-0.381 966 011 250 105 097 474 261 126 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 250 105 097 474 261 126(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.381 966 011 250 105 097 474 261 126(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.381 966 011 250 105 097 474 261 126| = 0.381 966 011 250 105 097 474 261 126


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.381 966 011 250 105 097 474 261 126.

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) 0.381 966 011 250 105 097 474 261 126 × 2 = 0 + 0.763 932 022 500 210 194 948 522 252;
  • 2) 0.763 932 022 500 210 194 948 522 252 × 2 = 1 + 0.527 864 045 000 420 389 897 044 504;
  • 3) 0.527 864 045 000 420 389 897 044 504 × 2 = 1 + 0.055 728 090 000 840 779 794 089 008;
  • 4) 0.055 728 090 000 840 779 794 089 008 × 2 = 0 + 0.111 456 180 001 681 559 588 178 016;
  • 5) 0.111 456 180 001 681 559 588 178 016 × 2 = 0 + 0.222 912 360 003 363 119 176 356 032;
  • 6) 0.222 912 360 003 363 119 176 356 032 × 2 = 0 + 0.445 824 720 006 726 238 352 712 064;
  • 7) 0.445 824 720 006 726 238 352 712 064 × 2 = 0 + 0.891 649 440 013 452 476 705 424 128;
  • 8) 0.891 649 440 013 452 476 705 424 128 × 2 = 1 + 0.783 298 880 026 904 953 410 848 256;
  • 9) 0.783 298 880 026 904 953 410 848 256 × 2 = 1 + 0.566 597 760 053 809 906 821 696 512;
  • 10) 0.566 597 760 053 809 906 821 696 512 × 2 = 1 + 0.133 195 520 107 619 813 643 393 024;
  • 11) 0.133 195 520 107 619 813 643 393 024 × 2 = 0 + 0.266 391 040 215 239 627 286 786 048;
  • 12) 0.266 391 040 215 239 627 286 786 048 × 2 = 0 + 0.532 782 080 430 479 254 573 572 096;
  • 13) 0.532 782 080 430 479 254 573 572 096 × 2 = 1 + 0.065 564 160 860 958 509 147 144 192;
  • 14) 0.065 564 160 860 958 509 147 144 192 × 2 = 0 + 0.131 128 321 721 917 018 294 288 384;
  • 15) 0.131 128 321 721 917 018 294 288 384 × 2 = 0 + 0.262 256 643 443 834 036 588 576 768;
  • 16) 0.262 256 643 443 834 036 588 576 768 × 2 = 0 + 0.524 513 286 887 668 073 177 153 536;
  • 17) 0.524 513 286 887 668 073 177 153 536 × 2 = 1 + 0.049 026 573 775 336 146 354 307 072;
  • 18) 0.049 026 573 775 336 146 354 307 072 × 2 = 0 + 0.098 053 147 550 672 292 708 614 144;
  • 19) 0.098 053 147 550 672 292 708 614 144 × 2 = 0 + 0.196 106 295 101 344 585 417 228 288;
  • 20) 0.196 106 295 101 344 585 417 228 288 × 2 = 0 + 0.392 212 590 202 689 170 834 456 576;
  • 21) 0.392 212 590 202 689 170 834 456 576 × 2 = 0 + 0.784 425 180 405 378 341 668 913 152;
  • 22) 0.784 425 180 405 378 341 668 913 152 × 2 = 1 + 0.568 850 360 810 756 683 337 826 304;
  • 23) 0.568 850 360 810 756 683 337 826 304 × 2 = 1 + 0.137 700 721 621 513 366 675 652 608;
  • 24) 0.137 700 721 621 513 366 675 652 608 × 2 = 0 + 0.275 401 443 243 026 733 351 305 216;
  • 25) 0.275 401 443 243 026 733 351 305 216 × 2 = 0 + 0.550 802 886 486 053 466 702 610 432;
  • 26) 0.550 802 886 486 053 466 702 610 432 × 2 = 1 + 0.101 605 772 972 106 933 405 220 864;
  • 27) 0.101 605 772 972 106 933 405 220 864 × 2 = 0 + 0.203 211 545 944 213 866 810 441 728;
  • 28) 0.203 211 545 944 213 866 810 441 728 × 2 = 0 + 0.406 423 091 888 427 733 620 883 456;
  • 29) 0.406 423 091 888 427 733 620 883 456 × 2 = 0 + 0.812 846 183 776 855 467 241 766 912;
  • 30) 0.812 846 183 776 855 467 241 766 912 × 2 = 1 + 0.625 692 367 553 710 934 483 533 824;
  • 31) 0.625 692 367 553 710 934 483 533 824 × 2 = 1 + 0.251 384 735 107 421 868 967 067 648;
  • 32) 0.251 384 735 107 421 868 967 067 648 × 2 = 0 + 0.502 769 470 214 843 737 934 135 296;
  • 33) 0.502 769 470 214 843 737 934 135 296 × 2 = 1 + 0.005 538 940 429 687 475 868 270 592;
  • 34) 0.005 538 940 429 687 475 868 270 592 × 2 = 0 + 0.011 077 880 859 374 951 736 541 184;
  • 35) 0.011 077 880 859 374 951 736 541 184 × 2 = 0 + 0.022 155 761 718 749 903 473 082 368;
  • 36) 0.022 155 761 718 749 903 473 082 368 × 2 = 0 + 0.044 311 523 437 499 806 946 164 736;
  • 37) 0.044 311 523 437 499 806 946 164 736 × 2 = 0 + 0.088 623 046 874 999 613 892 329 472;
  • 38) 0.088 623 046 874 999 613 892 329 472 × 2 = 0 + 0.177 246 093 749 999 227 784 658 944;
  • 39) 0.177 246 093 749 999 227 784 658 944 × 2 = 0 + 0.354 492 187 499 998 455 569 317 888;
  • 40) 0.354 492 187 499 998 455 569 317 888 × 2 = 0 + 0.708 984 374 999 996 911 138 635 776;
  • 41) 0.708 984 374 999 996 911 138 635 776 × 2 = 1 + 0.417 968 749 999 993 822 277 271 552;
  • 42) 0.417 968 749 999 993 822 277 271 552 × 2 = 0 + 0.835 937 499 999 987 644 554 543 104;
  • 43) 0.835 937 499 999 987 644 554 543 104 × 2 = 1 + 0.671 874 999 999 975 289 109 086 208;
  • 44) 0.671 874 999 999 975 289 109 086 208 × 2 = 1 + 0.343 749 999 999 950 578 218 172 416;
  • 45) 0.343 749 999 999 950 578 218 172 416 × 2 = 0 + 0.687 499 999 999 901 156 436 344 832;
  • 46) 0.687 499 999 999 901 156 436 344 832 × 2 = 1 + 0.374 999 999 999 802 312 872 689 664;
  • 47) 0.374 999 999 999 802 312 872 689 664 × 2 = 0 + 0.749 999 999 999 604 625 745 379 328;
  • 48) 0.749 999 999 999 604 625 745 379 328 × 2 = 1 + 0.499 999 999 999 209 251 490 758 656;
  • 49) 0.499 999 999 999 209 251 490 758 656 × 2 = 0 + 0.999 999 999 998 418 502 981 517 312;
  • 50) 0.999 999 999 998 418 502 981 517 312 × 2 = 1 + 0.999 999 999 996 837 005 963 034 624;
  • 51) 0.999 999 999 996 837 005 963 034 624 × 2 = 1 + 0.999 999 999 993 674 011 926 069 248;
  • 52) 0.999 999 999 993 674 011 926 069 248 × 2 = 1 + 0.999 999 999 987 348 023 852 138 496;
  • 53) 0.999 999 999 987 348 023 852 138 496 × 2 = 1 + 0.999 999 999 974 696 047 704 276 992;
  • 54) 0.999 999 999 974 696 047 704 276 992 × 2 = 1 + 0.999 999 999 949 392 095 408 553 984;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


5. 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:


0.381 966 011 250 105 097 474 261 126(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0111 11(2)

6. Positive number before normalization:

0.381 966 011 250 105 097 474 261 126(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.381 966 011 250 105 097 474 261 126(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0111 11(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0111 11(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 1111(2) × 2-2


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): -2

Mantissa (not normalized):
1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-2 + 2(11-1) - 1 =

(-2 + 1 023)(10) =

1 021(10)


10. 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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1021(10) =

011 1111 1101(2)


12. 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. 1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 1111 =

1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 1111


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1101

Mantissa (52 bits) =
1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 1111


Decimal number -0.381 966 011 250 105 097 474 261 126 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 1111

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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