Convert the Number 0.000 030 227 005 481 719 3 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations
Number 0.000 030 227 005 481 719 3(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 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: 0.000 030 227 005 481 719 3.
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) 0.000 030 227 005 481 719 3 × 2 = 0 + 0.000 060 454 010 963 438 6;
- 2) 0.000 060 454 010 963 438 6 × 2 = 0 + 0.000 120 908 021 926 877 2;
- 3) 0.000 120 908 021 926 877 2 × 2 = 0 + 0.000 241 816 043 853 754 4;
- 4) 0.000 241 816 043 853 754 4 × 2 = 0 + 0.000 483 632 087 707 508 8;
- 5) 0.000 483 632 087 707 508 8 × 2 = 0 + 0.000 967 264 175 415 017 6;
- 6) 0.000 967 264 175 415 017 6 × 2 = 0 + 0.001 934 528 350 830 035 2;
- 7) 0.001 934 528 350 830 035 2 × 2 = 0 + 0.003 869 056 701 660 070 4;
- 8) 0.003 869 056 701 660 070 4 × 2 = 0 + 0.007 738 113 403 320 140 8;
- 9) 0.007 738 113 403 320 140 8 × 2 = 0 + 0.015 476 226 806 640 281 6;
- 10) 0.015 476 226 806 640 281 6 × 2 = 0 + 0.030 952 453 613 280 563 2;
- 11) 0.030 952 453 613 280 563 2 × 2 = 0 + 0.061 904 907 226 561 126 4;
- 12) 0.061 904 907 226 561 126 4 × 2 = 0 + 0.123 809 814 453 122 252 8;
- 13) 0.123 809 814 453 122 252 8 × 2 = 0 + 0.247 619 628 906 244 505 6;
- 14) 0.247 619 628 906 244 505 6 × 2 = 0 + 0.495 239 257 812 489 011 2;
- 15) 0.495 239 257 812 489 011 2 × 2 = 0 + 0.990 478 515 624 978 022 4;
- 16) 0.990 478 515 624 978 022 4 × 2 = 1 + 0.980 957 031 249 956 044 8;
- 17) 0.980 957 031 249 956 044 8 × 2 = 1 + 0.961 914 062 499 912 089 6;
- 18) 0.961 914 062 499 912 089 6 × 2 = 1 + 0.923 828 124 999 824 179 2;
- 19) 0.923 828 124 999 824 179 2 × 2 = 1 + 0.847 656 249 999 648 358 4;
- 20) 0.847 656 249 999 648 358 4 × 2 = 1 + 0.695 312 499 999 296 716 8;
- 21) 0.695 312 499 999 296 716 8 × 2 = 1 + 0.390 624 999 998 593 433 6;
- 22) 0.390 624 999 998 593 433 6 × 2 = 0 + 0.781 249 999 997 186 867 2;
- 23) 0.781 249 999 997 186 867 2 × 2 = 1 + 0.562 499 999 994 373 734 4;
- 24) 0.562 499 999 994 373 734 4 × 2 = 1 + 0.124 999 999 988 747 468 8;
- 25) 0.124 999 999 988 747 468 8 × 2 = 0 + 0.249 999 999 977 494 937 6;
- 26) 0.249 999 999 977 494 937 6 × 2 = 0 + 0.499 999 999 954 989 875 2;
- 27) 0.499 999 999 954 989 875 2 × 2 = 0 + 0.999 999 999 909 979 750 4;
- 28) 0.999 999 999 909 979 750 4 × 2 = 1 + 0.999 999 999 819 959 500 8;
- 29) 0.999 999 999 819 959 500 8 × 2 = 1 + 0.999 999 999 639 919 001 6;
- 30) 0.999 999 999 639 919 001 6 × 2 = 1 + 0.999 999 999 279 838 003 2;
- 31) 0.999 999 999 279 838 003 2 × 2 = 1 + 0.999 999 998 559 676 006 4;
- 32) 0.999 999 998 559 676 006 4 × 2 = 1 + 0.999 999 997 119 352 012 8;
- 33) 0.999 999 997 119 352 012 8 × 2 = 1 + 0.999 999 994 238 704 025 6;
- 34) 0.999 999 994 238 704 025 6 × 2 = 1 + 0.999 999 988 477 408 051 2;
- 35) 0.999 999 988 477 408 051 2 × 2 = 1 + 0.999 999 976 954 816 102 4;
- 36) 0.999 999 976 954 816 102 4 × 2 = 1 + 0.999 999 953 909 632 204 8;
- 37) 0.999 999 953 909 632 204 8 × 2 = 1 + 0.999 999 907 819 264 409 6;
- 38) 0.999 999 907 819 264 409 6 × 2 = 1 + 0.999 999 815 638 528 819 2;
- 39) 0.999 999 815 638 528 819 2 × 2 = 1 + 0.999 999 631 277 057 638 4;
- 40) 0.999 999 631 277 057 638 4 × 2 = 1 + 0.999 999 262 554 115 276 8;
- 41) 0.999 999 262 554 115 276 8 × 2 = 1 + 0.999 998 525 108 230 553 6;
- 42) 0.999 998 525 108 230 553 6 × 2 = 1 + 0.999 997 050 216 461 107 2;
- 43) 0.999 997 050 216 461 107 2 × 2 = 1 + 0.999 994 100 432 922 214 4;
- 44) 0.999 994 100 432 922 214 4 × 2 = 1 + 0.999 988 200 865 844 428 8;
- 45) 0.999 988 200 865 844 428 8 × 2 = 1 + 0.999 976 401 731 688 857 6;
- 46) 0.999 976 401 731 688 857 6 × 2 = 1 + 0.999 952 803 463 377 715 2;
- 47) 0.999 952 803 463 377 715 2 × 2 = 1 + 0.999 905 606 926 755 430 4;
- 48) 0.999 905 606 926 755 430 4 × 2 = 1 + 0.999 811 213 853 510 860 8;
- 49) 0.999 811 213 853 510 860 8 × 2 = 1 + 0.999 622 427 707 021 721 6;
- 50) 0.999 622 427 707 021 721 6 × 2 = 1 + 0.999 244 855 414 043 443 2;
- 51) 0.999 244 855 414 043 443 2 × 2 = 1 + 0.998 489 710 828 086 886 4;
- 52) 0.998 489 710 828 086 886 4 × 2 = 1 + 0.996 979 421 656 173 772 8;
- 53) 0.996 979 421 656 173 772 8 × 2 = 1 + 0.993 958 843 312 347 545 6;
- 54) 0.993 958 843 312 347 545 6 × 2 = 1 + 0.987 917 686 624 695 091 2;
- 55) 0.987 917 686 624 695 091 2 × 2 = 1 + 0.975 835 373 249 390 182 4;
- 56) 0.975 835 373 249 390 182 4 × 2 = 1 + 0.951 670 746 498 780 364 8;
- 57) 0.951 670 746 498 780 364 8 × 2 = 1 + 0.903 341 492 997 560 729 6;
- 58) 0.903 341 492 997 560 729 6 × 2 = 1 + 0.806 682 985 995 121 459 2;
- 59) 0.806 682 985 995 121 459 2 × 2 = 1 + 0.613 365 971 990 242 918 4;
- 60) 0.613 365 971 990 242 918 4 × 2 = 1 + 0.226 731 943 980 485 836 8;
- 61) 0.226 731 943 980 485 836 8 × 2 = 0 + 0.453 463 887 960 971 673 6;
- 62) 0.453 463 887 960 971 673 6 × 2 = 0 + 0.906 927 775 921 943 347 2;
- 63) 0.906 927 775 921 943 347 2 × 2 = 1 + 0.813 855 551 843 886 694 4;
- 64) 0.813 855 551 843 886 694 4 × 2 = 1 + 0.627 711 103 687 773 388 8;
- 65) 0.627 711 103 687 773 388 8 × 2 = 1 + 0.255 422 207 375 546 777 6;
- 66) 0.255 422 207 375 546 777 6 × 2 = 0 + 0.510 844 414 751 093 555 2;
- 67) 0.510 844 414 751 093 555 2 × 2 = 1 + 0.021 688 829 502 187 110 4;
- 68) 0.021 688 829 502 187 110 4 × 2 = 0 + 0.043 377 659 004 374 220 8;
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:
0.000 030 227 005 481 719 3(10) =
0.0000 0000 0000 0001 1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010(2)
5. Positive number before normalization:
0.000 030 227 005 481 719 3(10) =
0.0000 0000 0000 0001 1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010(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 16 positions to the right, so that only one non zero digit remains to the left of it:
0.000 030 227 005 481 719 3(10) =
0.0000 0000 0000 0001 1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010(2) =
0.0000 0000 0000 0001 1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010(2) × 20 =
1.1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010(2) × 2-16
7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:
Sign 0 (a positive number)
Exponent (unadjusted): -16
Mantissa (not normalized):
1.1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010
8. Adjust the exponent.
Use the 11 bit excess/bias notation:
Exponent (adjusted) =
Exponent (unadjusted) + 2(11-1) - 1 =
-16 + 2(11-1) - 1 =
(-16 + 1 023)(10) =
1 007(10)
9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.
Use the same technique of repeatedly dividing by 2:
- division = quotient + remainder;
- 1 007 ÷ 2 = 503 + 1;
- 503 ÷ 2 = 251 + 1;
- 251 ÷ 2 = 125 + 1;
- 125 ÷ 2 = 62 + 1;
- 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) =
1007(10) =
011 1110 1111(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 52 bits, only if necessary (not the case here).
Mantissa (normalized) =
1. 1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010 =
1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010
12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:
Sign (1 bit) =
0 (a positive number)
Exponent (11 bits) =
011 1110 1111
Mantissa (52 bits) =
1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010
The base ten decimal number 0.000 030 227 005 481 719 3 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1110 1111 - 1111 1011 0001 1111 1111 1111 1111 1111 1111 1111 1111 0011 1010
Sign (1 bit):
Exponent (11 bits):
0
621
611
601
591
581
570
561
551
541
531
52
Mantissa (52 bits):
1
511
501
491
481
470
461
451
440
430
420
411
401
391
381
371
361
351
341
331
321
311
301
291
281
271
261
251
241
231
221
211
201
191
181
171
161
151
141
131
121
111
101
91
80
70
61
51
41
30
21
10
0
Convert to 64 bit double 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 64 bit double precision IEEE 754 floating point binary standard representation
How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard
Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
- 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 - 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:
- 1. Start with the positive version of the number:
|-31.640 215| = 31.640 215
- 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
- division = quotient + remainder;
- 31 ÷ 2 = 15 + 1;
- 15 ÷ 2 = 7 + 1;
- 7 ÷ 2 = 3 + 1;
- 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:
31(10) = 1 1111(2)
- 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
- 2) 0.280 43 × 2 = 0 + 0.560 86;
- 3) 0.560 86 × 2 = 1 + 0.121 72;
- 4) 0.121 72 × 2 = 0 + 0.243 44;
- 5) 0.243 44 × 2 = 0 + 0.486 88;
- 6) 0.486 88 × 2 = 0 + 0.973 76;
- 7) 0.973 76 × 2 = 1 + 0.947 52;
- 8) 0.947 52 × 2 = 1 + 0.895 04;
- 9) 0.895 04 × 2 = 1 + 0.790 08;
- 10) 0.790 08 × 2 = 1 + 0.580 16;
- 11) 0.580 16 × 2 = 1 + 0.160 32;
- 12) 0.160 32 × 2 = 0 + 0.320 64;
- 13) 0.320 64 × 2 = 0 + 0.641 28;
- 14) 0.641 28 × 2 = 1 + 0.282 56;
- 15) 0.282 56 × 2 = 0 + 0.565 12;
- 16) 0.565 12 × 2 = 1 + 0.130 24;
- 17) 0.130 24 × 2 = 0 + 0.260 48;
- 18) 0.260 48 × 2 = 0 + 0.520 96;
- 19) 0.520 96 × 2 = 1 + 0.041 92;
- 20) 0.041 92 × 2 = 0 + 0.083 84;
- 21) 0.083 84 × 2 = 0 + 0.167 68;
- 22) 0.167 68 × 2 = 0 + 0.335 36;
- 23) 0.335 36 × 2 = 0 + 0.670 72;
- 24) 0.670 72 × 2 = 1 + 0.341 44;
- 25) 0.341 44 × 2 = 0 + 0.682 88;
- 26) 0.682 88 × 2 = 1 + 0.365 76;
- 27) 0.365 76 × 2 = 0 + 0.731 52;
- 28) 0.731 52 × 2 = 1 + 0.463 04;
- 29) 0.463 04 × 2 = 0 + 0.926 08;
- 30) 0.926 08 × 2 = 1 + 0.852 16;
- 31) 0.852 16 × 2 = 1 + 0.704 32;
- 32) 0.704 32 × 2 = 1 + 0.408 64;
- 33) 0.408 64 × 2 = 0 + 0.817 28;
- 34) 0.817 28 × 2 = 1 + 0.634 56;
- 35) 0.634 56 × 2 = 1 + 0.269 12;
- 36) 0.269 12 × 2 = 0 + 0.538 24;
- 37) 0.538 24 × 2 = 1 + 0.076 48;
- 38) 0.076 48 × 2 = 0 + 0.152 96;
- 39) 0.152 96 × 2 = 0 + 0.305 92;
- 40) 0.305 92 × 2 = 0 + 0.611 84;
- 41) 0.611 84 × 2 = 1 + 0.223 68;
- 42) 0.223 68 × 2 = 0 + 0.447 36;
- 43) 0.447 36 × 2 = 0 + 0.894 72;
- 44) 0.894 72 × 2 = 1 + 0.789 44;
- 45) 0.789 44 × 2 = 1 + 0.578 88;
- 46) 0.578 88 × 2 = 1 + 0.157 76;
- 47) 0.157 76 × 2 = 0 + 0.315 52;
- 48) 0.315 52 × 2 = 0 + 0.631 04;
- 49) 0.631 04 × 2 = 1 + 0.262 08;
- 50) 0.262 08 × 2 = 0 + 0.524 16;
- 51) 0.524 16 × 2 = 1 + 0.048 32;
- 52) 0.048 32 × 2 = 0 + 0.096 64;
- 53) 0.096 64 × 2 = 0 + 0.193 28;
- We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:
0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)
- 6. Summarizing - the positive number before normalization:
31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)
- 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:
31.640 215(10) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24
- 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:
Sign: 1 (a negative number)
Exponent (unadjusted): 4
Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0
- 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:
Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
100 0000 0011(2)
- 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):
Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0
Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
- Conclusion:
Sign (1 bit) = 1 (a negative number)
Exponent (8 bits) = 100 0000 0011
Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100
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