- #1
Calu
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Is this proof that e is an irrational number valid?
e = ∑[itex]^{∞}_{n=0}[/itex] 1/n! = 1 + 1/1! + 1/2! + 1/3! + ... + 1/n! +...
Let e = a + b where
a = Sn = 1 + 1/1! + 1/2! + 1/3! + ... + 1/n!
b= 1/(n+1)! + 1/(n+2)! + 1/(n+3)! +...
Multiply both sides by (n!) giving e(n!) = a(n!) + b(n!)
For proof by contradiction, take e to be an integer, such that e = m/n where m and n are both positive integers and in their lowest terms.
We can take n to be a positive integer and as such, n! is a positive integer.
a(n!) = (1 + 1/1! + 1/2! + 1/3! + ... + 1/n!)(n!) = (n! + n!/1! + n!/2! + n!/3! + ... + n!/n!)
We can see a(n!) is positive integer as n > 1,2,3 ... ,n-1,n and as such each fraction in the parentheses will cancel down to produce a positive integer. The sum of positive integers is yet another positive integer.
We can note that according to our equation, e = a + b, b is simpy the difference of two positive integers, which in turn is a positive integer.
However, it can be shown that b(n!) is not a positive integer:
b(n!) = [1/(n+1)! + 1/(n+2)! + 1/(n+3)! +...](n!) = [n!/(n+1)! + n!/(n+2)! + n!/(n+3)! +...]
= 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + ...
We know n is a positive integer ∴ n ≥ 1, n+1 ≥ 2, n+2 > 2, n+3 >2.
So we have:
b(n!) < 1/2 + 1/22 + 1/23 +...
b(n!) < a/1-r
b(n!) < 0.5/0.5
b(n!) < 1
We know b(n!) > 0 as e - a ≠ 0
∴ 0 < b(n!) < 1 which means b(n!) cannot be an integer and contradicts the statement above, that b(n!) is the difference of two integers.
Therefore e must be irrational.
This was how it was taught to me, but I don't feel confident on the proof. I haven't proven that b is irrational, which would tell me that e is irrational, as e would become the summation of a rational and an irrational number multiplied by n!, where n! is an integer and a rational number. Can someone tell me if the proof is valid?
e = ∑[itex]^{∞}_{n=0}[/itex] 1/n! = 1 + 1/1! + 1/2! + 1/3! + ... + 1/n! +...
Let e = a + b where
a = Sn = 1 + 1/1! + 1/2! + 1/3! + ... + 1/n!
b= 1/(n+1)! + 1/(n+2)! + 1/(n+3)! +...
Multiply both sides by (n!) giving e(n!) = a(n!) + b(n!)
For proof by contradiction, take e to be an integer, such that e = m/n where m and n are both positive integers and in their lowest terms.
We can take n to be a positive integer and as such, n! is a positive integer.
a(n!) = (1 + 1/1! + 1/2! + 1/3! + ... + 1/n!)(n!) = (n! + n!/1! + n!/2! + n!/3! + ... + n!/n!)
We can see a(n!) is positive integer as n > 1,2,3 ... ,n-1,n and as such each fraction in the parentheses will cancel down to produce a positive integer. The sum of positive integers is yet another positive integer.
We can note that according to our equation, e = a + b, b is simpy the difference of two positive integers, which in turn is a positive integer.
However, it can be shown that b(n!) is not a positive integer:
b(n!) = [1/(n+1)! + 1/(n+2)! + 1/(n+3)! +...](n!) = [n!/(n+1)! + n!/(n+2)! + n!/(n+3)! +...]
= 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + ...
We know n is a positive integer ∴ n ≥ 1, n+1 ≥ 2, n+2 > 2, n+3 >2.
So we have:
b(n!) < 1/2 + 1/22 + 1/23 +...
b(n!) < a/1-r
b(n!) < 0.5/0.5
b(n!) < 1
We know b(n!) > 0 as e - a ≠ 0
∴ 0 < b(n!) < 1 which means b(n!) cannot be an integer and contradicts the statement above, that b(n!) is the difference of two integers.
Therefore e must be irrational.
This was how it was taught to me, but I don't feel confident on the proof. I haven't proven that b is irrational, which would tell me that e is irrational, as e would become the summation of a rational and an irrational number multiplied by n!, where n! is an integer and a rational number. Can someone tell me if the proof is valid?