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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?
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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