Convert the Number 0.199 099 987 745 285 034 14 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number. Detailed Explanations
Number 0.199 099 987 745 285 034 14(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)
The first steps we'll go through to make the conversion:
Convert to binary (to base 2) the integer part of the number.
Convert to binary the fractional part of the number.
1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.
Keep track of each remainder.
We stop when we get a quotient that is equal to zero.
- division = quotient + remainder;
- 0 ÷ 2 = 0 + 0;
2. Construct the base 2 representation of the integer part of the number.
Take all the remainders starting from the bottom of the list constructed above.
0(10) =
0(2)
3. Convert to binary (base 2) the fractional part: 0.199 099 987 745 285 034 14.
Multiply it repeatedly by 2.
Keep track of each integer part of the results.
Stop when we get a fractional part that is equal to zero.
- #) multiplying = integer + fractional part;
- 1) 0.199 099 987 745 285 034 14 × 2 = 0 + 0.398 199 975 490 570 068 28;
- 2) 0.398 199 975 490 570 068 28 × 2 = 0 + 0.796 399 950 981 140 136 56;
- 3) 0.796 399 950 981 140 136 56 × 2 = 1 + 0.592 799 901 962 280 273 12;
- 4) 0.592 799 901 962 280 273 12 × 2 = 1 + 0.185 599 803 924 560 546 24;
- 5) 0.185 599 803 924 560 546 24 × 2 = 0 + 0.371 199 607 849 121 092 48;
- 6) 0.371 199 607 849 121 092 48 × 2 = 0 + 0.742 399 215 698 242 184 96;
- 7) 0.742 399 215 698 242 184 96 × 2 = 1 + 0.484 798 431 396 484 369 92;
- 8) 0.484 798 431 396 484 369 92 × 2 = 0 + 0.969 596 862 792 968 739 84;
- 9) 0.969 596 862 792 968 739 84 × 2 = 1 + 0.939 193 725 585 937 479 68;
- 10) 0.939 193 725 585 937 479 68 × 2 = 1 + 0.878 387 451 171 874 959 36;
- 11) 0.878 387 451 171 874 959 36 × 2 = 1 + 0.756 774 902 343 749 918 72;
- 12) 0.756 774 902 343 749 918 72 × 2 = 1 + 0.513 549 804 687 499 837 44;
- 13) 0.513 549 804 687 499 837 44 × 2 = 1 + 0.027 099 609 374 999 674 88;
- 14) 0.027 099 609 374 999 674 88 × 2 = 0 + 0.054 199 218 749 999 349 76;
- 15) 0.054 199 218 749 999 349 76 × 2 = 0 + 0.108 398 437 499 998 699 52;
- 16) 0.108 398 437 499 998 699 52 × 2 = 0 + 0.216 796 874 999 997 399 04;
- 17) 0.216 796 874 999 997 399 04 × 2 = 0 + 0.433 593 749 999 994 798 08;
- 18) 0.433 593 749 999 994 798 08 × 2 = 0 + 0.867 187 499 999 989 596 16;
- 19) 0.867 187 499 999 989 596 16 × 2 = 1 + 0.734 374 999 999 979 192 32;
- 20) 0.734 374 999 999 979 192 32 × 2 = 1 + 0.468 749 999 999 958 384 64;
- 21) 0.468 749 999 999 958 384 64 × 2 = 0 + 0.937 499 999 999 916 769 28;
- 22) 0.937 499 999 999 916 769 28 × 2 = 1 + 0.874 999 999 999 833 538 56;
- 23) 0.874 999 999 999 833 538 56 × 2 = 1 + 0.749 999 999 999 667 077 12;
- 24) 0.749 999 999 999 667 077 12 × 2 = 1 + 0.499 999 999 999 334 154 24;
- 25) 0.499 999 999 999 334 154 24 × 2 = 0 + 0.999 999 999 998 668 308 48;
- 26) 0.999 999 999 998 668 308 48 × 2 = 1 + 0.999 999 999 997 336 616 96;
We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)
4. Construct the base 2 representation of the fractional part of the number.
Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:
0.199 099 987 745 285 034 14(10) =
0.0011 0010 1111 1000 0011 0111 01(2)
5. Positive number before normalization:
0.199 099 987 745 285 034 14(10) =
0.0011 0010 1111 1000 0011 0111 01(2)
The last steps we'll go through to make the conversion:
Normalize the binary representation of the number.
Adjust the exponent.
Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.
Normalize the mantissa.
6. Normalize the binary representation of the number.
Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:
0.199 099 987 745 285 034 14(10) =
0.0011 0010 1111 1000 0011 0111 01(2) =
0.0011 0010 1111 1000 0011 0111 01(2) × 20 =
1.1001 0111 1100 0001 1011 101(2) × 2-3
7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:
Sign 0 (a positive number)
Exponent (unadjusted): -3
Mantissa (not normalized):
1.1001 0111 1100 0001 1011 101
8. Adjust the exponent.
Use the 8 bit excess/bias notation:
Exponent (adjusted) =
Exponent (unadjusted) + 2(8-1) - 1 =
-3 + 2(8-1) - 1 =
(-3 + 127)(10) =
124(10)
9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.
Use the same technique of repeatedly dividing by 2:
- division = quotient + remainder;
- 124 ÷ 2 = 62 + 0;
- 62 ÷ 2 = 31 + 0;
- 31 ÷ 2 = 15 + 1;
- 15 ÷ 2 = 7 + 1;
- 7 ÷ 2 = 3 + 1;
- 3 ÷ 2 = 1 + 1;
- 1 ÷ 2 = 0 + 1;
10. Construct the base 2 representation of the adjusted exponent.
Take all the remainders starting from the bottom of the list constructed above.
Exponent (adjusted) =
124(10) =
0111 1100(2)
11. Normalize the mantissa.
a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.
b) Adjust its length to 23 bits, only if necessary (not the case here).
Mantissa (normalized) =
1. 100 1011 1110 0000 1101 1101 =
100 1011 1110 0000 1101 1101
12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:
Sign (1 bit) =
0 (a positive number)
Exponent (8 bits) =
0111 1100
Mantissa (23 bits) =
100 1011 1110 0000 1101 1101
The base ten decimal number 0.199 099 987 745 285 034 14 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 1100 - 100 1011 1110 0000 1101 1101
Sign (1 bit):
Exponent (8 bits):
0
301
291
281
271
261
250
240
23
Mantissa (23 bits):
1
220
210
201
190
181
171
161
151
141
130
120
110
100
90
81
71
60
51
41
31
20
11
0
Convert to 32 bit single precision IEEE 754 binary floating point representation standard
A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)
The latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation
How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard
Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:
- 1. If the number to be converted is negative, start with its the positive version.
- 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
- 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
- 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
- 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
- 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
- 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 - 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
- 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.
Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:
- 1. Start with the positive version of the number:
|-25.347| = 25.347
- 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
- division = quotient + remainder;
- 25 ÷ 2 = 12 + 1;
- 12 ÷ 2 = 6 + 0;
- 6 ÷ 2 = 3 + 0;
- 3 ÷ 2 = 1 + 1;
- 1 ÷ 2 = 0 + 1;
- We have encountered a quotient that is ZERO => FULL STOP
- 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:
25(10) = 1 1001(2)
- 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
- #) multiplying = integer + fractional part;
- 1) 0.347 × 2 = 0 + 0.694;
- 2) 0.694 × 2 = 1 + 0.388;
- 3) 0.388 × 2 = 0 + 0.776;
- 4) 0.776 × 2 = 1 + 0.552;
- 5) 0.552 × 2 = 1 + 0.104;
- 6) 0.104 × 2 = 0 + 0.208;
- 7) 0.208 × 2 = 0 + 0.416;
- 8) 0.416 × 2 = 0 + 0.832;
- 9) 0.832 × 2 = 1 + 0.664;
- 10) 0.664 × 2 = 1 + 0.328;
- 11) 0.328 × 2 = 0 + 0.656;
- 12) 0.656 × 2 = 1 + 0.312;
- 13) 0.312 × 2 = 0 + 0.624;
- 14) 0.624 × 2 = 1 + 0.248;
- 15) 0.248 × 2 = 0 + 0.496;
- 16) 0.496 × 2 = 0 + 0.992;
- 17) 0.992 × 2 = 1 + 0.984;
- 18) 0.984 × 2 = 1 + 0.968;
- 19) 0.968 × 2 = 1 + 0.936;
- 20) 0.936 × 2 = 1 + 0.872;
- 21) 0.872 × 2 = 1 + 0.744;
- 22) 0.744 × 2 = 1 + 0.488;
- 23) 0.488 × 2 = 0 + 0.976;
- 24) 0.976 × 2 = 1 + 0.952;
- We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
- 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:
0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)
- 6. Summarizing - the positive number before normalization:
25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)
- 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:
25.347(10) =
1 1001.0101 1000 1101 0100 1111 1101(2) =
1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
1.1001 0101 1000 1101 0100 1111 1101(2) × 24
- 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:
Sign: 1 (a negative number)
Exponent (unadjusted): 4
Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101
- 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:
Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
1000 0011(2)
- 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):
Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101
Mantissa (normalized): 100 1010 1100 0110 1010 0111
- Conclusion:
Sign (1 bit) = 1 (a negative number)
Exponent (8 bits) = 1000 0011
Mantissa (23 bits) = 100 1010 1100 0110 1010 0111
Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
1 - 1000 0011 - 100 1010 1100 0110 1010 0111
Available Base Conversions Between Decimal and Binary Systems
Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):
1. Integer -> Binary
2. Decimal -> Binary
3. Binary -> Integer
4. Binary -> Decimal