# Why Is the Number e Considered Irrational?

• VietDao29
In summary, the conversation is about proving that e is irrational and it does not make sense. e is not rational and you truncate the series after n+1 terms to get an integer plus a term that's definitely not an integer.
VietDao29
Homework Helper
My book does not make sense to me. Here is what it says:
I know that:
$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n}, 0 < \theta < 1$$
If e is rational then $$e = \frac{m}{n}; m, n \in Z$$
And the greatest common factor of m, n is 1.
$$\Leftrightarrow 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n} = \frac{m}{n}$$
Multiple both sides by n!, the right side is an integer, while the left side is an integer plus $\frac{\theta}{n}$
This makes the contradiction, therefore e is irrational.
My question is: Why they say $$e = \frac{m}{n}; m, n \in Z$$, so that they can multiple this by n! and get an integer. Can $$e = \frac{p}{q}; p, q \in Z$$?
Is there a better way of proving this?
Thanks a lot,
Viet Dao,

Last edited:
there may not be a single line of truth there in the process. e is not rational. you use the taylor series inappropriately, what is theta?
actually the exact definition of e is
$$e=\lim_{h\rightarrow 0}(1+h)^{\frac{1}{h}}=\lim_{h\rightarrow \infty}(1+\frac{1}{h})^h=2.718281828 . . . ..$$

The use of Taylor series is correct. Think of the theta term as an "explicit" error term.

This might be a little less confusing if you changed the order slightly. First assume that e=m/n. You have no control over the m and n in this assumption, only that they exist. Next, since you know

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

you truncate this series after n+1 terms (note the first term is indexed by 0), the size of the denominator, to get your

$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n},\ 0 < \theta < 1$$

etc...

In this order you are truncating based on the denominator of the assumed rational form of e. The way you had it arranged it appears that you truncate at some n, then e just happens to have that same n in the denominator, which would be unlikely (this isn't really what they're doing, but I'm taking a guess that it's what's troubling you). Does that make sense?

You are correct in why they wanted to truncate after n+1 terms, so they could clear the denominators and get an integer plus a term that's definitely not an integer.

If you like, you can replace e=m/n above by e=p/q. Then you'd truncate after p+1 terms, etc. No real change.

shmoe said:
...The way you had it arranged it appears that you truncate at some n, then e just happens to have that same n in the denominator...
This still troubles me...
Say you have:
$$\sum_{n = 0}^{4} \frac{1}{n!} = \frac{65}{24}$$ The denominator is 24, not 4.
$$(\sum_{n = 0}^{4} \frac{1}{n!}) + \frac{1}{4!4} = \frac{87}{32}$$. Theta is 1. The denominator is 32, not 4.
So I still don't know why they assume that the denominator is n, which in this case is 4...
Viet Dao,

Ok, if you happen to 'know' e=m/4 where m is an integer, then it would look like:

$$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{\theta}{4!4}=e=\frac{m}{4}$$

where $$0<\theta<1$$, this inequality is key. This is just:

$$\frac{65}{24}+\frac{\theta}{4!4}=\frac{m}{4}$$

Multiply by 4!=24:

$$65+\frac{\theta}{4}=6m$$

and you have your contradiction. One of the points I was hoping to make is that we are free to truncate the series for e wherever we like. After we assume that e is rational, we then choose to truncate at a convenient spot. If e=m/4, we truncate after 5 terms (the 1/4! term).

Ah, okay. I get it. Thanks.
Viet Dao,

## What is an irrational number?

An irrational number is a real number that cannot be expressed as a ratio of two integers. This means that it cannot be written as a fraction in the form of a/b, where a and b are integers.

## How is E defined as an irrational number?

The irrational number E, also known as Euler's number, is defined as the base of the natural logarithm. In decimal form, it is approximately 2.71828.

## Why is E considered an important irrational number in mathematics?

E is considered an important irrational number in mathematics because it appears in many mathematical equations and has numerous applications in fields such as calculus, statistics, and physics. It also has interesting properties, such as being its own derivative and being the limit of (1 + 1/n)^n as n approaches infinity.

## How is E different from a rational number?

A rational number can be expressed as a finite or recurring decimal, while E cannot be written as a decimal with a pattern that repeats. Rational numbers can also be written as a fraction, while E cannot. Additionally, E is a transcendental number, meaning it is not the root of any non-zero polynomial with rational coefficients, while rational numbers are algebraic numbers.

## Is there a way to approximate E?

Yes, E can be approximated using various methods, such as the Taylor series or the continued fraction expansion. One common approximation is 2.71828, but using more terms in the approximation can lead to a more accurate value.

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