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Topic: Quadruple with (a, b, c, d)
Replies: 2   Last Post: Feb 2, 2013 11:33 PM

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quasi

Posts: 10,188
Registered: 7/15/05
Re: Quadruple with (a, b, c, d)
Posted: Feb 2, 2013 7:43 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

"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 (a-b, b-c, c-d, d-a).
>
>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 4-space
>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://board-2.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^(n-1)}*{(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.

quasi



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