First I find the cube root of 12 and 18
Cube root of 12 = 2.28942849
Cube root of 18 = 2.62074139
Then I add the two cube roots
Cube root of 12 * Cube root of 18 = 6
Therefore,
Cube root of 12 * Cube root of 18 = Cube root of 216
Can you please explain this:
2.28942849 * 2.28942849 * 2.62074139 * 2.62074139 * 2.62074139 = 216
I’m sure we’ve done this already, but:
x=x – I think we all agreed that one, didn’t we? - (whatever something is, that’s what it is)
from that:
x-x=0 – (whatever something is, removing it results in its removal)
and
x/x=1 – whatever something is, that’s what it is - exactly one instance of itself
So…
x-x = 0 = (x/x)-1
or
π-π = 0 = (π/π)-1
Whatever π is, removing it results in its removal; this is the same as removing one from the number of instances π (whatever it is) is of itself.
What am I missing?
Sure, but it doesn’t need to be. We know the answer right from the start: it will be 0 followed by an infinite series of 0s. Actually going through with the calculation would be pointless, because there’s no conceivable way we could ever get anything but 0 by subtracting a digit from itself.
Neither of these statements is true.
Sure, but what I think he’s driving at, after all this, is just that the one way you can’t prove it is longhand; you have to infer it logically, rather than calculate it mechanically.
But, as others have said, this is true for integers, which still have all those decimals after them, they just happen to be filled with zeroes.
your math is so terrible it’s bolstering my confidence =)
Those are of course approximations to the true values. But fair enough; let’s run with that.
You’re multiplying them actually — which is good, as that’s what was in my example.
Certainly true, though I would have passed through this equality on my way to the one above, rather than vice-versa.
It doesn’t equal that, actually. Probably you left out a factor of 2.28942849 on the left.
I think someone should get Kozmik a computer algebra system like Mathematica, so that he can see that computers are perfectly capable of symbolic manipulation and are not limited to computations in decimal notation.
From Koskik’s link:
“In computability theory and computer programming, a crucial part of the definition of an algorithm is that it must terminate—that is, produce its answer after running for a finite number of steps. Whether this number is or isn’t too large for practical execution of the algorithm on a real computer is the subject of computational complexity theory.”
It isn’t! What I was referring to was the Turing Halting Problem, which is linked to from the link you gave.
The Turing Halting Problem applies to certain algorithms, whereby it is impossible to predict whether or not they will terminate for a particular input. The only way to get a partial answer is to put an input in and let it run.
If it terminates, then you know it will terminate for that input. If it doesn’t terminate, you DON’T know whether or not it will eventually if you only wait a little bit longer, and there is no way to find out. However long you wait, just because it hasn’t terminated doesn’t mean that it definitely won’t if you have a couple of millenia more patience! Note that this is not the same as being stuck in a loop - the algorithm isn’t repeating itself. This is a true example of an uncomputable problem.
No it isn’t! You’re getting hung up on the idea that “π” is irrational and so can only be approximately represented using integers, and so “can’t” be represented in a computer! Numbers in a computer are arrays of charged capacitors, or patches of magnetised material. They only stand for integers because we chose that they do. They could just as easily stand for irrationals, or negative numbers, or imaginary numbers, or the abstract “x”.
Just because you can’t write π as the ratio of two integers doesn’t make the subtraction or irrationals intractable, no more than the lack of negative numbers in Roman numerals would make negative algebra intractable. Truly intractable problems are interesting, but π - π = π / π - 1 IS NOT one of them!
(Fun fact, 22 π[sup]4[/sup] = 2143. Exactly. Who’d a thunk it? Substitute the fourth root of 2143/22 for π into π - π = π / π - 1 and it all becomes so easy…)
Actually, it’s not exact. Pi is a transcendental number, which means it is not the root of any polynomial with integer coefficients. The fourth root of 2143/22, on the other hand, is algebraic (the opposite of transcendental), being a root of the polynomial 22x[sup]4[/sup]-2143.
Well that’s spoilt my day! Pesky Ramanujan and his approximations… you did provoke me into reading about transcendental numbers though. More edumacation for me!
Hey, sorry about that, matt! To make it up to you, here’s another approximation involving pi (and e):
Interestingly, e[sup]pi*sqrt(163)[/sup] is very nearly an integer; it is approximately 262,537,412,640,768,743.99999999999925.
Isn’t revealing this a violation of the mathematicians’ code?
(For my next trick, I’ll raise e to the i pi power – without a net!)