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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]
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,
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]
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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