[MOVED] Working in different bases

mandm87

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Aug 30, 2007
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I am learning how to do math problems using numbers of bases other than 10. How do you solve this problem?
100 base two - 11 base two

I understand the basics but I am confused now that I have to borrow.
 
100<sub>2</sub> = (1 x 2<sup>2</sup>) + (0 x 2<sup>1</sup>) + (0 x 2<sup>0</sup>)

11<sub>2</sub> = (1 x 2<sup>1</sup>) + (1 x 2<sup>0</sup>)

100<sub>2</sub> - 11<sub>2</sub> = (1 x 2<sup>2</sup>) - [(1 x 2<sup>1</sup>) + (1 x 2<sup>0</sup>)] = 1<sub>10</sub> = 1<sub>2</sub>
 
Actually, when adding or subtracting in any base, you can follow the same mechanics as those used in base 10. In your problem 100 - 11, you just have to remember that 1-1 = 0 and 10 - 1 = 1.
.100
-011


You have to do some borrowing. You borrow 1 from the leftmost digit in the minuend leaving it a 0. This makes the middle number a 10. But we have to borrow 1 from it making the middle number 1 and the rightmost number 10.
Now reading from left to right in the minuend the first column is 0, the second column is 1 and the third column is 10.
Now, do the subtracting. Starting at the rightmost column, 10-1=1. The rightmost number in the remainder is 1.
In the middle column we have 1-1=0. The second digit from the right in the remainder is zero.
In the leftmost column we have 0-0=0. Our remainder is 001 or simply 1.
 
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