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Chapter 8: Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8 Lecture Notes
Rational Numbers and Irrational Numbers
Winter 2017
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
8.1: Rational Numbers
8.2: Irrational Numbers
Chapter 8: Rational Numbers and Irrational Numbers
The Set of Rational Numbers
Definition 8.1.1: a rational number is any number of the form
m
,
n
with m, n ∈ Z and n 6= 0. So every integer m is an example of a
rational number, just take n = 1. Some examples of non-integer
rational numbers are
1 −2 15 45
, , .
,
3 3 7 21
Actually the last two numbers represent the same rational number,
if you ‘cancel the 3.’
m1
m2
Definition 8.1.2: the two rational numbers
and
are equal
n1
n2
if m1 · n2 = m2 · n1 .
Notation: the set of all rational numbers is denoted by Q.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
8.1: Rational Numbers
8.2: Irrational Numbers
Chapter 8: Rational Numbers and Irrational Numbers
Operations with Rational Numbers
Addition:
m1 m2
m1 · n2 + n1 · m2
+
=
.
n1
n2
n1 · n2
Multiplication:
m1
n1
m2
n2
=
m1 · m2
.
n1 · n2
For example,
2
7
14
2
7
31
=
and
+
= .
3
5
15
3
5
15
Definition 8.1.5: x is the multiplicative inverse of y if x · y = 1.
m
n
Theorem 8.1.6: if m 6= 0 the multiplicative inverse of
is .
n
m
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Polynomials and Rational Roots
Suppose a0 , a1 , . . . , an are integers, and an 6= 0. The expression
an x n + an−1 x n−1 + · · · + a2 x 2 + a1 x + a0
is called a polynomial with integral coefficients. A number x0 is a
root of the polynomial if substituting x = x0 into the polynomial
and simplifying, results in the value zero.
4
is a root of the polynomial 25x 2 − 10x − 8 since
5
2
4
4
25
− 10
− 8 = 16 − 8 − 8 = 0.
5
5
Example 1:
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Example 2
The polynomial x 5 + x − 1 has no rational roots.
m
is a root and gcd(m, n) = 1. Then
Solution: suppose
n
gcd(m, n5 ) = 1 and gcd(n, m5 ) = 1 as well. (Why?) Observe
m 5
n
+
m
− 1 = 0 ⇔ m5 + m · n4 − n5 = 0.
n
So
I
m(m4 + n4 ) = n5 ⇒ m | n5 ⇒ m = ±1.
I
m5 = n(n4 − mn3 ) ⇒ n | m5 ⇒ n = ±1.
m
Thus the only possible values for
are ±1, neither of which
n
satisfies the polynomial.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
The Rational Roots Theorem
We can generalize the previous two examples.
m
is a rational root of the polynomial
Theorem 8.1.9: if
n
ak x k + ak−1 x k−1 + · · · + a2 x 2 + a1 x + a0
with integer coefficients, and m and n are relatively prime, then
m | a0 and n | ak .
Proof: as in Example 2, substitute x = m/n, set the polynomial
equal to zero, and multiply through by nk to obtain
ak mk + ak−1 mk−1 n + · · · + a1 mnk−1 + a0 nk = 0.
This equation can be rearranged in two ways:
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
1. m(ak mk−1 + ak−1 mk−2 n + · · · + a1 nk−1 ) = −a0 nk .
Since m divides the left side, m divides the right side. Since
gcd(m, nk ) = 1, m | −a0 . This means m = −1 or m | a0 ;
either way, m | a0 .
2. ak mk = −(ak−1 mk−1 + · · · + a1 mnk−2 + a0 nk−1 ) n.
Since n divides the right side, n divides the left side. Since
gcd(n, mk ) = 1, n | ak ,
which completes the proof.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Example 3
Find all rational roots, if any, of the polynomial
2x 3 − 5x 2 + 14x − 35.
m
Solution: let x = , in lowest terms. By Theorem 8.1.9 we must
n
have
m | 35 and n | 2.
This means
m = ±1, ± 5, ± 7, ± 35, and n = ±1, ± 2,
giving sixteen possibilities for x :
1
5
7
35
x = ±1, ± , ± 5, ± , ± 7, ± , ± 35, ± .
2
2
2
2
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Of these sixteen possibilities, only x =
5
is a root of
2
2x 3 − 5x 2 + 14x − 35,
as you can check.
Aside: as we shall see, the cubic polynomial does have two other
roots, but they are not rational, they are complex. Check that
2x 3 − 5x 2 + 14x − 35 = (2x − 5)(x 2 + 7).
√
Consequently, the other two roots are x = ± 7 i, where i 2 = −1.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
The Set of Real Numbers
Let R denote the set of real numbers, the set of numbers that you
work with when doing calculus. We shall not define exactly what
the real numbers are, but you should be aware √
that√R contains
many numbers that are not rational, e.g. π, e, 2, 3, sin(π/7),
etc. etc
Definition: a real number that is not a rational number is said to
be irrational.
To show a real number x is irrational you have to show that there
are no integers m, n such that
x=
m
.
n
In some cases, this can be quite difficult.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
√
8.1: Rational Numbers
8.2: Irrational Numbers
2 Is Irrational
One well-known example of an
√ irrational number,
√ going all the way
back to the Pythagoreans, is 2. To show that 2 is irrational, we
assume to the contrary that there are relatively prime numbers m
and n such that
√
m
m2
2=
⇔ 2 = 2 ⇔ 2 n2 = m2 .
n
n
Since 2 divides the left side of this last equation, we must have
2 | m2 . Since 2 is prime, 2 | m. This means m = 2k, for some
natural number k. Then
2n2 = m2 = (2k)2 = 4k 2 ⇔ n2 = 2k 2 .
So, similarly, 2 | n. But then m and n have a common factor 2,
contradicting the assumption that they are relatively prime.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
√
8.1: Rational Numbers
8.2: Irrational Numbers
p Is Irrational
√
Theorem 8.2.6: if p is prime, then p is irrational.
√
Proof: we mimic the proof for 2. Suppose that there are
relatively prime numbers m and n such that
√
m2
m
⇔ p = 2 ⇔ p n2 = m2 .
p=
n
n
Since p divides the left side of this last equation, we must have
p | m2 . Since p is prime, p | m. This means m = p k, for some
natural number k. Then
p n2 = m2 = (pk)2 = p 2 k 2 ⇔ n2 = p k 2 .
So, similarly, p | n. But then m and n have a common factor p,
contradicting that they are relatively prime.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Example 4
All of
√ √ √ √ √ √ √ √
2, 3, 5, 7, 11, 13, 17, 19 etc.
are irrational. The square roots of some composite numbers are
irrational as well: for example
√ √ √
6, 8, 10
are all irrational. But some composite numbers have rational
square roots, for example
√
√
√
√
4 = 2, 9 = 3, 36 = 6, 144 = 12,
to list a few.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Square Roots and Rational Numbers
Lemma 8.2.7: a natural number other than 1 is a perfect square
if and only if every prime number in its canonical factorization
occurs to an even power.
Proof: let the canonical factorization of n be n = p1α1 p2α2 · · · pkαk ,
then n2 = p12α1 p22α2 · · · pk2αk , which must be the canonical
factorization of n2 , by the Fundamental Theorem of Arithmetic.
Thus a perfect square, n2 , has even powers of all its prime divisors.
On the other hand, if m = p12α1 p22α2 · · · pk2αk , then m = n2 with
n = p1α1 p2α2 · · · pkαk .
Theorem 8.2.8: if the square root of a natural number is rational,
then the square root is an integer.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Proof: let N 6= 1 be a natural number; and suppose
√
N=
some natural numbers m, n. Then n2 N = m2 ; consequently
(r1γ1 r2γ2
2
· · · rvγv ) p1α1 p2α2
· · · pkαk
=
q1β1 q2β2
· · · quβu
2
m
, for
n
,
in terms of canonical factorizations, for some primes pa , qb , rc . So
r12γ1 r22γ2 · · · rv2γv p1α1 p2α2 · · · pkαk = q12β1 q22β2 · · · qu2βu .
We need to show that each αi is even. By the uniqueness of prime
factorizations,
αi = 2βb , or αi = 2βb − 2γc ,
for some numbers b and c. In either case, αi is even.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Example 5
Show that
√
√
7 is irrational.
√
√
m
Solution: suppose 3 + 7 is a rational number r = . Then
n
√
√
3+ 7 = r
√
√
⇒ 3 = r− 7
√
⇒ 3 = r 2 − 2r 7 + 7
√
r2 + 4
⇒ 7 =
2r
2
√
m + 4n2
⇒ 7 =
.
2mn
This means
3+
√
7 is rational, contradicting Theorem 8.2.6.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla
Chapter 8: Rational Numbers and Irrational Numbers
8.1: Rational Numbers
8.2: Irrational Numbers
Example 6
Show that
√
3
5 is irrational.
√
m
Solution: suppose 3 5 is a rational number r = . Then
n
m3
5 = 3 ⇒ 5 n3 = m3 .
n
If you consider the canonical factorizations of m and n into a
product of primes, then you can see that the exponent of 5 on the
left side must be congruent to 1, mod 3, but the exponent of 5 on
the right side
must be congruent to 0, mod 3. This contradiction
√
3
shows that 5 cannot be a rational number.
Chapter 8 Lecture Notes Rational Numbers and Irrational Numbers
MAT246H1S Lec0101 Burbulla