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Einstein Tue 14 Jan 16:37:50 2020 No.11301491 View Reply Original Report

Quoted By: >>11301497 >>11301502 >>11301528 >>11301593

Can I express every element of N as 2^a*3^b where a and b are elements of Q ?

Anonymous

Anonymous Tue 14 Jan 2020 16:44:16 No.11301497 Report

Quoted By:

>>11301491
i dont do caggot theory but if Q is all the natural numbers and N is also all the natural numbers sure

Anonymous

Anonymous Tue 14 Jan 2020 16:46:20 No.11301502 Report

Quoted By:

>>11301491
No. 5 is not representable this way. We write a*ln2 + b*ln3 = ln5. We would like b = (ln5 - a*ln2)/ln3 to be in Q given that a is in Q. Well, I guess I don't know that this is irrational, but it certainly should be. I wonder if we can prove it.

Anonymous

Anonymous Tue 14 Jan 2020 17:07:39 No.11301528 Report

Quoted By: >>11301530

>>11301491
Suppose that
5 = 2^((k-m)/n) * 3^((u-v)/w)
then
5^(nw) = 2^(kw - mw) * 3^(un - vn)
5^(nw) * 2^(mw) * 3^(vn) = 2^(kw) * 3^(un)
The LHS is divisble by 5 but the RHS isn't.

Einstein

Einstein Tue 14 Jan 2020 17:11:57 No.11301530 Report

Quoted By:

>>11301528
k qed thread closed thanks fag

Anonymous

Anonymous Tue 14 Jan 2020 18:03:28 No.11301593 Report

Quoted By:

>>11301491
you ain’t sneaky

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