- #1

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we know e (exponential) is a irrational number....

how can we prove it??

how can we prove it??

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- Thread starter newton1
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- #1

- 152

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we know e (exponential) is a irrational number....

how can we prove it??

how can we prove it??

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- #2

emu

One way yields an infinite series of non-repeating rational numbers. The sum is therefore irrational.

Try proving the sqrt(5) is irrational.

- #3

HallsofIvy

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One standard method is to use the Taylor's series (which may be what emu meant): e= 1+ 1/2 + 1/6+ ...+ 1/n!+ ...

If j is any positve integer then e*j!= integer+ 1/(q+1)+ 1/(q+1)(q+2)+ ... which is not an integer so e cannot be written as a fraction with denominator j for any j.

A theorem I saw years ago was this: If c> 0, and there exist a function f(x), continuous on [0,c], positive on (0,c) and such that f(x) and its iterated anti-derivatives can be taken to be integer valued at both 0 and c, the c is irrational!

Taking f(x)= sin(x) in this theorem shows that pi is irrational.

It can also be used to prove: If c is a positive number other than 1 and ln(c) is rational, then c is irrational.

Since e is a positive number, not equal to 1, and ln(e)= 1 is rational, it follows that e is irrational.

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- #5

Paradox

- #6

Hurkyl

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e*j!= integer+ 1/(q+1)+ 1/(q+1)(q+2)+ ... which is not an integer

That's not obvious... I don't see why the infinite sequence there cannot add up to an integral value.

Hurkyl

- #7

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It's quite simple to prove that 1/(n+1)!<e-An<1/(n!*n);

Let's suppose e is rational, so it's equal to p/q, where p and q are integers.

1/(n+1)!<e-(1+1/2!+...+1/n!)<1/(n!*n); | *n!;

1/(n+1)<n!*p/q-n!*(1+1/2!+...+1/n!)<1/n;

But between 1/(n+1) and 1/n is no integer...

n!*p/q must be an integer because for n big enough n! is a multiple of q;

So e is not rational...

QED

- #8

climbhi

- #9

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thank you....:)

- #10

Hurkyl

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That bit isn't obvious either.... but the ordinary taylor remainder formula gives e/(n+1)! for that term which is sufficient for the proof. Don't tell me how to get that end of the inequality, it'd be a good exercise to figure it out myself!

Hurkyl

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