**#Problem 8**

Prove that e is irrational

**Solution scheme and approach**

e=1+1/1!+2/2!+3/3!+.....(i)

Let e is rational so e=p/q

multiplying equation number (i) by q! we get

q!*e=q!+q!/1!+q!/2!+q!/3!+....+q/q!+ other terms

Now as q!+q!/1!+...+q!/q! is an integer and q!*e is also an integer

So we can say other terms should be an integer. Let I signifies the rest terms whose summation is an integer.

I=q![1/(q+1)!+1/(q+2)!+1/(q+3)!+...]

=1/(q+1)+1/(q+1)*(q+2)+1/(q+1)*(q+2)*(q+3)+.....

<1/(q+1)+1/(q+1)^2+1/(q+1)^3+....=1/(q+1)/[1-1/(q+1)]=1/q<1

=>I< 1

Hence contradiction.

**So, e should be an irrational number. Q.E.D.**

[This solution has been proposed by Fourier]

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