#### 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]

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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