And Square Root is the Power to HALF......One on TWO:?:oo:
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And Square Root is the Power to HALF......One on TWO:?:oo:
Sq Root is the Power to 0.5.....Same thing
The first post said "Use exactly four 9s and any number of arithmethic or higher level math operations." If the operation can be expressed by symbols without numeric terms it seems to be allowed: log (x), !, etc.
The issue here is whether to allow the symbol for square root because by convention the exponent 2 is omitted, while cubic root, for example, is not allowed because the exponent 3 must be written. The 9th root of nine, however, would have to be allowed, unquestionably, but it uses two of the four 9s allowed and is useless anyway.
We need to ask the third umpire.:D
:clap3: He got game.
I have a way to do it but I am not going to.
Consider
ln e = 1,
e/e = 1,
ln pi / ln pi = 1
pi / pi = 1
We can have number 1 without using any 9s. :2tsup:
But I won’t use it because it defects the purpose of this thread.
It is a bit like the stuff lawyers do. Legitimate but very very dodgy. :D
Here is another one
cos(pi) = -1
cos(pi) * cos(pi) = 1
:D
I beg to disagree.:) It is not a legitimate way because the "operation" is not performed on the four nines, it is performed on other numeric entities, albeit represented by symbols (pi, e, etc.). Have you found how to legitimately get 1 from only one 9?Quote:
Originally Posted by Wongo
No bro.Quote:
Have you found how to legitimately get 1 from only one 9?
I have been trying to do exactly that. If I could crack that I could crack 67. No luck so far.:doh:
infinite root of 9
Infinite root of 9 doesn't make it 1.
I think the real solution does exist though. The truth is out there.
Almost.Quote:
Originally Posted by Frank&Earnest
Zelk
yes it does... you could write it as lim (x->infinite) of x root of 9, if you want to be that way...:DQuote:
Infinite root of 9 doesn't make it 1.
limit is an approximation.:cool:
What is 9 divided by infinity then? Don't tell me it is 0.
:D
I guess you are one of those who still refuse to accept that 0.99999... recurrent is =1, then. You are fighting a rearguard battle, but if that's what lifts your boat...:p I'm not a good enough mathematician to follow you down that path.
Just to amuse me: assuming that you could get 1 from one 9, how do you get 67? It still baffles me.
Hang on mate!!
0.99999... is not 1. Why is it so hard to understand?
I have more than 1 way to do it but it is pointless so I won't show you. That will keep you thinking.... :UQuote:
Just to amuse me: assuming that you could get 1 from one 9, how do you get 67? It still baffles me.