how to prove that G is irrational given the fact that G=1 - 1/(3^2) + 1/(5^2) - 1/(7^2) +.....,

Question

asnaseer · Answer

I think you could prove it by contradiction as follows:

1. lets assume that the expression is rational, which means it can be written as a fraction. lets call this fraction $\displaystyle\frac{p}{q}$ where p and q are integers and $\displaystyle\frac{p}{q}$ is in its simplest form.

2. this would imply:$$\frac{p}{q}=1-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+...$$therefore:$$p=q-\frac{q}{3^2}+\frac{q}{5^2}-\frac{q}{7^2}+...$$now, since p is an integer, this implies q must be exactly divisible by $3^2,5^2,7^2,...$. which is an infinite list of factors, therefore q must be infinity, therefore the original expression cannot be written as $\displaystyle\frac{p}{q}$, therefore it must be irrational.

anonymous · Answer

this is not correct, because q doesn't have to be divided by those, the sum could still add up to p, like 2=1+1/2+1/4+1/8+.... 1 are not divisible by 2,4,8,...., hence the sum equal 0

asnaseer · Answer

good point

anonymous · Answer

srry my bad, the sum equal 2 :(

asnaseer · Answer

yes - I assumed it was a typo :)

asnaseer · Answer

ok - I guess this needs more thought...