E is an irrational number

  • Thread starter VietDao29
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  • #1
VietDao29
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My book does not make sense to me. Here is what it says:
I know that:
[tex]e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n}, 0 < \theta < 1[/tex]
If e is rational then [tex]e = \frac{m}{n}; m, n \in Z[/tex] :confused:
And the greatest common factor of m, n is 1.
[tex]\Leftrightarrow 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n} = \frac{m}{n}[/tex]
Multiple both sides by n!, the right side is an integer, while the left side is an integer plus [itex]\frac{\theta}{n}[/itex]
This makes the contradiction, therefore e is irrational.
My question is: Why they say [tex]e = \frac{m}{n}; m, n \in Z[/tex], so that they can multiple this by n! and get an integer. Can [tex]e = \frac{p}{q}; p, q \in Z[/tex]?
Is there a better way of proving this?
Thanks a lot,
Viet Dao,
 
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Answers and Replies

  • #2
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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
[tex]e=\lim_{h\rightarrow 0}(1+h)^{\frac{1}{h}}=\lim_{h\rightarrow \infty}(1+\frac{1}{h})^h=2.718281828 . . . ..[/tex]
 
  • #3
shmoe
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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

[tex]e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... [/tex]

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

[tex]e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} + \frac{\theta}{n!n},\ 0 < \theta < 1[/tex]

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.
 
  • #4
VietDao29
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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:
[tex]\sum_{n = 0}^{4} \frac{1}{n!} = \frac{65}{24}[/tex] The denominator is 24, not 4.
[tex](\sum_{n = 0}^{4} \frac{1}{n!}) + \frac{1}{4!4} = \frac{87}{32}[/tex]. 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... :cry:
Viet Dao,
 
  • #5
shmoe
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Ok, if you happen to 'know' e=m/4 where m is an integer, then it would look like:

[tex]1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{\theta}{4!4}=e=\frac{m}{4}[/tex]

where [tex]0<\theta<1[/tex], this inequality is key. This is just:

[tex]\frac{65}{24}+\frac{\theta}{4!4}=\frac{m}{4}[/tex]

Multiply by 4!=24:

[tex]65+\frac{\theta}{4}=6m[/tex]

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).
 
  • #6
VietDao29
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Ah, okay. I get it. Thanks. :smile:
Viet Dao,
 

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