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# Irrational Numbers

## 1.

Calculus++ Light## 2. Irrational Numbers

## 3.

Question 1. The Dirichlet function is defined1, if x is rational

as follows f ( x )

0, if x is irrational

Is this function even or odd or neither?

Is this function periodic? If yes, find a period

of this function.

Solution. If x is a rational number, so is – x.

k k

.

n

n

If x is a irrational number, so is – x.

Hence f (– x) = f (x), and therefore the Dirichlet

function is even.

## 4.

The sum of two rational numbers is a rationalk

k

k

n

k

n

1

2

1

2

2

1

number:

.

n1 n2

n1n2

The sum of a rational and an irrational number

is an irrational number.

Let x be a rational number, let y be an

irrational number, and let us assume that

z = x + y is a rational number.

Then y = z + (– x) is also a rational number.

Contradiction!

Hence, the sum of a rational and an irrational

number is an irrational number.

## 5.

Therefore f (x + y) = f (x) for any rationalnumber y.

Thus, the Dirichlet function is periodic.

Any rational number is a period of this

function.

However, unlike trigonometric functions

sin(x) or cos(x), the Dirichlet function does

not have minimal (or principal) period T.

## 6.

Question 2. The numbers 2 and 3 areirrational. Show that the number 2 3

is irrational too.

Solution. We have 2 3

3 2 3 2

3 2

1

.

3 2

3 2

3 2

If 2 3 is a rational number, then

1

m

n

2 3

3 2

m

2 3 n

1

1 n m

3 2 3 2

2

2

2 m n

## 7.

n mis a rational number.

2

2m n

2

2

Contradiction!!!

Therefore our assumption was incorrect and

2 3 is an irrational number.

## 8.

Question 3. Letf ( x) x x x x x ,

and denote g ( x) f 1 ( x).

1

2

2

1

3

3

1

4

4

1

5

5

Find a general formula for the second

derivative of inverse function, g (x), and

calculate g (0).

1

Solution. We know that g ( x)

.

The chain rule yields

f ( g ( x))

1

g ( x)

f ( g ( x)) g ( x).

2

( f ( g ( x)))

f ( g ( x))

.

g ( x)

3

( f ( g ( x)))

## 9.

f ( x) x x x x xf ( g ( x))

g ( x)

.

3

( f ( g ( x)))

1

2

2

1

3

3

1

4

4

1

5

5

Since f (0) = 0, we have g (0) = 0.

f ( x) 1 x x x x f ( g (0)) 1

2

3

f ( x) 1 2 x 3x 4 x f ( g (0)) 1

2

g (0) 1.

3

4

## 10.

Question. Which of the following conditionsimply that a real number x is rational?

I. x is rational

II. x2 and x5 are rational

III. x2 and x4 are rational

a) I only b) II only c) I and II only

d) I and III only e) II and III only

m

Solution: If x is rational, then x .

2

n

m

Therefore x 2 is also rational.

n

Counterexample to III: 2 is irrational, but

2

2

2 and

2

4

2 4 are rational.

2

## 11.

Let now x2 and x5 be rational:k

m

5

2

and x .

x

l

n

If m = 0, then x2 = 0, x5 = 0, and x = 0 is a

rational number.

x5

k n2

.

In all other cases x 2 2

2

(x )

l m

Therefore x is a rational number.

a) I only b) II only c) I and II only

d) I and III only e) II and III only

## 12.

Calculus++Also known as

Hysterical Calculus

## 13.

Question 1. Show that 2 is irrational.Solution. Any integer number n is either even,

n = 2k, or odd, n = 2k + 1, where k is another

integer number.

The square of an odd number is odd

(2k 1) 4k 4k 1 2(2k 2k ) 1.

2

2

2

Hence n2 can be even only if n is even.

Analogously, the square of an even number is

even: (2k)2 = 4 k2 = 2(2 k2).

Hence n2 can be odd only if n is odd.

That is n2 is even (odd), if and only if n is

even (odd).

## 14.

Let us now assume that2 is a rational

k

number, that is 2 , where, k and n, do

n

not have common factors.

In particular, either both k and n are odd, or

only one of them is even.

2

k

Then 2 2

n

k 2n .

2

2

That is, k2 is even, and hence k is also even:

k = 2m, where m is another integer number.

2

2

2

2

But then 2n 4m n 2m .

That is, n2 is even, and hence n is also even.

Contradiction!

## 15.

Thus, our assumption that 2 is a rationalnumber leads to a contradiction, and hence

this number is irrational.

Remark. Using a similar argument one can

show that 3 2 is an irrational number.

To show that 3 is an irrational number, note

that any integer number n is either divisible

by 3: n = 3k,

or n = 3k +1,

or n = 3k +2.

## 16. Higher derivatives

Notations for n-th order derivatives:n

d

f (x) or

n

dx

f

(n)

( x).

The following properties are often useful for

calculating high-order derivatives:

n

n

n

d f ( x) d g ( x)

d

( f ( x) g ( x))

n

n

n

dx n n dx

dx n k

d x

d x

if

k

<

n,

0

,

n!,

n

n

dx

dx

n k

d x

k!

k n

if

k

>

n.

and

x

,

k

dx

(k n)!

## 17.

Question 5. Find the n-th derivative of then

function

x

f ( x)

1 x

.

Solution. Recall the formula for the sum of a

geometrical series

n

1 x

1 x x x

.

1 x

n

n

x 1 1

1 1 x

Hence f ( x)

1 x

1 x 1 x

1

2

n 1

(1 x x x ).

1 x

2

n 1

## 18.

Thereforen

n

n

d

d

1

d

2

n 1

f ( x) n

n (1 x x x )

n

dx

dx 1 x dx

1

d 1

0

2

dx 1 x (1 x)

2

d

1

2

d

1

2

3

2

dx 1 x dx (1 x) (1 x)

3

2 3

d

2

d 1

4

3

3

dx 1 x dx (1 x) (1 x)

4

2 3 4

d 2 3

d

1

5

4

4

(1 x)

dx 1 x dx (1 x)

## 19.

nd

n!

Thus

f ( x)

.

n

n 1

dx

(1 x)