e is irrational


From the Taylor seriesMathworldPlanetmath for ex we know the following equation:

e=k=01k!. (1)

Now let us assume that e is rational. This would there are two natural numbersMathworldPlanetmath a and b, such that:

e=ab.

This yields:

b!e.

Now we can write e using (1):

b!e=b!k=01k!.

This can also be written:

b!e=k=0bb!k!+k=b+1b!k!.

The first sum is obviously a natural number, and thus

k=b+1b!k!

must also be . Now we see:

k=b+1b!k!=1b+1+1(b+1)(b+2)+<k=1(1b+1)k=1b.

Since 1b1 we conclude:

0<k=b+1b!k!<1.

We have also seen that this is an integer, but there is no integer between 0 and 1. So there cannot exist two natural numbers a and b such that e=ab, so e is irrational.

Title e is irrational
Canonical name EIsIrrational
Date of creation 2013-03-22 12:33:02
Last modified on 2013-03-22 12:33:02
Owner mathwizard (128)
Last modified by mathwizard (128)
Numerical id 13
Author mathwizard (128)
Entry type Theorem
Classification msc 11J82
Classification msc 11J72
Related topic ErIsIrrationalForRinmathbbQsetminus0
Related topic NaturalLogBase