if v 1 = 1 + 10 3 + 10 2 3 3 and v 2 = 7 + 4 10 3 + 10 2 3 3 let ω 1 = v 1 − k and let ω 2 = v 2 − l where k , l ∈ Z How do we ''shift'' v 1 , v 2 by some integers k , l so that ω 1 ω 2 ∈ Z ?

Question

if       v          1        =            1      +              10        3            +                        10          2                3              3   and       v          2        =            7      +      4              10        3            +                        10          2                3              3  let       ω          1        =      v          1        −  k and let       ω          2        =      v          2        −  l where   k  ,  l  ∈      Z  How do we &#039;&#039;shift&#039;&#039;       v          1        ,      v          2       by some integers   k  ,  l so that       ω          1            ω          2        ∈      Z  ?

luthersavage6lm · Accepted Answer

w    1        w    2    =            (      57      −      k      −      7      l      )      +              10                              1            3                              (      11      −      k      −      l      )      +              10                              2            3                              (      18      −      k      −      4      l      )        3  Now if the above is an integer, clearly the irrational terms must become zero and other terms in the numerator must be multiple of   3.

pin1ta4r3k7b · Accepted Answer

Set   α  =            10      3      , so that       α    3    =  10. Multiplying out and simplifying,                              ω          1                          ω          2                            =                  1          9                (        1        +        α        +                  α          2                −        3        k        )        (        7        +        4        α        +                  α          2                −        3        l        )                                  =                  1          9                                      (                          (        57        −        21        k        −        3        l        +        9        k        l        )        +        (        21        −        12        k        −        3        l        )        α        +        (        12        −        3        k        −        3        l        )                  α          2                                      )                                   .            Now the numbers   1  ,  α  ,      α    2   are linearly independent over the rationals. So the only way this expression can be a rational number is if the coefficients of   α and       α    2   are zero. This gives you two equations which you can solve to find   k  ,  l. If   k  ,  l are not integers then your problem has no solution; but it turns out that   k  =  1,   l  =  3. If you now substitute back into the above, and if       ω    1        ω    2   is not an integer, then again your problem has no solution; but it turns out that       ω    1        ω    2    =  6.

if v_1=(1+root(3)(10)+root(3)(10^2))/3 and v_2=(7+4root(3)(10)+root(3)(10^2))/3 let w_1=v_1−k and let w_2=v_2−l where k,l in Z How do I ''shift'' v_1,v_2 by some integers k,l so that w_1w_2 in Z?

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