Question

Use the following outline to prove that e is an irrational number

1. If\({\rm{e}}\)were rational, then it would be of the form\({\rm{e = p / q}}\), where\({\rm{p}}\)and\({\rm{ q}}\)are positive integers and\({\rm{q > 2}}\). Use Taylor's Formula to write

\(\begin{aligned}{}\frac{{\rm{p}}}{{\rm{q}}}{\rm{ = e = 1 + }}\frac{{\rm{1}}}{{{\rm{1!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2!}}}}{\rm{ + L + }}\frac{{\rm{1}}}{{{\rm{q!}}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\\{\rm{ = }}{{\rm{s}}_{\rm{q}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\end{aligned}\)

where\({\rm{0 < z < 1}}\).

2. Show that\({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right)\)is an integer.
3. Show that\({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < 1}}\).
4. Use parts (b) and (c) to deduce that \(e\) is irrational.

Short answer

1. Hence the given equation is\(\frac{{\rm{p}}}{{\rm{q}}}{\rm{ = e = }}{{\rm{s}}_{\rm{q}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\;\;\;{\rm{( where 0 < z < 1)}}\) proved.
2. The given equation\({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = integer }}\)is proved.
3. Hence the given equation\({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < 1}}\)is proved.
4. Hence \({\rm{e is an irrational number e}} \ne \frac{{\rm{p}}}{{\rm{q}}}\) is proved.

Step-by-step solution

Concept Introduction

Any real number that cannot be represented as the quotient of two integers is an irrational number. There is no number that equals the square root of 2 among integers or fractions, for example.

Use Taylor’s Formula to prove \(\frac{{\rm{p}}}{{\rm{q}}}{\rm{ = e = }}{{\rm{s}}_{\rm{q}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\)

a)

If\({\rm{e}}\)were rational, it would take the form\({\rm{e = }}\frac{{\rm{p}}}{{\rm{q}}}\), where p and q are positive integers and\({\rm{q > 2}}\)respectively.

\({\rm{e = }}\frac{{\rm{p}}}{{\rm{q}}}\)

Let's pretend we have,

\(\begin{aligned}{}{\rm{f(x) = }}{{\rm{e}}^{\rm{x}}}\\{{\rm{f}}'}{\rm{(x) = }}{{\rm{e}}^{\rm{x}}} \Rightarrow {{\rm{f}}^{\rm{n}}}{\rm{(x) = }}{{\rm{e}}^{\rm{x}}}\\ \Rightarrow {\rm{f(0) = }}{{\rm{e}}^{\rm{0}}}{\rm{ = 1}}\\{{\rm{f}}^{\rm{n}}}{\rm{(0) = }}{{\rm{e}}^{\rm{0}}}{\rm{ = 1}}\end{aligned}\)

The Taylor polynomial of\({{\rm{n}}^{{\rm{th}}}}\)degree for\({\rm{f(x)}}\)about\({\rm{x = a}}\)is given by

\({{\rm{T}}_{\rm{n}}}{\rm{(x) = f(a) + }}{{\rm{f}}'}{\rm{(a)}}\frac{{{\rm{(x - a)}}}}{{{\rm{1!}}}}{\rm{ + }}{{\rm{f}}^{''}}{\rm{(a)}}\frac{{{{{\rm{(x - a)}}}^{\rm{2}}}}}{{{\rm{2!}}}}{\rm{ + }}{{\rm{f}}^{'''}}{\rm{(a)}}\frac{{{{{\rm{(x - a)}}}^{\rm{3}}}}}{{{\rm{3!}}}}{\rm{ + \ldots + }}{{\rm{f}}^{\rm{n}}}{\rm{(a)}}\frac{{{{{\rm{(x - a)}}}^{\rm{n}}}}}{{{\rm{n!}}}}\)

When the remainder is calculated as follows:

\({{\rm{R}}_{\rm{n}}}{\rm{(x) = }}{{\rm{f}}^{{\rm{(n + 1)}}}}{\rm{(z)}}\frac{{{{{\rm{(x - a)}}}^{{\rm{n + 1}}}}}}{{{\rm{(n + 1)!}}}}\)

\(z\)is the number between x and a.

\({\rm{f(x)}}\)is a function defined by

\({\rm{f(x) = }}{{\rm{T}}_{\rm{n}}}{\rm{(x) + }}{{\rm{R}}_{\rm{n}}}{\rm{(x)}}\)

As a result, the Taylor polynomial function\({{\rm{e}}^{\rm{x}}}\)with\({{\rm{q}}^{{\rm{th}}}}\)degree and its remainder around\({\rm{x = a}}\)is given as

\(\begin{aligned}{}{\rm{f(x) = }}{{\rm{T}}_{\rm{n}}}{\rm{(x) + }}{{\rm{R}}_{\rm{n}}}{\rm{(x)}}{{\rm{e}}^{\rm{x}}}{\rm{ = f(a) + }}{{\rm{f}}'}{\rm{(a)}}\frac{{{\rm{(x - a)}}}}{{{\rm{1!}}}}{\rm{ + }}{{\rm{f}}^{''}}{\rm{(a)}}\frac{{{{{\rm{(x - a)}}}^{\rm{2}}}}}{{{\rm{2!}}}}{\rm{ + }}{{\rm{f}}^{'''}}{\rm{(a)}}\frac{{{{{\rm{(x - a)}}}^{\rm{3}}}}}{{{\rm{3!}}}}{\rm{ + \ldots + }}{{\rm{f}}^{\rm{n}}}{\rm{(a)}}\frac{{{{{\rm{(x - a)}}}^{\rm{q}}}}}{{{\rm{q!}}}}\\{\rm{ + }}{{\rm{f}}^{{\rm{(q + 1)}}}}{\rm{(z)}}\frac{{{{{\rm{(x - a)}}}^{{\rm{q + 1}}}}}}{{{\rm{(q + 1)!}}}}\;\;\;{\rm{ (where z lies between x and a ) }}\end{aligned}\)

Taking x=1 and a=0, we have

\(\begin{aligned}{}{{\rm{e}}^{\rm{1}}}{\rm{ = f(0) + }}{{\rm{f}}'}{\rm{(0)}}\frac{{{\rm{(1 - 0)}}}}{{{\rm{1!}}}}{\rm{ + }}{{\rm{f}}^{''}}{\rm{(0)}}\frac{{{{{\rm{(1 - 0)}}}^{\rm{2}}}}}{{{\rm{2!}}}}{\rm{ + }}{{\rm{f}}^{'''}}{\rm{(0)}}\frac{{{{{\rm{(1 - 0)}}}^{\rm{3}}}}}{{{\rm{3!}}}}{\rm{ + \ldots + }}{{\rm{f}}^{\rm{n}}}{\rm{(1)}}\frac{{{{{\rm{(1 - 0)}}}^{\rm{q}}}}}{{{\rm{q!}}}}\\{\rm{ + }}{{\rm{f}}^{{\rm{(q + 1)}}}}{\rm{(z)}}\frac{{{{{\rm{(1 - 0)}}}^{{\rm{q + 1}}}}}}{{{\rm{(q + 1)!}}}}\;\;\;{\rm{( where 0 < z < 1)}}\\{{\rm{e}}^{\rm{1}}}{\rm{ = 1 + 1 \times }}\frac{{\rm{1}}}{{{\rm{1!}}}}{\rm{ + 1 \times }}\frac{{\rm{1}}}{{{\rm{2!}}}}{\rm{ + 1 \times }}\frac{{\rm{1}}}{{{\rm{3!}}}}{\rm{ + \ldots + 1 \times }}\frac{{\rm{1}}}{{{\rm{q!}}}}\\\;\;\;{\rm{ + }}{{\rm{e}}^{\rm{z}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{(q + 1)!}}}}{\rm{;}}\;\;\;\;\;\;{\rm{( where 0 < z < 1)}}\\\left. {{\rm{e = 1 + }}\frac{{\rm{1}}}{{{\rm{1!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{3!}}}}{\rm{ + \ldots + }}\frac{{\rm{1}}}{{{\rm{q!}}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}{\rm{;}}\;\;\;\;\;\;{\rm{ (where 0 < z < 1}}} \right)\\{\rm{e = }}{{\rm{s}}_{\rm{q}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\;\;\;{\rm{( where 0 < z < 1)}}\end{aligned}\)

With \({{\rm{s}}_{\rm{q}}}{\rm{ = 1 + }}\frac{{\rm{1}}}{{{\rm{1!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{3!}}}}{\rm{ + \ldots + }}\frac{{\rm{1}}}{{{\rm{q!}}}}\)

Consequently, we have

\(\frac{{\rm{p}}}{{\rm{q}}}{\rm{ = e = }}{{\rm{s}}_{\rm{q}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\;\;\;{\rm{( where 0 < z < 1)}}\)

Therefore, the given statement \(\frac{{\rm{p}}}{{\rm{q}}}{\rm{ = e = }}{{\rm{s}}_{\rm{q}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\;\;\;{\rm{( where 0 < z < 1)}}\) is proved .

prove \({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right)\) is an integer

b)

Consider the given values and simplify,

\(\begin{aligned}{}\frac{{\rm{p}}}{{\rm{q}}}{\rm{ = e = }}{{\rm{s}}_{\rm{q}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\;\;\;\;\;\;\;\;\;{\rm{( where 0 < z < 1)}}\\{\rm{ }}{{\rm{s}}_{\rm{q}}}{\rm{ = 1 + }}\frac{{\rm{1}}}{{{\rm{1!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{3!}}}}{\rm{ + \ldots + }}\frac{{\rm{1}}}{{{\rm{q!}}}}\end{aligned}\)

We are fully aware of this.

\(\begin{aligned}{}{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = q!e - q!}}{{\rm{s}}_{\rm{q}}}\\{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = q!e - q!}}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{1!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{3!}}}}{\rm{ + \ldots + }}\frac{{\rm{1}}}{{{\rm{p!}}}}} \right)\\{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = q! \times }}\frac{{\rm{p}}}{{\rm{q}}}{\rm{ - q!}}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{1!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2!}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{3!}}}}{\rm{ + \ldots + }}\frac{{\rm{1}}}{{{\rm{q!}}}}} \right)\\{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = }}\frac{{{\rm{pq(q - 1)!}}}}{{\rm{q}}}{\rm{ - }}\left( {{\rm{q! + }}\frac{{{\rm{q!}}}}{{{\rm{1!}}}}{\rm{ + }}\frac{{{\rm{q!}}}}{{{\rm{2!}}}}{\rm{ + }}\frac{{{\rm{q!}}}}{{{\rm{3!}}}}{\rm{ + \ldots + }}\frac{{{\rm{q!}}}}{{{\rm{q!}}}}} \right)\\{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = p(q - 1)! - }}\left( {{\rm{q! + }}\frac{{{\rm{q!}}}}{{{\rm{1!}}}}{\rm{ + }}\frac{{{\rm{q!}}}}{{{\rm{2!}}}}{\rm{ + }}\frac{{{\rm{q!}}}}{{{\rm{3!}}}}{\rm{ + \ldots + 1}}} \right)\end{aligned}\)

The first term is a whole number, and the second term is also a whole number. Consequently, we have

\(\begin{aligned}{}{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = integer - integer }}\\{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = integer }}\\{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = integer }}\end{aligned}\)

(Proved.)

Therefore,\({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = integer }}\)is proved.

Prove that \({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < 1}}\)

c)

Let us evaluate,

We've got,

\(\begin{aligned}{}{\rm{e = }}{{\rm{s}}_{\rm{q}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\;\;\;{\rm{( where 0 < z < 1)}}\\\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\;\;\;{\rm{( where 0 < z < 1)}}\\ \Rightarrow {\rm{0 < }}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < }}\frac{{{{\rm{e}}^{\rm{1}}}}}{{{\rm{(q + 1)!}}}}\\{\rm{0 < }}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < }}\frac{{\rm{e}}}{{{\rm{(q + 1)!}}}}\\{\rm{0 \times q! < }}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{q! < }}\frac{{{\rm{eq!}}}}{{{\rm{(q + 1)!}}}}\\{\rm{0 < }}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{q! < }}\frac{{\rm{e}}}{{{\rm{(q + 1)}}}}\\{\rm{0 < }}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{q! < }}\frac{{\rm{e}}}{{{\rm{(q + 1)}}}}{\rm{ < }}\frac{{\rm{3}}}{{{\rm{(q + 1)}}}}\\{\rm{0 < }}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{q! < }}\frac{{\rm{3}}}{{{\rm{(q + 1)}}}}\end{aligned}\)

We have\({\rm{q > 2}}\)since,

\(\begin{aligned}{}{\rm{0 < q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < }}\frac{{\rm{3}}}{{{\rm{(q + 1)}}}}\\{\rm{0 < q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < }}\frac{{\rm{3}}}{{{\rm{(2 + 1)}}}}\\{\rm{0 < q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < 1}}\;\;\\\;{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < 1}}\\{\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < 1}}\end{aligned}\)

Therefore, \({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < 1}}\) is proved.

Prove that \({\rm{e is an irrational number}}\)

d)

Consider the equation and simplify,

\({\rm{e = }}\frac{{\rm{p}}}{{\rm{q}}}{\rm{ = }}{{\rm{s}}_{\rm{q}}}{\rm{ + }}\frac{{{{\rm{e}}^{\rm{z}}}}}{{{\rm{(q + 1)!}}}}\;\;\;{\rm{( where 0 < z < 1)}}\) (1)

As a result, we have

\({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ = integer }}\) (2)

We also have

\({\rm{q!}}\left( {{\rm{e - }}{{\rm{s}}_{\rm{q}}}} \right){\rm{ < 1}}\) (3)

Relationships (2) and (3) are incompatible.

As a result, the assumed relationship (1) is false.

As a result, we have

\({\rm{e}} \ne \frac{{\rm{p}}}{{\rm{q}}}\)

Therefore, \({\rm{e is an irrational number e}} \ne \frac{{\rm{p}}}{{\rm{q}}}\) is proved.