Prove that e is an Irrational Number

The number e is irrational as it cannot be written as a ratio of two integers. Its approximated value is 2.72. That is,

e ≈ 2.72.

So e is an irrational number between 2 and 3. The number e is regarded as a mathematical constant (known as Euler’s number or Napier’s constant). Also, e is the base of natural logarithms. In this post, we will show that e is not a rational number.

e is an Irrational Number Proof

Question: Prove that e is irrational.

Proof:

The number e can be expressed as follows:

$e = 1 + \dfrac{1}{1!} +\dfrac{1}{2!} + \dfrac{1}{3!} + \cdots$.

That is,

$e=\sum\limits_{n=0}^\infty \dfrac{1}{n!}$.

Let us put

$S_k=\sum\limits_{n=0}^k \dfrac{1}{n!}$.

Step 1: We claim that 0 < e-Sk < $\frac{1}{k \cdot k!}$.

Note that $e-S_k=\sum\limits_{i=1}^\infty \dfrac{1}{(k+i)!}$. Now, we calculate that

0 < e-S_k < $\dfrac{1}{(k+1)!} \left( 1+\dfrac{1}{k+1} + \dfrac{1}{(k+1)^2}+\cdots \right)$

⇒ 0 < e-S_k < $\dfrac{1}{(k+1)!} \left( \dfrac{1}{1-\frac{1}{k+1}} \right)$

⇒ 0 < e-S_k < $\dfrac{1}{k \cdot k!}$

This proves our claim.

Step 2: Getting a contradiction.

For a contradiction, assume that e is a rational number. So we can write $e =\dfrac{p}{q}$ for some positive integers p and q co-prime to each other.

Putting k = q in step 1, we deduce that

0 < e-S_q < $\dfrac{1}{q \cdot q!}$

⇒ 0 < q! (e-S_q) < $\dfrac{1}{q}$ …(I)

Observe that q! S_q is an integer. Also, q! e is an integer by assumption, because e=p/q. So (I) implies that q! (e-S_q) is an integer lying between 0 and 1, which is a contradiction. So our assumption was wrong.

This proves that e is an irrational number.

Also Read:

• Product of three consecutive integers is divisible by 6
• Square of an odd integer is of the form 8n+1
• Is 3/2 an Integer, Rational Number?

e lies between 2 and 3

Prove that 2 < e < 3.

Answer:

As e = $1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots$, it is obvious that e > 2. To show e<3, we will use the fact

2^n-1 < n! for n≥3.

Therefore,

e = $1 + \dfrac{1}{1!}+ \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots$

⇒ e < $1 + \left( 1+ \dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots \right)$

⇒ e < $1 + \dfrac{1}{1-\frac{1}{2}}$

⇒ e < 3.

So the number e lies between 2 and 3, that is, 2<e<3.

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