MHB Proof of the Irrationality of e - attributed to Joseph Fourier

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(NOTE: Maybe this post belongs in the Number Theory Forum? Apologies if it is wrongly located!)

I am reading Julian Havil's book, "The Irrationals: The Story of the Numbers You Can't Count On"

In Chapter 4: Irrationals, Old and New, Havil gives a proof of the irrationality of e which was attributed to Joseph Fourier ... the proof is by contradiction and reads as follows:View attachment 3571In the above proof, Havil writes:

"Then

$$n!e = n! \frac{m}{n} = (n-1)! m$$

$$= ( n! + \frac{n!}{1!} + \frac{n!}{2!} + \frac{n!}{3!} + \frac{n!}{4!} + \ ... \ ... \ \frac{n!}{n!} ) + \ ... \ ... \ + R$$

which makes $$R \ne 0$$ the difference between two integers and so an integer itself."Can someone please explain how the expression above demonstrates that R is an integer?

Peter***EDIT***

oh! just saw the answer I think ... I was thrown by the ellipses ( that is the ... ... ) after the term in parentheses (brackets) in the expression:

$$= ( n! + \frac{n!}{1!} + \frac{n!}{2!} + \frac{n!}{3!} + \frac{n!}{4!} + \ ... \ ... \ \frac{n!}{n!} ) + \ ... \ ... \ + R$$

I thought there was some term there ... but did not know what ... but now believe that there is nothing there ...

That is I think maybe it should read:

$$n!e = ( n! + \frac{n!}{1!} + \frac{n!}{2!} + \frac{n!}{3!} + \frac{n!}{4!} + \ ... \ ... \ \frac{n!}{n!} ) + R$$

BUT ... why did Havil include the ... ... in the expression?

... just a typo? ... or is there actually a term there?
 
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Hi Peter,

I'm not sure why he wrote the ... between the expansion and R, but you got the idea.
 
Fallen Angel said:
Hi Peter,

I'm not sure why he wrote the ... between the expansion and R, but you got the idea.

Thanks Fallen Angel ... glad you can confirm it is mysterious ... or probably a typo ...

Peter
 
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