Binary numbers (fundamental) question

  • #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
 

Answers and Replies

  • #2
84
0
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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