who_loves_maths
I wanna be a nebula too!!
- Joined
- Jun 8, 2004
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- HSC
- 2005
^ yes. this was my argument exactly - look at my post at the top of this page.
there is a correction that need to be made to your argument, however. you do not need to prove your 'a' is irrational - simply not an integer will suffice. ('seems' almost too easy )
so far, there are, very 'simply', two ways that can be used to prove no_arg wrong conclusively:
1) Prove [ln(pi +1) -2]/(1 - ln(pi)) is non-integral.
or,
2) Prove [6 - ln(pi +1)]/ln(pi) is non-integral.
Both seem like the blindingly obvious since you can't simplify them arithmetically into integers ... but i suppose the point is to prove that now (otherwise i assume no_arg will remain unsatisfied).
there is a correction that need to be made to your argument, however. you do not need to prove your 'a' is irrational - simply not an integer will suffice. ('seems' almost too easy )
so far, there are, very 'simply', two ways that can be used to prove no_arg wrong conclusively:
1) Prove [ln(pi +1) -2]/(1 - ln(pi)) is non-integral.
or,
2) Prove [6 - ln(pi +1)]/ln(pi) is non-integral.
Both seem like the blindingly obvious since you can't simplify them arithmetically into integers ... but i suppose the point is to prove that now (otherwise i assume no_arg will remain unsatisfied).
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