RATIONAL AND IRRATIONAL NUMBERS
DEFINITION:
Rational numbers are all numbers of the form
EXAMPLE:
p
, where p and q are integers and q 6= 0.
q
1
5
50
, − , 2, 0, , etc.
2
3
10
NOTATIONS:
N = all natural numbers, that is, 1, 2, 3, . . .
Z = all integer numbers, that is, 0, ±1, ±2, ±3, . . .
Q = all rational numbers
R = all real numbers
DEFINITION:
A number which is not rational is said to be irrational.
EXAMPLE: We prove that
that is
√
2 is irrational. Assume to the contrary that
√
2 is rational,
√
p
2= ,
q
where p and q are integers and q =
6 0. Moreover, let p and q have no common divisor > 1. Then
p2
⇒ 2q 2 = p2 .
(1)
q2
Since 2q 2 is even, it follows that p2 is even. Then p is also even (in fact, if p is odd, then p2 is
odd). This means that there exists k ∈ Z such that
2=
p = 2k.
(2)
Substituting (2) into (1), we get
2q 2 = (2k)2
⇒
2q 2 = 4k 2
⇒
q 2 = 2k 2 .
Since 2k 2 is even, it follows that q 2 is even. Then q is also even. This is a contradiction. EXERCISE SET: Prove that the following numbers are irrational:
√
√
√
1. 3 4
5∗ . 2 + 3
√
√
√
6∗∗ . 2 + 3 3
2. 6
1√
3.
2+5
7∗∗∗ . sin 1◦
3
1
1
1
4∗ . log5 2
8∗∗∗ . e = 2 + + + . . . +
+ ...
2! 3!
n!
1
SOLUTIONS
1. We prove that
√
3
4 is irrational. Assume to the contrary that
√
p
3
4= ,
q
√
3
4 is rational, that is
where p and q are integers and q 6= 0. Moreover, let p and q have no common divisor > 1. Then
4=
p3
q3
⇒
4q 3 = p3 .
(1)
Since 4q 3 is even, it follows that p3 is even. Then p is also even (in fact, if p is odd, then p3 is odd). This means
that there exists k ∈ Z such that
p = 2k.
(2)
Substituting (2) into (1), we get
4q 3 = (2k)3
4q 3 = 8k 3
⇒
⇒
q 3 = 2k 3 .
Since 2k 3 is even, it follows that q 3 is even. Then q is also even. This is a contradiction. √
√
2. We prove that 6 is irrational. Assume to the contrary that 6 is rational, that is
√
p
6= ,
q
where p and q are integers and q 6= 0. Moreover, let p and q have no common divisor > 1. Then
6=
p2
q2
⇒
6q 2 = p2 .
(1)
Since 6q 2 is even, it follows that p2 is even. Then p is also even (in fact, if p is odd, then p2 is odd). This means
that there exists k ∈ Z such that
p = 2k.
(2)
Substituting (2) into (1), we get
6q 2 = (2k)2
⇒
6q 2 = 4k 2
⇒
3q 2 = 2k 2 .
Since 2k 2 is even, it follows that 3q 2 is even. Then q is also even (in fact, if q is odd, then 3q 2 is odd). This is
a contradiction. 1√
1√
3. We prove that
2 + 5 is irrational. Assume to the contrary that
2 + 5 is rational, that is
3
3
1√
p
2+5= ,
3
q
where p and q are integers and q 6= 0. Then
√
Since
√
2 is irrational and
2=
3(p − 5q)
.
q
3(p − 5q)
is rational, we obtain a contradiction. q
4∗ . We prove that log5 2 is irrational. Assume to the contrary that log5 2 is rational, that is
log5 2 =
p
,
q
where p and q are integers and q 6= 0. Then
5p/q = 2
⇒
5p = 2q .
Since 5p is odd and 2q is even, we obtain a contradiction. 2
5∗ . We prove that
√
2+
√
3 is irrational. Assume to the contrary that
√
√
2+
3=
√
2+
√
3 is rational, that is
p
,
q
where p and q are integers and q 6= 0. Then
√
√
2+
√ 2
p2
3 = 2
q
⇒
√ √
p2
2+2 2 3+3= 2
q
⇒
√
p2
5+2 6= 2
q
⇒
√
6=
p2 − 5q 2
.
2q 2
p2 − 5q 2
is rational, we obtain a contradiction. 2q 2
√
√
√
√
6∗∗ . We prove that 2 + 3 3 is irrational. Assume to the contrary that 2 + 3 3 is rational, that is
Since
6 is irrational and
√
√
3
2+
3=
p
,
q
where p and q are integers and q 6= 0. It follows that
√
3
3=
p √
− 2,
q
hence
3=
p √
− 2
q
3
=
√
p3
p √ 2 √ 3
p3
p
p2 √
p2 √
2 −
2 = 3 −3 2 2+6 −2 2
−3 2 2+3
3
q
q
q
q
q
q
2
3
√
p
p
p
= 3 +6 − 2 3 2 +2 .
q
q
q
We can rewrite this as
√
Since
√
2 is irrational and
p3
p
+6 −3
3
p3 + 6pq 2 − 3q 3
q
q
2=
=
.
2
3p2 q + 2q 3
p
3 2 +2
q
p3 + 6pq 2 − 3q 3
is rational, we obtain a contradiction. 3p2 q + 2q 3
7∗∗∗ . We prove that sin 1◦ is irrational. Assume to the contrary that sin 1◦ is rational. Then cos2 1◦ and
cos 2◦ are also rational, since
cos2 1◦ = 1 − sin2 1◦
cos 2◦ = cos2 1◦ − sin2 1◦ .
and
Similarly, cos 4◦ , cos 8◦ , cos 16◦ , and cos 32◦ are rational, since
cos 4◦ = 2 cos2 2◦ − 1,
cos 8◦ = 2 cos2 4◦ − 1,
cos 16◦ = 2 cos2 8◦ − 1,
and
cos 32◦ = 2 cos2 16◦ − 1.
On the other hand, we have
√
3
= cos 30◦ = cos(32◦ − 2◦ ) = cos 32◦ cos 2◦ + sin 32◦ sin 2◦
2
= cos 32◦ cos 2◦ + 2 cos 16◦ sin 16◦ sin 2◦
= cos 32◦ cos 2◦ + 4 cos 16◦ cos 8◦ sin 8◦ sin 2◦
= cos 32◦ cos 2◦ + 8 cos 16◦ cos 8◦ cos 4◦ sin 4◦ sin 2◦
= cos 32◦ cos 2◦ + 16 cos 16◦ cos 8◦ cos 4◦ cos 2◦ sin2 2◦
= cos 32◦ cos 2◦ + 64 cos 16◦ cos 8◦ cos 4◦ cos 2◦ cos2 1◦ sin2 1◦ .
√
The right-hand side is rational. One can prove that
3
is irrational. We obtain a contradiction. 2
3
1
1
1
8∗∗∗ . We prove that 2 + + + . . . +
+ . . . is irrational. Assume to the contrary that this number is
2! 3!
n!
rational, that is
1
1
1
p
= 2 + + + ... +
+ ...,
q
2! 3!
n!
where p and q are integers and q 6= 0. We multiply both sides by qn! with
n > q.
We get
1
1
1
pn! = qn! 2 + + + . . . +
+ ...
2! 3!
n!
1
1
1
1
1
1
= qn! 2 + + + . . . +
+ qn!
+
+
+ ...
2! 3!
n!
(n + 1)! (n + 2)! (n + 3)!
1
1
1
1
1
1
= qn! 2 + + + . . . +
+q
+
+
+ ... ,
2! 3!
n!
n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)
so
1
1
1
1
1
1
+
+
+ ... .
=q
pn! − qn! 2 + + + . . . +
2! 3!
n!
n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)
1
1
1
are integer. If we prove that
Note that pn! and qn! 2 + + + . . . +
2! 3!
n!
1
1
1
q
+
+
+ . . . < 1,
n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)
we obtain a contradiction. To this end we observe that
1
1
<
,
(n + 1)(n + 2)
(n + 1)2
1
1
<
,....
(n + 1)(n + 2)(n + 3)
(n + 1)3
By this and a formula of geometric progression we have
1
1
1
1
1
1
q
+
+
+ ... < q
+
+
+
.
.
.
n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)
n + 1 (n + 1)2
(n + 1)3
1
=q
1
(n + 1) 1 −
n+1
1
=q
n+1
n+1−
n+1
1
=q
n+1−1
q
= ,
n
which is < 1 by (1). 4
(1)