-0.381 966 011 250 105 097 474 346 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 250 105 097 474 346(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
-0.381 966 011 250 105 097 474 346(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:

|-0.381 966 011 250 105 097 474 346| = 0.381 966 011 250 105 097 474 346


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: 0.381 966 011 250 105 097 474 346.

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) 0.381 966 011 250 105 097 474 346 × 2 = 0 + 0.763 932 022 500 210 194 948 692;
  • 2) 0.763 932 022 500 210 194 948 692 × 2 = 1 + 0.527 864 045 000 420 389 897 384;
  • 3) 0.527 864 045 000 420 389 897 384 × 2 = 1 + 0.055 728 090 000 840 779 794 768;
  • 4) 0.055 728 090 000 840 779 794 768 × 2 = 0 + 0.111 456 180 001 681 559 589 536;
  • 5) 0.111 456 180 001 681 559 589 536 × 2 = 0 + 0.222 912 360 003 363 119 179 072;
  • 6) 0.222 912 360 003 363 119 179 072 × 2 = 0 + 0.445 824 720 006 726 238 358 144;
  • 7) 0.445 824 720 006 726 238 358 144 × 2 = 0 + 0.891 649 440 013 452 476 716 288;
  • 8) 0.891 649 440 013 452 476 716 288 × 2 = 1 + 0.783 298 880 026 904 953 432 576;
  • 9) 0.783 298 880 026 904 953 432 576 × 2 = 1 + 0.566 597 760 053 809 906 865 152;
  • 10) 0.566 597 760 053 809 906 865 152 × 2 = 1 + 0.133 195 520 107 619 813 730 304;
  • 11) 0.133 195 520 107 619 813 730 304 × 2 = 0 + 0.266 391 040 215 239 627 460 608;
  • 12) 0.266 391 040 215 239 627 460 608 × 2 = 0 + 0.532 782 080 430 479 254 921 216;
  • 13) 0.532 782 080 430 479 254 921 216 × 2 = 1 + 0.065 564 160 860 958 509 842 432;
  • 14) 0.065 564 160 860 958 509 842 432 × 2 = 0 + 0.131 128 321 721 917 019 684 864;
  • 15) 0.131 128 321 721 917 019 684 864 × 2 = 0 + 0.262 256 643 443 834 039 369 728;
  • 16) 0.262 256 643 443 834 039 369 728 × 2 = 0 + 0.524 513 286 887 668 078 739 456;
  • 17) 0.524 513 286 887 668 078 739 456 × 2 = 1 + 0.049 026 573 775 336 157 478 912;
  • 18) 0.049 026 573 775 336 157 478 912 × 2 = 0 + 0.098 053 147 550 672 314 957 824;
  • 19) 0.098 053 147 550 672 314 957 824 × 2 = 0 + 0.196 106 295 101 344 629 915 648;
  • 20) 0.196 106 295 101 344 629 915 648 × 2 = 0 + 0.392 212 590 202 689 259 831 296;
  • 21) 0.392 212 590 202 689 259 831 296 × 2 = 0 + 0.784 425 180 405 378 519 662 592;
  • 22) 0.784 425 180 405 378 519 662 592 × 2 = 1 + 0.568 850 360 810 757 039 325 184;
  • 23) 0.568 850 360 810 757 039 325 184 × 2 = 1 + 0.137 700 721 621 514 078 650 368;
  • 24) 0.137 700 721 621 514 078 650 368 × 2 = 0 + 0.275 401 443 243 028 157 300 736;
  • 25) 0.275 401 443 243 028 157 300 736 × 2 = 0 + 0.550 802 886 486 056 314 601 472;
  • 26) 0.550 802 886 486 056 314 601 472 × 2 = 1 + 0.101 605 772 972 112 629 202 944;
  • 27) 0.101 605 772 972 112 629 202 944 × 2 = 0 + 0.203 211 545 944 225 258 405 888;
  • 28) 0.203 211 545 944 225 258 405 888 × 2 = 0 + 0.406 423 091 888 450 516 811 776;
  • 29) 0.406 423 091 888 450 516 811 776 × 2 = 0 + 0.812 846 183 776 901 033 623 552;
  • 30) 0.812 846 183 776 901 033 623 552 × 2 = 1 + 0.625 692 367 553 802 067 247 104;
  • 31) 0.625 692 367 553 802 067 247 104 × 2 = 1 + 0.251 384 735 107 604 134 494 208;
  • 32) 0.251 384 735 107 604 134 494 208 × 2 = 0 + 0.502 769 470 215 208 268 988 416;
  • 33) 0.502 769 470 215 208 268 988 416 × 2 = 1 + 0.005 538 940 430 416 537 976 832;
  • 34) 0.005 538 940 430 416 537 976 832 × 2 = 0 + 0.011 077 880 860 833 075 953 664;
  • 35) 0.011 077 880 860 833 075 953 664 × 2 = 0 + 0.022 155 761 721 666 151 907 328;
  • 36) 0.022 155 761 721 666 151 907 328 × 2 = 0 + 0.044 311 523 443 332 303 814 656;
  • 37) 0.044 311 523 443 332 303 814 656 × 2 = 0 + 0.088 623 046 886 664 607 629 312;
  • 38) 0.088 623 046 886 664 607 629 312 × 2 = 0 + 0.177 246 093 773 329 215 258 624;
  • 39) 0.177 246 093 773 329 215 258 624 × 2 = 0 + 0.354 492 187 546 658 430 517 248;
  • 40) 0.354 492 187 546 658 430 517 248 × 2 = 0 + 0.708 984 375 093 316 861 034 496;
  • 41) 0.708 984 375 093 316 861 034 496 × 2 = 1 + 0.417 968 750 186 633 722 068 992;
  • 42) 0.417 968 750 186 633 722 068 992 × 2 = 0 + 0.835 937 500 373 267 444 137 984;
  • 43) 0.835 937 500 373 267 444 137 984 × 2 = 1 + 0.671 875 000 746 534 888 275 968;
  • 44) 0.671 875 000 746 534 888 275 968 × 2 = 1 + 0.343 750 001 493 069 776 551 936;
  • 45) 0.343 750 001 493 069 776 551 936 × 2 = 0 + 0.687 500 002 986 139 553 103 872;
  • 46) 0.687 500 002 986 139 553 103 872 × 2 = 1 + 0.375 000 005 972 279 106 207 744;
  • 47) 0.375 000 005 972 279 106 207 744 × 2 = 0 + 0.750 000 011 944 558 212 415 488;
  • 48) 0.750 000 011 944 558 212 415 488 × 2 = 1 + 0.500 000 023 889 116 424 830 976;
  • 49) 0.500 000 023 889 116 424 830 976 × 2 = 1 + 0.000 000 047 778 232 849 661 952;
  • 50) 0.000 000 047 778 232 849 661 952 × 2 = 0 + 0.000 000 095 556 465 699 323 904;
  • 51) 0.000 000 095 556 465 699 323 904 × 2 = 0 + 0.000 000 191 112 931 398 647 808;
  • 52) 0.000 000 191 112 931 398 647 808 × 2 = 0 + 0.000 000 382 225 862 797 295 616;
  • 53) 0.000 000 382 225 862 797 295 616 × 2 = 0 + 0.000 000 764 451 725 594 591 232;
  • 54) 0.000 000 764 451 725 594 591 232 × 2 = 0 + 0.000 001 528 903 451 189 182 464;

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:


0.381 966 011 250 105 097 474 346(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 1000 00(2)

6. Positive number before normalization:

0.381 966 011 250 105 097 474 346(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 1000 00(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:


0.381 966 011 250 105 097 474 346(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 1000 00(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 1000 00(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0110 0000(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 0110 0000


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 0110 0000 =

1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0110 0000


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 0110 0000


Decimal number -0.381 966 011 250 105 097 474 346 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 0110 0000

» Convert decimal -0.381 966 011 250 105 097 474 268 to 64 bit double precision IEEE 754 binary floating point representation standard

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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