PROOFS OF IRRATIONALITY 1. Closure of Irrational Numbers In

PROOFS OF IRRATIONALITY

NEIL MAKUR

Abstract. We start by looking at some basic properties regarding operations with rational

and irrational numbers. We then go on to show that certain radicals are irrational. Next,

we state and prove a criterion for irrationality and use it to prove that e is irrational. After

this, we look at alternate proofs of this fact, and show that e

is irrational for all nonzero

rationals r. Then, we see that π and π

are irrational, before going on to prove the same

for cos(1). We look at other irrational values related to the trigonometric functions next,

and prove that ln(r) is irrational for positive rational r 6= 1. We follow this up by looking

at other logarithms, and proving that cosh(1) is irrational. We then discuss transcendence,

and use the Lindemann–Weierstrass theorem to reprove many of our earlier statements. We

end with posing open problems in irrationality.

1. Closure of Irrational Numbers

In this section, we shall look at some basic theorems regarding operations with rational

and irrational numbers. All numbers in this paper will be assumed to be real unless they

are deﬁned during our discussion of transcendence. We will also assume all denominators to

be nonzero.

Theorem 1.1. If r ∈ Q, and a 6∈ Q, then the following are true.

(1) r + a 6∈ Q.

(2) ra 6∈ Q.

(3) −a 6∈ Q.

(4) a

−1

6∈ Q.

Proof. Our methods for all of these will be proof by contradiction.

(1) If we assume that r + a ∈ Q, we will let it be r

∈ Q. Then, a = r

− r. Because Q

is additively closed, this means that a ∈ Q, a contradiction.

(2) If we assume that ra ∈ Q, we will let it be r

∈ Q. Then, a =

. Because Q is

multiplicatively closed, this means that a ∈ Q, a contradiction.

(3) If we assume that −a ∈ Q, we will let it be

∈ Q. Then, a = −

, meaning that it

is rational, which is a contradiction.

(4) If we assume that a

−1

∈ Q, we will let it be

∈ Q. Then, a =

, meaning that it is

rational, which is a contradiction.



2. Irrational Radicals

2.1. Square Roots of Primes. It is well known that

√

2 is irrational. We will present the

classic proof here.

Date: July 2020.

2 NEIL MAKUR

Proposition 2.1.

√

2 is irrational.

Proof. Let us assume that

√

2 =

, where a and b are relatively prime integers. Then, 2 =

or 2b

= a

. Now, if a was odd, then a

= (2k + 1)

= 4k

+ 4k + 1 is also odd, so a must

be even. We will write a = 2k. Now, we have that 2b

= (2k)

= 4k

. Dividing by 2 gives

= 2k

. Therefore, b must be even. However, this contradicts our assumption that a and

b are relatively prime. Therefore,

√

2 is irrational. 

We can also show this for

√

Proposition 2.2.

√

3 is irrational.

Proof. Let us suppose that

√

3 =

where a and b are relatively prime integers. We can say

that 3b

= a

. Now, if we assume that 3 - a, then (as 3|a

), 3|a, which is a contradiction.

Therefore, we can write a = 3k, and get 3b

= (3k)

= 9k

. We can divide by 3 to get

= 3k

, so we can write b = 3l. This contradicts our assumption that a and b are relatively

prime, so

√

3 is irrational. 

We can generalize this proofs easily to any prime. We shall ﬁrst prove a lemma.

Lemma 2.3. If p|a

for a prime p, then p|a.

Proof. Let us suppose that p - a. Because p|a · a, and p - a, we have that p|a. This is a

contradiction, and so p|a. 

Theorem 2.4. If p is a prime, then

√

p is irrational.

Proof. Let us suppose that

√

p =

where a and b are relatively prime integers. Then,

= a

. We write a = pk to get pb

= (pk)

= p

. Dividing by p gives b

= pk

, and so

we can write b = pl, meaning that a and b are not relatively prime, which is a contradiction.

Therefore,

√

p 6∈ Q. 

2.2. More Interesting Square Roots. We want to show that other roots are irrational.

We will show this for

√

6 below.

Proposition 2.5.

√

6 is irrational.

Proof. Let us suppose that

√

6 =

where a and b are relatively prime integers. Then,

= a

. We can write a = 2k to get 6b

= 4k

. Thus, 3b

= 2k

, so we can write b = 2l.

This contradicts our assumption that gcd(a, b) = 1, so

√

6 6∈ Q. 

Note that we could have said a = 3k, and used this to show that

√

6 is irrational. We can

generalize the proof of this.

Theorem 2.6. If p and q are distinct primes, then

√

pq is irrational.

Proof. Let us suppose that

√

pq =

where a and b are integers with gcd(a, b) = 1. Then,

pqb

= a

, so we can write a = pk to get pqb

= p

. Therefore, qb

= pk

, and (as q 6= p is

prime) we can say b = pl. This means that gcd(a, b) 6= 1, which is a contradiction. 

PROOFS OF IRRATIONALITY 3

2.3. Sums of Square Roots. We want to show that sums of square roots are also irrational.

We will prove this for an example ﬁrst.

Proposition 2.7.

√

2 +

√

3 is irrational.

Proof. Let us suppose that

√

2 +

√

3 =

for integers a and b such that

is fully reduced.

Then,



√

2 +

√



. This is equal to

√

+ 2

√

3 +

√

= 2 + 2

√

6 + 3 = 5 + 2

√

Thus,

√

6 =

−5b

, which means that

√

6 ∈ Q. This is a contradiction, so

√

2 +

√

3 6∈ Q. 

We can now see how to easily generalize this proof.

Theorem 2.8. If p and q are distinct primes, then

√

p +

√

q is irrational.

Proof. Let us suppose that

√

p +

√

q =

, where we have completely reduced fraction

with

a, b ∈ Z. Then,



√

p +

√



= p+q+2

√

pq =

. However, this implies that

√

pq is rational,

which is a contradiction. Therefore,

√

p +

√

q is irrational. 

2.4. More General Roots. We will show that 2

1/n

is irrational for integers n ≥ 2.

Theorem 2.9. 2

1/n

is irrational for n ∈ Z ≥ 2.

Proof. If we assume that 2

1/n

where a, b ∈ Z are relatively prime, then we can say

= a

. Thus, we can write a = 2k and derive b

= 2

n−1

. This means that b is also

even, creating a contradiction. 

We can in fact prove that 2

a/b

is irrational for positive integers a and b such that gcd(a, b) =

1, with a 6= 0 and b 6= 1. To do this, we shall use the rational root theorem.

Lemma 2.10 (Rational Root Theorem). If a fully reduced fraction

is a root of a

n−1

+ ··· + a

x + a

, then p|a

, and q|a

Proof. Let us suppose that

is a root. Then, we can say a

n−1

+···+a

= 0.

Multiplying by q

gives a

+ a

n−1

q + ···+ a

n−1

+ a

= 0. We can rearrange this

to get a

+ a

n−1

q + ··· + a

n−1

= −a

. Note that the left hand side is divisible

by p. Because gcd(p, q) = 1, this means that p|a

. Similarly, we have that a

n−1

q + ···+

n−1

+ a

= −a

. Because q divides the left hand side, it must divide a

. 

Theorem 2.11. For a fully reduced fraction

, where b 6= 1 and a 6= 0, 2

a/b

is irrational.

Proof. We only need to prove this for

> 0, because the reciprocal will also be irrational.

Therefore, we will assume

> 0. We can see that 2

a/b

is a root of x

− 2

. Therefore,

the rational roots are of the form ±2

for k ≤ a. Note that, if b is odd, the roots cannot

be negative, and if b is even, then if x is a root, so is −x, so we only need to check the

positive cases. If this is a root, then (2

)

− 2

= 0. This means that a = kb, or that

= k,

contradicting the assumption that

is a fully reduced fraction with b 6= 1. 

3. An Irrationality Criterion

Here we will prove the following theorem.

Theorem 3.1. A real number α is irrational if and only if, for any ε > 0, there exist integers

p and q such that

0 < |qα −p| < ε.

4 NEIL MAKUR

3.1. The Dirichlet Approximation Theorem. To prove our desired criterion, we need

to use the Dirichlet Approximation Theorem. We will state and prove it here.

Theorem 3.2 (Dirichlet Approximation Theorem). Let N be a positive integer, and let α

be any real number. Then, there exist integers p and q with 1 ≤ q ≤ N such that

|qα − p| <

Proof. We will consider the set A = {a

, a

, . . . , a

} where we deﬁne a

= iα −biαc. Clearly,

A ⊂ [0, 1). Because |A| = N + 1, there is some positive k ≤ N so that



k−1



contains two

elements of A. We will call these two elements a

and a

where i > j. Thus, |a

− a

| <

Using our deﬁnition of a

, this means that |iα−biαc−jα+bjαc| <

. We can rearrange this

to get |(i − j)α −(biαc − bjαc) | <

. Thus, we can choose q = i − j, and p = biαc − bjαc

to prove the theorem. 

3.2. The Irrationality Criterion. Now that we have proven the Dirichlet Approximation

Theorem, we are ready to prove the irrationality criterion.

Proof of Theorem 3.1. Let us suppose that α 6∈ Q, and let ε > 0. We shall choose N such

that

< ε. By the Dirichlet Approximation Theorem, there are integers p and q with

1 ≤ q ≤ N such that |qα − p| <

< ε. Because α is irrational, this must be nonzero. Let

us now assume that α ∈ Q. We will write α =

for relatively prime integers r and s. Now,

if we have

, where p, q ∈ Z, then



−



qr − ps



≥

Multiplying by q gives

|qα − p| ≥

Thus, if we choose

> s, we get |qα − p| ≥

> ε. The only case we have disregarded is if

. In this case, |qα − p| = 0. Thus, if α is rational, there are not integers p and q such

that 0 < |qα − p| < ε for all ε > 0. 

3.3. Applications of the Irrationality Criterion. We shall use the irrationality criterion

to show that e is irrational.

Theorem 3.3. e is irrational.

Proof. We know that e =

∞

n=0

. We will let N ∈ Z

such that

< ε. We will let

q = N!, and p be

n=0

. Then,

|qe − p| =



∞

n=0

− N!

n=0



= N!

∞

n=N+1

∞

n=N+1

(N + 1)(N + 2) ···(n − 1)(n)

PROOFS OF IRRATIONALITY 5

≤

∞

n=N+1

(N + 1)

n−N

Now, we may use the fact that this is a geometric series to get that this is equal to

N + 1

1 − 1/(N + 1)

This is equal to

N + 1

< ε.

We can see that |qe − p| 6= 0, so, by Theorem 3.1, e is irrational. 

Theorem 3.1 can also be used to show that ζ(3) is irrational. This proof can be found

in [RSb, Chapter 4].

4. e and its Powers

4.1. Alternate Proofs of e’s Irrationality. We have already shown that e is irrational

using the irrationality criterion. We will present two alternate proofs here.

Second Proof of Theorem 3.3. We know that e =

∞

n=0

. If we assume that e =

for

relatively prime integers a and b, then we can say that n!be = n!a for every n > 0. We can

see that our right hand side is an integer. However, our left side is equal to

k=0

+ b

∞

k=n+1

We can see, however, that the second sum is equal to

(n + 1)

(n + 1)(n + 2)

+ ··· .

We can see that this sum is greater than

n+1

. We can also see that, by comparison with

geometric series, that this is less than

. Thus, for large n, this is not an integer, which is a

contradiction. 

Third Proof of Theorem 3.3. Let us assume that e =

where p and q are relatively prime

integers. Then, we can write

+ ··· +

+ ··· .

Now, we will rearrange this to get

−

− ··· −

(q + 1)!

(q + 2)!

+ ··· .

We can multiply by q! to get

pq!

−

− ··· −

(q + 1)!

(q + 2)!

+ ··· .

6 NEIL MAKUR

It is easy to see that the left hand side is an integer. We can also see, by comparison with

geometric series, that our right hand side is less than

. It is also clearly positive. Thus,

this means that there is an integer n such that 0 < n <

, which is a contradiction. 

Remark 4.1. It is not hard to see that these proofs are very similar. One can in fact modify

one proof to obtain the other.

4.2. The Irrationality of e

. To show that e

is irrational, we will use a similar technique

that we used for e.

Theorem 4.2. e

is irrational.

Proof. If we assume that e

where a, b ∈ Z and gcd(a, b) = 1, then be = ae

−1

. We can

use the series e =

∞

k=0

and e

−1

∞

k=0

(−1)

, and multiply by n!. We can see that n!be

has an integer part, and a leftover term of

n!b



(n + 1)!

(n + 2)!

+ ···



We know that this term is between

n+1

and

. Similarly, n!ae

−1

has a leftover part of

(−1)

n+1

n!a



(n + 1)!

−

(n + 2)!

± ···



Now, if n is even, then this is larger than −

n+1

, but smaller than

−

n + 1

(n + 1)

∓ ··· = −

n + 1



1 + 1/(n + 1)



= −

n + 1



1 −

n + 2



< 0.

However, this means that, for large n, n!ae

−1

is slightly smaller than an integer, while n!be

is slightly larger than an integer. This is a contradiction, and so e

is irrational. 

This idea can also be used to show that e

is irrational, see [AZHE10, Chapter 8].

4.3. Integer Powers of e. We shall prove the following theorem.

Theorem 4.3. For integers s 6= 0, e

is irrational.

To do this, we shall need to make use of a lemma.

Lemma 4.4. For some n ≥ 1, let

(x) =

(1 − x)

Then, the following are true.

(1) f

(x) can be written as

k=n

where c

∈ Z.

(2) For 0 < x < 1, 0 < f

(x) <

(3) For all k > 0, f

(k)

(0) and f

(k)

(1) are integers.

Proof. Part (1) is clear, and part (2) can be checked by basic diﬀerentiation, so we will prove

only part (3). Note that, f

(k)

(0) is 0 unless n ≤ k ≤ 2n. Because we are plugging in 0, we

only need to worry about the constant term. The constant term for these are

, which is

an integer in the range. Now, notice that f

(x) = f

(1 − x), so f

(k)

(x) = (−1)

(k)

(1 − x),

which means that f

(k)

(1) = (−1)

(k)

(0), which is an integer. 

PROOFS OF IRRATIONALITY 7

We are now able to prove our theorem.

Proof of Theorem 4.3. Let us suppose that e

for relatively prime integers a and b. We

will let n be determined later, and we will let

F (x)

= s

f(x) − s

2n−1

(x) + s

2n−2

(x) ∓ ··· + f

(2n)

(x),

where we write f(x) for f

(x) from the above lemma. Because f

(k)

(x) = 0 for k > 2n, we

can write

F (x) = s

f(x) − s

2n−1

(x) + s

2n−2

(x) ∓ ··· .

This allows us to say

(x) = −sF (x) + s

2n+1

f(x).

Now, let us look at

F (x)). The product rule gives

F (x)) = se

F (x) + e

(x) = se

F (x) + e

(−sF (x) + s

2n+1

f(x))

= s

2n+1

f(x).

This allows us to ﬁnd

= b

2n+1

f(x)dx = b (e

F (x)) |

= be

F (1) − bF (0) = aF (1) − bF (0).

Part (3) of our lemma implies that this is an integer. Part (2) of the lemma gives us bounds,

which allow us to say that

0 < N <

dx =

− 1).

We can assume that s is positive, and the closure properties will cover the case of s < 0. We

will also let n be such that n! > as

2n+1

. Thus,

N <

2n+1

< 1.

This is a contradiction, and so e

is irrational. 

4.4. Rational Powers of e. We have shown that for n ∈ Z\{0}, e

is irrational. For

n ∈ R\{0}, this is clearly false (for example, n = ln(2)). We wish to determine this for the

rationals.

Theorem 4.5. If r is a nonzero rational number, then e

is irrational.

Proof. We will write r =

for p, q ∈ Z and gcd(p, q) = 1. Let us suppose that e

p/q

for

relatively prime integers a and b. This means that e

. This means that e

is rational,

a contradiction. 

8 NEIL MAKUR

5. Pi, Trigonometric, Logarithmic, and Hyperbolic Values

5.1. π and π

’s Irrationality. We will ﬁrst prove the well-known theorem that π is irra-

tional.

Theorem 5.1. π is irrational.

Proof. We will let π =

where a and b are integers with gcd(a, b) = 1. We deﬁne

f(x)

(a − bx)

Note that we can make an analog of Lemma 4.4 into a lemma based around f(x). Proving

the exact details of this analog is left as an exercise for the reader. We will now deﬁne

F (x) = f(x) − f

(2)

(x) + f

(4)

(x) ∓ ··· .

Notice that we can have this inﬁnite series because the derivatives beyond 2n vanish. We

also have that F

(x) = −F (x) + f(x). Basic calculus gives that

(x) sin(x) − F (x) cos(x)) = F

(x) sin(x) + F

(x) cos(x) − F

(x) cos(x) + F (x) sin(x)

= F

(x) sin(x) + F (x) sin(x) = (−F (x) + f(x)) sin(x) + F (x) sin(x) = f(x) sin(x).

Thus,

f(x) sin(x)dx = (F

(x) sin(x) − F (x) cos(x)) |

= F (π) + F (0).

This is an integer. However, we can also establish a bound on this for 0 < x < π:

0 < f(x) sin(x) < f(x) <

Thus, our integral (which is an integer) is arbitrarily small for large enough n, and so we

have a contradiction. Thus, π is irrational. 

Corollary 5.2. For positive integers n, π

1/n

is irrational.

Proof. Let us suppose that π

1/n

is rational, say

. Then, we have that

= π. This is a

contradiction, and so π

1/n

is irrational. 

Now, we will do the same for π

. When we write f(x), we will mean f

(x) from Lemma

4.4.

Theorem 5.3. π

is irrational.

Proof. Let us assume π

where a, b ∈ Z, and are relatively prime. We deﬁne

F (x)

= b



f(x) − π

2n−2

(2)

(x) + π

2n−4

(4)

(x) ∓ ···



Note that this satisﬁes F

(x) = −π

F (x) + b

2n+2

f(x). We can get that

(x) sin(πx) − πF (x) cos(πx)) = F

(x) sin(πx) + π

F (x) sin(πx)

= b

2n+2

f(x) sin(πx) = π

f(x) sin(πx).

Therefore,

= π

f(x) sin(πx)dx =



−1

(x) sin(πx) − F (x) cos(πx)



= F (1) + F(0).

PROOFS OF IRRATIONALITY 9

We can see that N ∈ Z. Now, let us choose n so large such that

< 1. Our bounds on

f(x) allow us to say that

0 < N = π

f(x) sin(πx)dx < πa

sin(πx)dx =

< 1.

This is a contradiction. 

5.2. The Irrationality of cos(1). Here we will prove that cos(1) is irrational using the

irrationality criterion.

Proposition 5.4. cos(1) is irrational.

Proof. We know that cos(1) =

∞

n=0

(−1)

(2n)!

. We will choose N such that

< ε. We will let

q = N! and p = N!

n=0

(−1)

(2n)!

. Then, we have that

|q cos(1) − p| =



∞

n=0

(−1)

(2n)!

− N!

n=0

(−1)

(2n)!



∞

n=N+1

(−1)

(2n)!

≤

∞

n=N+1

(−1)

(N + 1)

2n−N

≤

∞

n=N+1

(N + 1)

2n−N

(N + 1)

N+2

1 − 1/(N + 1)

(N + 1)

N+2

(N + 1)

− 1

(N + 1)

N + 2

< ε.

Therefore, by Theorem 3.1, cos(1) is irrational. 

This proof can be easily modiﬁed to show that sin(1) is irrational, and is left as an exercise

to the reader.

5.3. Other Trigonometric Values. Many values of trigonometric functions, such as cos





and sin





are irrational. We will show these here.

Proposition 5.5. cos





is irrational.

Proof. By the triple angle formula, we have that cos





= 4 cos





− 3 cos





. Writing

x for cos





gives

= 4x

− 3x, or 8x

− 6x − 1 = 0. The rational roots of these must be

±1, ±

, ±

, or ±

. Direct substitution gives that none of these are actual roots. Thus, as

cos





is a root of this, cos





is irrational. 

We can prove that sin





is irrational 2 diﬀerent ways.

Proposition 5.6. sin





is irrational.

First Proof of Proposition 5.6. From the triple angle formula for sin, we have that sin





3 sin





−4 sin





, meaning that sin





is a root of

= 3x − 4x

, or 8x

−6x + 1 = 0.

We can once again use the rational root theorem to show that this has no rational roots,

and so sin





is irrational. 

Second Proof of Proposition 5.6. We have that, from the cosine double angle formula, cos





1 − 2 sin





. Thus, if sin





were rational, then cos





would be too. This is a contra-

diction, so sin





must be irrational. 

10 NEIL MAKUR

5.4. Inverse Trigonometric Functions. We will prove a result regarding the irrationality

of inverse trigonometric functions.

Theorem 5.7. For every odd integer n ≥ 3,

cos

−1



√



is irrational.

Recall that, letting n = 9, this is useful in the solution of Hilbert’s Third Problem. In the

proof, we shall make use of a lemma.

Lemma 5.8. cos((k + 1)ϕ) = 2 cos(kϕ) cos(ϕ) − cos((k − 1)ϕ).

Proof. Recall that cos(α + β) = cos(α) cos(β) − sin(α) sin(β), and that cos(α − β) =

cos(α) cos(β) + sin(α) sin(β). Adding these gives that

cos(α + β) + cos(α − β) = cos(α) cos(β) −sin(α) sin(β) + cos(α) cos(β) + sin(α) sin(β)

= 2 cos(α) cos(β).

If we substitute θ = α + β and φ = α − β, then α =

θ+φ

, and β =

θ−φ

. This gives the

formula

cos(θ) + cos(φ) = 2 cos



θ + φ



cos



θ − φ



Now, if we let θ = (k + 1)ϕ, and φ = (k − 1)ϕ, we get that

cos((k + 1)ϕ) + cos((k − 1)ϕ) = 2 cos(kϕ) cos(ϕ),

which is the same thing as saying that

cos((k + 1)ϕ) = 2 cos(kϕ) cos(ϕ) − cos((k − 1)ϕ).



Proof of Theorem 5.7. We shall deﬁne ϕ

= cos

−1



√



. We wish to show that

cos(kϕ

) =

√

where A

is an integer such that n - A

. This can be done with induction on k. For k = 0, 1,

this is true with A

= 1. We can use our formula to show this for k + 1:

cos((k + 1)ϕ

) = 2

√

−

k−1

√

k−1

− nA

k−1

√

k+1

√

k+1

It is clear that, if n - A

, then n - A

k+1

. Now, let us assume that

for relatively

prime integers k and l. Then, we have that lϕ

= kπ. Taking the cosine of both sides gives

√

= cos(lϕ

) = cos(kπ) = ±1. Therefore,

√

= ±A

. It is clear that l 6= 0, and that this

doesn’t work for l = 1. Therefore, we can assume that l ≥ 2. In this case, n|

√

. From the

fact that

√

, we have that n|A

, which is a contradiction. 

PROOFS OF IRRATIONALITY 11

5.5. Irrationality of Natural Logarithms. It is well known that ln(2) is irrational. We

shall prove this here.

Proposition 5.9. ln(2) is irrational.

Proof. Let us assume that ln(2) =

with a, b ∈ Z and gcd(a, b) = 1. Then, 2 = e

a/b

However, we know that e

a/b

must be irrational, so this is a contradiction. 

It is clear that there is nothing about 2 in this situation, and so we can prove this for

(almost) any rational number.

Theorem 5.10. If r ∈ Q

\{1}, then ln(r) is irrational.

Proof. If we assume that ln(r) =

where a and b are integers with gcd(a, b) = 1, then we can

say that r = e

a/b

. However, this means that r must be irrational, which is a contradiction. 

5.6. Irrationality of Logarithms with Integer Bases. A well known irrational number

is log

(3). We shall prove that this is irrational.

Proposition 5.11. log

(3) is irrational.

Proof. Let us suppose that log

(3) =

with a, b ∈ Z and gcd(a, b) = 1. Then, 3 = 2

a/b

, or

= 2

. By the Fundamental Theorem of Arithmetic, this is only possible if a = b = 0.

Therefore, log

(3) is irrational. 

It is clear that there is nothing special about 3 here, so we will prove this generally.

Proposition 5.12. If n ∈ Z is not a power of 2, then log

(n) is irrational.

Proof. Let us write n = 2

a such that a is odd. Now, let us assume that log

(n) =

where b

and c are relatively prime positive integers. Then, 2

= 2

. This is only possible if either

a = 1, or c = 0, which contradict our initial conditions. Therefore, log

(n) is irrational. 

We will show that the same is true for 10.

Proposition 5.13. If n ∈ Z is not a power of 10, then log

(n) is irrational.

Proof. Let us write n = 2

a such that a is relatively prime to 10. If we assume that

log

(n) =

where b and c are relatively prime positive integers, then we have that 10

. If a 6= 1, then this is impossible. If a = 1, then we have that p

c = p

c = b.

However, this means that p

= p

, or that n is a power of 10, which is a contradiction. 

This may lead us to believe that this is true for every integer. However, this is false.

Counterexample. log

(16) =

5.7. Irrationality of cosh(1). Recall that cosh(x) is deﬁned as

cosh(x) =

+ e

−x

and that sinh(x) is deﬁned as

sinh(x) =

− e

−x

Remark 5.14. If the reader is familiar with the complex deﬁnitions of sin(x) and cos(x), they

may notice that cosh(t) = cos(it), and that sinh(t) = −i sin(it), where i =

√

−1.

12 NEIL MAKUR

Thus, cosh(1) =

e+e

−1

. We wish to show that this is irrational.

Proposition 5.15. cosh(1) is irrational.

Proof. We have that

cosh(1) =

e + e

−1

∞

n=0

∞

n=0

(−1)

∞

n=0

1 + (−1)



+ ···



+ ··· =

∞

n=0

(2n)!

To show that this is irrational, we shall use our irrationality criterion. Let us choose N such

that

< ε, and let q = N!, p =

n=0

(2n)!

. Therefore,

|q cosh(1) − p| =



∞

n=0

(2n)!

− N!

n=0

(2n)!



∞

n=N+1

(2n)!

∞

n=N+1

(N + 1)(N + 2) ···(2n)

≤

∞

n=N+1

(N + 1)

2n−N

(N + 1)

N+2

1 − 1/(N + 1)

(N + 1)

N+2

(N + 1)

− 1

(N + 1)

N + 2

< ε.

Thus, by Theorem 3.1, cosh(1) is irrational. 

This proof is easily modiﬁed to show that sinh(1) is irrational, and we leave it as an

exercise to the reader.

6. Transcendence

In this paper, we have looked at irrationality. We will now turn to transcendence.

Deﬁnition 6.1. A number α is said to be algebraic if it is the root of some nonzero poly-

nomial with integer (or equivalently, rational) coeﬃcients.

Example. The number i =

√

−1 is a root of x

+ 1, and is therefore algebraic.

Example. A rational number

is a root of x −

(and bx − a), and is therefore algebraic.

Example.

√

2 is a root of x

− 2, and is therefore algebraic.

Remark 6.2. The algebraic numbers form a ﬁeld (i.e. they are closed under addition, sub-

traction, multiplication, and division). The simplest proof of this requires Abstract Algebra.

We refer the reader to [RSa] to obtain the necessary background. If α and β are algebraic

over Q, then we have that [Q(α, β) : Q] is ﬁnite. We shall deﬁne this number to be n.

Therefore, the numbers α + β, (α + β)

, . . . , (α + β)

n+1

are linearly dependent. This gives us

a nonzero polynomial with root α + β. The same is true for αβ and α

−1

Deﬁnition 6.3. A number is said to be transcendental if it is not algebraic.

Our main tool will be the Lindemann–Weierstrass theorem.

Theorem 6.4 (The Lindemann–Weierstrass Theorem). If u 6= 0 is algebraic, then e

transcendental.

PROOFS OF IRRATIONALITY 13

The proof of this theorem is beyond the scope of this paper. We refer the reader to [Wik20]

for a proof.

This theorem will allow us to easily prove many transcendence (and hence, irrationality)

results.

Corollary 6.5. e is transcendental.

Proof. Take u to be 1. 

Corollary 6.6. e

is transcendental for rational r.

Proof. Rational numbers are algebraic. Therefore, take u = r. 

Corollary 6.7. π is transcendental.

Proof. Let us assume that π is algebraic. Then, we have that iπ is algebraic. This means

that, by Theorem 6.4, e

iπ

= −1 is transcendental, a contradiction. 

Corollary 6.8. π

is transcendental for nonzero integers n.

Proof. Let us assume that n > 0. Then, if π

is a root of p(x), π is a root of p(x

), which

is a contradiction. Now, because the algebraic numbers form a ﬁeld, if π

−n

is algebraic for

n > 0, then π

would be algebraic, which is a contradiction. 

Corollary 6.9. π

a/b

is transcendental for rational

6= 0.

Proof. If π

a/b

were algebraic, then



a/b



= π

would also be algebraic. This is a contradic-

tion. 

Corollary 6.10. sin(n) and cos(n) are transcendental for nonzero algebraic numbers n.

Proof. We have that e

= cos(n) + i sin(n). Now, we know that sin(n) =

1 − cos

(n).

Thus, if cos(n) were transcendental, then so is cos

(n) and 1 − cos

(n). This means that

1 − cos

(n) = sin(n) is also transcendental. We can also swap sin(n) and cos(n) to get

that sin(n) is transcendental if and only if cos(n) is as well. Now, if sin(n) is algebraic, so

is cos(n), and therefore, cos(n) + i sin(n) is algebraic. Because we are assuming that n is

algebraic, this is a contradiction, and so sin(n) and cos(n) are transcendental. 

Corollary 6.11. sin

−1

(n) and cos

−1

(n) are transcendental for algebraic nonzero n.

Proof. Let φ = sin

−1

(n). Then, sin(φ) and cos(φ) are algebraic. If we assume that φ

is algebraic, then e

iφ

= cos(φ) + i sin(φ) is transcendental by the Lindemann–Weierstrass

Theorem. However, this is an algebraic number, so φ must be transcendental. The same is

true for φ

= cos

−1

(n). 

Corollary 6.12. cosh(n) and sinh(n) are transcendental for algebraic nonzero n.

Proof. Let us suppose that n is algebraic. Then, in is also algebraic. Thus, sin(in) =

i sinh(n) and cos(in) = cosh(n) are transcendental. Clearly, if i sinh(n) is transcendental, so

is sinh(n). 

Corollary 6.13. ln(n) is transcendental for nonzero algebraic n.

Proof. Let us suppose that ln(n) is algebraic. Then, we have that e

ln(n)

= n is transcendental

by Theorem 6.4, which is a contradiction. 

14 NEIL MAKUR

Another important theorem is the Gelfond–Schneider Theorem.

Theorem 6.14 (The Gelfond–Schneider Theorem). If a 6= 0, 1 is algebraic, and b is an

irrational algebraic number, than a

is transcendental.

For a proof, we refer the reader to [War12]. This allows us to prove a few more facts.

Corollary 6.15.

√

is transcendental.

Proof. Take a = b =

√

2 (which we already know is irrational). 

Corollary 6.16. e

is transcendental.

Proof. We apply Euler’s Formula to get that

= e

−i·iπ



iπ



−i

= (−1)

−i

so let a = −1 and b = −i. Therefore, e

is transcendental. 

7. Open Problems in Irrationality

Despite the fact that almost any person knows what an irrational number is, there are

still many questions about which numbers are irrational.

Question 7.1. Is e + π irrational?

Question 7.2. Is e · π irrational?

Question 7.3. Is π

irrational?

None of these answers are known, but are all believed to be “yes”. However, there is some

work done towards the solutions of the ﬁrst two questions.

Proposition 7.4. At least one of e · π and e + π is irrational.

Proof. It is clear that both e and π are roots of (x−e)(x −π) = x

−(e+ π)x +e·π. Because

both e and π are transcendental, at least one of e + π and e · π must be irrational. 

References

[AZHE10] Martin Aigner, G¨unter M Ziegler, Karl H Hofmann, and Paul Erdos. Proofs from the Book.

Springer, 2010.

[Niv61] Ivan Niven. Numbers: rational and irrational, volume 1. Random House New York, 1961.

[Niv85] Ivan Niven. Irrational numbers, volume 11. American Mathematical Soc., 1985.

[Niv00] Ivan Niven. A simple proof that π is irrational. In Pi: A Source Book, pages 276–276. Springer,

2000.

[RSa] Simon Rubinstein-Salzedo. Abstract Algebra. Self Published.

[RSb] Simon Rubinstein-Salzedo. Inﬁnite Series. Self Published.

[War12] Matthijs Warrens. On irrational and transcendental numbers. 2012.

[Wik20] Wikipedia contributors. Lindemann–weierstrass theorem, 2020. [Online; accessed 30-July-2020 ].