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Happiness is doing what is right.
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Piscis Babelis est parvus, flavus, et hiridicus, et est probabiliter insolitissima raritas in toto mundo.
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systematically
"I'm not back."  Bill Harding, Twister
Originally posted by Capablanca Kepler:
Take any four integers a,b,c,d.So we have(a,b,c,d) now change this to
(ab,bc,cd,da),Continue this process,Is it necessary that we will get (0,0,0,0) at the end of process?In how many steps?(ab means absolute value of the difference between a and b)
[ December 23, 2003: Message edited by: Capablanca Kepler ]
systematically
"I'm not back."  Bill Harding, Twister
Originally posted by Jim Yingst:
I can say fairly definitely that if we include all rational numbers, or all integers (unbounded) (rather than just integers within particular bounds) then there is no maximum bound on the number of steps. Given any 4 rational numbers a, b, c, d, which have n cycles we can always find another set a', b', c', d' (bounded in [0, 1] if you like) which has n + 1 cycles. So we could also find another set a'', b'', c'', d'' with n + 2. By induction we can generate a set of 4 rational numbers with any number of cycles desired.
systematically
"I'm not back."  Bill Harding, Twister
systematically
"I'm not back."  Bill Harding, Twister
Originally posted by Jim Yingst:
One problem with squaring both sides of an equation is that you can create additional solutions that aren't solutions of the original equation. E.g.
x  1 = 1
has one solution, x = 2. But if we square both sides, we get
x^2  2x + 1 = 1
x = 0, 2
When we square both sides of an equation we're obligated to check our solution(s) to see if they're still valid for the original equation.
I fear this is a problem for your solution.
let y1=y4=0; y3=1, .......................(5)
systematically
MH
"I'm not back."  Bill Harding, Twister
"I'm not back."  Bill Harding, Twister
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