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Q: Is the number 0.1111... Repeats forever there for its irrational true or false?

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There are an infinite number of values between 0.01 and 0.027 for example .011 .0111 .01111 .01111 .011111 etc. , .012, .0121, .0122, etc.

In any base other than 2 (ie other than binary): 10101010 + 01111 = 10102121 In base 2 (binary): 10101010 + 01111 = 10111001

10101010 + 01111 = 10111001 (170+15 = 185)

with hexidecimal you neednt use as many chartictors to represent a number. in binary 15 would be 01111 where as in hex it would be E resulting in much quicker coding times

with hexidecimal you neednt use as many chartictors to represent a number. in binary 15 would be 01111 where as in hex it would be E resulting in much quicker coding times

Multiply by 16. ANSWER: The answer is bcd 11 = 3 111 = 7 1111 i= F 01111 =E and so foirth

3 bit stuffing is required

10,101,010 + 01,111 = 10,102,121Although you didn't say so, we suspect that you may have meantthe two numbers in the question to be binary (base-2) numbers.If so, then| 0 | 0 | 0 | 0= (170)100 | | | | = (15)10Their sum is (185)10

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1 = 00001, 2 = 00010, 3 = 00011, 4 = 00100, 5 = 00101,6 = 00110, 7 = 00111, 8 = 01000, 9 = 01001, 10 = 01010, 11 = 01011, 12 = 01100, 13 = 01101, 14 = 01110, 15 = 01111, 16 = 10000, 17 = 10001, 18 = 10010, 19 = 10011, 20 = 10100.

0 = 00000 1 = 00001 2 = 00010 3 = 00011 4 = 00100 5 = 00101 6 = 00110 7 = 00111 8 = 01000 9 = 01001 10 = 01010 11 = 01011 12 = 01100 13 = 01101 14 = 01110 15 = 01111 16 = 10000 17 = 10001 18 = 10010 19 = 10011 20 = 10100

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