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Re: Quadruple with (a, b, c, d)
Posted:
Feb 2, 2013 11:33 PM


"mina_world" wrote: > >Hello teacher~ > >Suppose not all four integers a, b, c, d are equal. >Start with (a, b, c, d) and repeatedly replace (a, b, c, d) >by (ab, bc, cd, da). > >Then at least one number of the quadruple will eventually >become arbitrarily large. > > >Solution: > >Let P_n = (a_n, b_n, c_n, d_n) be the quadruple after >n iterations. > >Then a_n + b_n + c_n + d_n = 0 for n >= 1. > >A very importand function for the point P_n in 4space >is the square of its distance from the origin (0,0,0,0), >which is (a_n)^2 + (b_n)^2 + (c_n)^2 + (d_n)^2. > >.... >omission >For reference, text copy with jpg. >http://board2.blueweb.co.kr/user/math565/data/math/olilim.jpg >... > >for n >=2, >(a_n)^2 + (b_n)^2 + (c_n)^2 + (d_n)^2 >>= {2^(n1)}*{(a_1)^2 + (b_1)^2 + (c_1)^2 + (d_1)^2}. > >The distance of the points P_n from the origin increases >without bound, which means that at least one component must >become arbitrarily large > > >Hm, my question is... > >I know that a_n + b_n + c_n + d_n = 0 >and >I know that (a_n)^2 + (b_n)^2 + (c_n)^2 + (d_n)^2 increases >without bound. > >I can't understand that "at least one component must become >arbitrarily large". I need your logical explanation or proof.
Let
S_a = {a_n  n in N} S_b = {b_n  n in N} S_c = {c_n  n in N} S_d = {d_n  n in N}
The goal is to show that at least one of the sets
S_a, S_b, S_c, S_d
is unbounded above.
Since the set
{(a_n)^2 + (b_n)^2 + (c_n)^2 + (d_n)^2  n in N}
is unbounded, it follows that at least one of the sets
S_a, S_b, S_c, S_d
is unbounded.
Without loss of generality, assume S_a is unbounded.
If S_a is unbounded above, we're done, so assume instead that S_a is bounded above.
Since S_a is unbounded, S_a must be unbounded below.
Then
a_n + b_n + c_n + d_n = 0 for all n in N
=> b_n + c_n + d_n = (a_n) for all n in N
=> {b_n + c_n + d_n  n in N} is unbounded above
=> at least one of S_b, S_c, S_d is unbounded above
so at least one of S_a, S_b, S_c, S_d is unbounded above, as was to be shown.
Oh, thank you very much for your detailed explanation.



