proof

[MHENK]

How would I go about proving that

ln(3)
------
ln(6)

is irrational?

[wossname]

There's a good chance I'm wrong here, but isn't it impossible to prove that a number is irrational? If you prove it then you have obviously found a pattern in the digits and ergo it isn't irrational. But if you use a computer to search for patterns in the digits then you would have to run the algorithm for an infinitely long time and have a number that was infinitely accurate. You'd have to stop running it at some arbitrary point. 99.999999999999999999999999% certain is still not mathematical proof.

I'm talking from a coder's perspective though. A math jockey would probably know better.

[MHENK]

I believe that to prove that this is irrational, i would start by assuming that it IS rational. Then, manipulating the numbers and applying algebra SHOULD lead to a contradiction, which would prove that it is NOT rational, therefore it IS irrational.

Assuming that ln(3)/ln(6) is rational implies that there exist integers a and b, such that ln(3)/ln(6) = a/b.

I have no idea where to go from here, but I think my idea of a contradiction is the right way to go... somehow i need to manipulate the formula ln(3)/ln(6) = a/b into something that is a contradiction...

[MHENK]

I'm not sure if this is any progress, but now I've gotten to the point where I have

3^a=6^b

where a and b are both integers, not equal to zero.

If there is any way any of you know how to prove that this is NOT true, please let me know...

[MHENK]

Okay, i THINK i have it, see if you agree with me.

I worked the equation out and got

3^a=6^b.

3^a=(2^b)(3^b)

This cannot be true, because every integer splits uniquely into a product of primes, and if statement 2 above were true, there would be a number that split into 2 different products of primes. Therefore 3^a<>6^b. Therefore, by contradiction, ln(3)/ln(6) is irrational.

Look right to anyone???

[nicktheman]

Ok assume ln3/ln6 = a/b
Then 3/6 = (e^a)/(e^b)
e^b = 2(e^a)
b = a*ln2
ln2 IS irrational, therefore a and b cannot BOTH be rational

I'm not sure how strong this proof is but i dont have long and cant htink of a better 1 in 5 mins

hope this helps

[MarcoA]

Suppose that ln(3)/ln(6) is rational i.e.:

ln(3)/ln(6)=x/y (x,y natural numbers)
Then we have:

3^y = 6^x

i.e. a pair number is equal to an odd number, then ln(3)/ln(6) is irrational