Is this proof valid? (e is irrational)

[Calu]

Is this proof that e is an irrational number valid?

e = ∑^{∞}_{n=0} 1/n! = 1 + 1/1! + 1/2! + 1/3! + ... + 1/n! +...

Let e = a + b where

a = S_n = 1 + 1/1! + 1/2! + 1/3! + ... + 1/n!

b= 1/(n+1)! + 1/(n+2)! + 1/(n+3)! +...

Multiply both sides by (n!) giving e(n!) = a(n!) + b(n!)

For proof by contradiction, take e to be an integer, such that e = m/n where m and n are both positive integers and in their lowest terms.

We can take n to be a positive integer and as such, n! is a positive integer.

a(n!) = (1 + 1/1! + 1/2! + 1/3! + ... + 1/n!)(n!) = (n! + n!/1! + n!/2! + n!/3! + ... + n!/n!)

We can see a(n!) is positive integer as n > 1,2,3 ... ,n-1,n and as such each fraction in the parentheses will cancel down to produce a positive integer. The sum of positive integers is yet another positive integer.

We can note that according to our equation, e = a + b, b is simpy the difference of two positive integers, which in turn is a positive integer.

However, it can be shown that b(n!) is not a positive integer:

b(n!) = [1/(n+1)! + 1/(n+2)! + 1/(n+3)! +...](n!) = [n!/(n+1)! + n!/(n+2)! + n!/(n+3)! +...]
= 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + ...

We know n is a positive integer ∴ n ≥ 1, n+1 ≥ 2, n+2 > 2, n+3 >2.

So we have:

b(n!) < 1/2 + 1/2² + 1/2³ +...

b(n!) < a/1-r
b(n!) < 0.5/0.5
b(n!) < 1

We know b(n!) > 0 as e - a ≠ 0
∴ 0 < b(n!) < 1 which means b(n!) cannot be an integer and contradicts the statement above, that b(n!) is the difference of two integers.

Therefore e must be irrational.



This was how it was taught to me, but I don't feel confident on the proof. I haven't proven that b is irrational, which would tell me that e is irrational, as e would become the summation of a rational and an irrational number multiplied by n!, where n! is an integer and a rational number. Can someone tell me if the proof is valid?

 

[mfb]

The order and some steps are a bit messy, but the general idea is valid.

I haven't proven that b is irrational

That follows from the proof, but you don't need it for the contradiction.

 

[Calu]

The order and some steps are a bit messy, but the general idea is valid.

That follows from the proof, but you don't need for the contradiction.

Ahh, okay, I see. Thank you very much for the confirmation. Anything I can do to tidy it up a bit?

 

[mfb]

- don't use n for two different things (as index in the sum and as denominator for e afterwards)
- start with the assumption e=m/n, otherwise your split e=a+b is not well-defined (where are we supposed to do it?)

We can see a(n!) is positive integer as n > 1,2,3 ... ,n-1,n

n>n?

We can note that according to our equation, e = a + b, b is simpy the difference of two positive integers, which in turn is a positive integer.

You cannot tell "positive" from the difference, the difference of two positive integers can be negative. "Integer" is sufficient here, you can show b>0 afterwards.

b(n!) < a/1-r

"a" was used for something else before, "a" and "r" are not introduced and the denominator needs brackets (or a proper fraction: $\frac{a}{1-r}$).

We know b(n!) > 0 as e - a ≠ 0

The first part does not follow from the second part.

 

[Calu]

- don't use n for two different things (as index in the sum and as denominator for e afterwards)
- start with the assumption e=m/n, otherwise your split e=a+b is not well-defined (where are we supposed to do it?)

If I start with the assumption e=$\frac{p}{q}$ can I state that p,q must either both be positive or both be negative?

I'll start with my assumption in the future.

n>n?

This was meant to be ≥, apologies.

You cannot tell "positive" from the difference, the difference of two positive integers can be negative. "Integer" is sufficient here, you can show b>0 afterwards.

That's the same thought I had.

"a" was used for something else before, "a" and "r" are not introduced and the denominator needs brackets (or a proper fraction: $\frac{a}{1-r}$).

How do I create a proper fraction? I couldn't find it in the list of symbols.

The first part does not follow from the second part.

Would b(n!) > 0 as e(n!) - a(n!) ≠ 0 be correct?

Dividing by (n!):

b > 0 as e - a ≠ 0
Thanks for the help.

 

[mfb]

If I start with the assumption e=$\frac{p}{q}$ can I state that p,q must either both be positive or both be negative?

You can even assume p and q to be positive, as you know e is positive (or if you don't know this, it is easy to see based on the sum), but it does not matter as long as q is positive and that is always possible.

How do I create a proper fraction? I couldn't find it in the list of symbols.

This forum supports LaTeX.

Would b(n!) > 0 as e(n!) - a(n!) ≠ 0 be correct?

Is -4.3(n!) > 0 because 2.7(n!) - 7(n!) ≠ 0?

 

[Calu]

You can even assume p and q to be positive, as you know e is positive (or if you don't know this, it is easy to see based on the sum), but it does not matter as long as q is positive and that is always possible.


This forum supports LaTeX.


Is -4.3(n!) > 0 because 2.7(n!) - 7(n!) ≠ 0?

No. In what way do we know b > 0?

Also I found a much more concise explanation here: http://cazelais.disted.camosun.bc.ca/250/e-irrational.pdf. However some of the notation is confusing me. I believe R_n is the value I have labelled b in my original post, but I'm unsure as to why it takes a value such that 0 < R_n < $\frac{3}{(n+1)!}$

And I'm also unsure what "Choose n > max(b,3)" means.

 

[mfb]

No. In what way do we know b > 0?

In the same we can know a>0: it is a sum of positive numbers.

but I'm unsure as to why it takes a value such that 0 < R_n < $\frac{3}{(n+1)!}$

You can get this in the same way you found 1/n! as upper limit - it is just a bit more general as it covers the case n=0 (which is not needed afterwards).

And I'm also unsure what "Choose n > max(b,3)" means.

n is chosen to be larger than b, and at least 4.