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Well I just read about ^^ (2^^2 = 2^2)
I was wondering, it was said that 3^^3 = 3^(3^3) BUT
^^ is a series of exponentials, so the answer should be 3^3^3, and PEDMA says you read from left to right, so it would be (3^3)^3
3^(3^3) = 3^27
(3^3)^3 = 3^(3*3) = 3^9
I'd say that's a pretty big difference, what do you think?
where did you read about it? ^ is right associative, and you never see x^ab as (x^a)^b
x^ab = (x^a)*b
I was saying that if you write:
2^2^2 should it be (2^2)^2 or 2^(2^2)
lets type it out explicitely:Quote:
Originally posted by alkatran
x^ab = (x^a)*b
I was saying that if you write:
2^2^2 should it be (2^2)^2 or 2^(2^2)
xab=(x^a)^b
^ is right associative, which means that you put parentesis around the rightmost ^ first so say:
2^2^2^2 = 2^(2^(2^2))
left associative operators work the other way around:
2-2-2-2=((2-2)-2)-2
Ok thanks.
Anybody else find it sortof funny that left associative operators can be rearranged any which way and still get the same answer?
2*3/4 = 3/4*2 = (2*3)/4 = 2*(3/4)
no it doesn't work that way every time. * is not associative only /. try doing that with two / operators and you'll see what i mean.
3/2/4 =1.5/4 = .375
3/4/2 = .75/2 = .375
/4/2*3 = 1/4/2*3 = .125*3 = .375
Seems pretty consistant to me.
errr.. you're treating them as factors.. I thought you were talking about commutation of left associative operators..
3^3^3 = 3^(3^3) <> (3^3)^3
The rule is, the highest power gets calculated first.
(2^3+4^2)^2^4
If you think about this problem. You should do 2^4 first. Then (...)^16.