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Binary numbers (fundamental) question

  1. Sep 11, 2007 #1
    I have 2 questions:

    FIRST
    I read from John H. Mathews’ Numerical Methods using MATLAB, 3E that the number
    1563 can be expanded in form of 10-base expansion as:

    (1 × 10[tex]^{3}[/tex]) − (5 × 10[tex]^{2}[/tex]) + (6 × 10[tex]^{1}[/tex]) + (3 × 10[tex]^{0}[/tex]) …............................................... (1)

    Which results 1000 − 500 + 60 + 3 equal to 563.

    But the number 1563 should be expanded in form of 10-base expansion as:

    1563 = 1000 + 500 + 60 + 3

    Which means:

    (1 × 10[tex]^{3}[/tex]) + (5 × 10[tex]^{2}[/tex]) + (6 × 10[tex]^{1}[/tex]) + (3 × 10[tex]^{0}[/tex]) …............................................... (2)

    Hence, (1) ≠ (2) or 1563 ≠ 563.

    So, can anyone explain me about these?

    SECOND
    If the number 1563 is expressible in expanded form of 10-base expansion as:

    1563 = 1000 + 500 + 60 + 3
    =(1 × 10[tex]^{3}[/tex]) − (5 × 10[tex]^{2}[/tex]) + (6 × 10[tex]^{1}[/tex]) + (3 × 10[tex]^{0}[/tex])

    how can I express the number 1563 in expanded form of 2-base expansion?

    Please advance
     
  2. jcsd
  3. Sep 11, 2007 #2
    I would have assumed the "-(5x10^2)" to be a typo and your (2) equation to be correct.

    If you understand why the base 10 expansion is correct, then you should be able to expand it into base 2 if I simply gave you the following list:

    2^0 = 1
    2^1 = 2
    2^2 = 4
    2^3 = 8
    2^4 = 16
    2^5 = 32
    2^6 = 64
    2^7 = 128
    2^8 = 256
    2^9 = 512
    2^10 = 1024
     
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