Doing binary arithmetic in Windows Calculator and

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The discussion centers on binary arithmetic using Windows Calculator, specifically the multiplication of B11111111 by B11111111, which results in B1111111000000001. This outcome is explained as correct, with B11111111 equating to 255, making the multiplication yield 62025 in decimal. The detailed breakdown shows how the binary result reflects the mathematical principles of binary multiplication, including the role of placeholder zeros. A user also shares a related observation about binary representations of integers composed of nines, noting a pattern of trailing ones. The conversation highlights common misconceptions in binary multiplication and the importance of understanding binary arithmetic fundamentals.
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I was doing some conversions from binary to decimal and vice versa today using Windows Calculator and I noticed the following, if I multiply B11111111 and B11111111 I get the following: B1111111000000001. Uhhh...great! What's going on here? It looks like it's rolling over the 8 LSB when it shouldn't, or something...:confused:
 
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No, it's just giving you the right answer, as remarkable as that might seem!

B11111111 is just 28-1= 255. B11111111 times B11111111 is the same as 255(255)= 62025 which, in binary, is B1111111000000001.

In more detail: B11111111 is, as I said, 28-1. B11111111 times B11111111 is (28- 1)(28- 1)= (28)2- 2(28)+ 1= 216- 29+ 1. 216- 29= 29(216- 9-1)= 29(27-1). That "27-1" is the reason for the 7 "1"s leading the number. That 29 is the reason it is followed by 9 "0"s- or would be if it were not for that final "+ 1".
 
HallsofIvy said:
In more detail: B11111111 is, as I said, 28-1. B11111111 times B11111111 is (28- 1)(28- 1)= (28)2- 2(28)+ 1= 216- 29+ 1. 216- 29= 29(216- 9-1)= 29(27-1). That "27-1" is the reason for the 7 "1"s leading the number. That 29 is the reason it is followed by 9 "0"s- or would be if it were not for that final "+ 1".

Nice explanation, HallsofIvy. You may be interested in another binary pattern I discovered recently: integers made up of n digits of 9s have binary representations that have exactly n digits of trailing 1s (e.g., 9,999,999 is 100110001001011001111111 in binary). I proved it here: http://www.exploringbinary.com/nines-in-binary/ .
 
Thanks for the explanation! After thinking about it, I see the flaw in my reasoning. I was expecting B11111111 * B11111111 to become B1111111111111111, remembering that when doing it out longhand the partial product of the multiplicand is always 1 if it and the multiplier are 1. Since there are no zeros in either of them, I expected the result to have no zeros also. However, I was forgetting about the "placeholder" zeros that accumulate as the partial products are shifted left.
 
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