2 that are not
prime are called composite.
Math 312, Lecture 2
September 11, 2015
Divisibility
We say that an integer a divides another integer b, if b/a is an integer.
That is, b = an for some integer n. Here we assume that a 6= 0, but allow
b to be an arbitrary integer.
The following expressions all mean the same thing:
a divides b,
a evenly divides b,
b is divisible by a,
b is a multiple of a.
We say that an integer n is a prime if n > 2 and n is not divisible by any
positive integer, other than 1 and n itself. Integers n > 2 that are not
prime are called composite.
For example, 2, 3, 5, 7, 11, 13 are all primes, but 6 = 2 · 3 and 9 = 3 · 3 are
composite.
Math 312, Lecture 2
September 11, 2015
Theorem: Every integer n > 2 is either a prime or a product of two
or more primes.
Math 312, Lecture 2
September 11, 2015
Theorem: Every integer n > 2 is either a prime or a product of two
or more primes.
This is readily proved by strong induction on n. (Check!)
Math 312, Lecture 2
September 11, 2015
Theorem: Every integer n > 2 is either a prime or a product of two
or more primes.
This is readily proved by strong induction on n. (Check!)
Note however that inductive proof does not give an algorithm for writing a
given integer as a product of primes.
Math 312, Lecture 2
September 11, 2015
Theorem: Every integer n > 2 is either a prime or a product of two
or more primes.
This is readily proved by strong induction on n. (Check!)
Note however that inductive proof does not give an algorithm for writing a
given integer as a product of primes.
Such algorithms are in short supply and are of crucial importance in
cryptography.
Math 312, Lecture 2
September 11, 2015
Euclid of Alexandria, approx. 300 BC
Math 312, Lecture 2
September 11, 2015
Euclid of Alexandria, approx. 300 BC
The theorem on the previous slide
Math 312, Lecture 2
September 11, 2015
Euclid of Alexandria, approx. 300 BC
The theorem on the previous slide is
stated and proved in Euclid’s
“Elements”,
Math 312, Lecture 2
September 11, 2015
Euclid of Alexandria, approx. 300 BC
The theorem on the previous slide is
stated and proved in Euclid’s
“Elements”, arguably the most
influential mathematics book ever
written. This book was used
Math 312, Lecture 2
September 11, 2015
Euclid of Alexandria, approx. 300 BC
The theorem on the previous slide is
stated and proved in Euclid’s
“Elements”, arguably the most
influential mathematics book ever
written. This book was used as a
textbook since 300 B.C. into the
20th century.
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer.
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that n = qa + r and 0 6 r 6 a − 1. Moreover
the integers q and r with these properties are uniquely defined.
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that n = qa + r and 0 6 r 6 a − 1. Moreover
the integers q and r with these properties are uniquely defined.
Remarks: (1) q is usually called the quotient, and r is called the remainder.
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that n = qa + r and 0 6 r 6 a − 1. Moreover
the integers q and r with these properties are uniquely defined.
Remarks: (1) q is usually called the quotient, and r is called the remainder.
(2) n is divisible by a if and only if r = 0.
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that n = qa + r and 0 6 r 6 a − 1. Moreover
the integers q and r with these properties are uniquely defined.
Remarks: (1) q is usually called the quotient, and r is called the remainder.
(2) n is divisible by a if and only if r = 0. This follows from uniqueness of
q and r .
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that n = qa + r and 0 6 r 6 a − 1. Moreover
the integers q and r with these properties are uniquely defined.
Remarks: (1) q is usually called the quotient, and r is called the remainder.
(2) n is divisible by a if and only if r = 0. This follows from uniqueness of
q and r .
(2) The division algorithm is a theorem, not an algorithm.
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that n = qa + r and 0 6 r 6 a − 1. Moreover
the integers q and r with these properties are uniquely defined.
Remarks: (1) q is usually called the quotient, and r is called the remainder.
(2) n is divisible by a if and only if r = 0. This follows from uniqueness of
q and r .
(2) The division algorithm is a theorem, not an algorithm. One of the
algorithms for finding q and r (for given n and a)
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that n = qa + r and 0 6 r 6 a − 1. Moreover
the integers q and r with these properties are uniquely defined.
Remarks: (1) q is usually called the quotient, and r is called the remainder.
(2) n is divisible by a if and only if r = 0. This follows from uniqueness of
q and r .
(2) The division algorithm is a theorem, not an algorithm. One of the
algorithms for finding q and r (for given n and a) is called “long division”.
It usually assumes that n > 0.
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that n = qa + r and 0 6 r 6 a − 1. Moreover
the integers q and r with these properties are uniquely defined.
Remarks: (1) q is usually called the quotient, and r is called the remainder.
(2) n is divisible by a if and only if r = 0. This follows from uniqueness of
q and r .
(2) The division algorithm is a theorem, not an algorithm. One of the
algorithms for finding q and r (for given n and a) is called “long division”.
It usually assumes that n > 0.
(3) Existence of p and q is usually proved by using the well-ordering
principle.
Math 312, Lecture 2
September 11, 2015
The division algorithm
Theorem: Let a > 1 be an integer. For integer n there exist two
integers, q and r such that n = qa + r and 0 6 r 6 a − 1. Moreover
the integers q and r with these properties are uniquely defined.
Remarks: (1) q is usually called the quotient, and r is called the remainder.
(2) n is divisible by a if and only if r = 0. This follows from uniqueness of
q and r .
(2) The division algorithm is a theorem, not an algorithm. One of the
algorithms for finding q and r (for given n and a) is called “long division”.
It usually assumes that n > 0.
(3) Existence of p and q is usually proved by using the well-ordering
principle.
n
(4) Note that for n > 0 the division algorithm is equivalent to writing as
a
r
a mixed fraction q + .
a
Math 312, Lecture 2
September 11, 2015
Examples of division with remainder
1. n = 30 and a = 7. What are q and r in this case?
Math 312, Lecture 2
September 11, 2015
Examples of division with remainder
1. n = 30 and a = 7. What are q and r in this case?
Answer: q = 4 and r = 2, 30 = 4 · 7 + 2.
Math 312, Lecture 2
September 11, 2015
Examples of division with remainder
1. n = 30 and a = 7. What are q and r in this case?
Answer: q = 4 and r = 2, 30 = 4 · 7 + 2.
2. n = 100 and a = 20. What are q and r ?
Math 312, Lecture 2
September 11, 2015
Examples of division with remainder
1. n = 30 and a = 7. What are q and r in this case?
Answer: q = 4 and r = 2, 30 = 4 · 7 + 2.
2. n = 100 and a = 20. What are q and r ?
Answer: q = 5 and r = 0, 100 = 5 · 20 + 0.
Math 312, Lecture 2
September 11, 2015
Examples of division with remainder
1. n = 30 and a = 7. What are q and r in this case?
Answer: q = 4 and r = 2, 30 = 4 · 7 + 2.
2. n = 100 and a = 20. What are q and r ?
Answer: q = 5 and r = 0, 100 = 5 · 20 + 0.
3. n = −17 and a = 4. What are q and r ?
Math 312, Lecture 2
September 11, 2015
Examples of division with remainder
1. n = 30 and a = 7. What are q and r in this case?
Answer: q = 4 and r = 2, 30 = 4 · 7 + 2.
2. n = 100 and a = 20. What are q and r ?
Answer: q = 5 and r = 0, 100 = 5 · 20 + 0.
3. n = −17 and a = 4. What are q and r ?
Answer: q = −5 and r = 3, −17 = (−5) · 4 + 3.
Math 312, Lecture 2
September 11, 2015