-29.931 663 088 704 072 204 551 133 979 64 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -29.931 663 088 704 072 204 551 133 979 64(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
-29.931 663 088 704 072 204 551 133 979 64(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:

|-29.931 663 088 704 072 204 551 133 979 64| = 29.931 663 088 704 072 204 551 133 979 64


2. First, convert to binary (in base 2) the integer part: 29.
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;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

29(10) =

1 1101(2)


4. Convert to binary (base 2) the fractional part: 0.931 663 088 704 072 204 551 133 979 64.

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) 0.931 663 088 704 072 204 551 133 979 64 × 2 = 1 + 0.863 326 177 408 144 409 102 267 959 28;
  • 2) 0.863 326 177 408 144 409 102 267 959 28 × 2 = 1 + 0.726 652 354 816 288 818 204 535 918 56;
  • 3) 0.726 652 354 816 288 818 204 535 918 56 × 2 = 1 + 0.453 304 709 632 577 636 409 071 837 12;
  • 4) 0.453 304 709 632 577 636 409 071 837 12 × 2 = 0 + 0.906 609 419 265 155 272 818 143 674 24;
  • 5) 0.906 609 419 265 155 272 818 143 674 24 × 2 = 1 + 0.813 218 838 530 310 545 636 287 348 48;
  • 6) 0.813 218 838 530 310 545 636 287 348 48 × 2 = 1 + 0.626 437 677 060 621 091 272 574 696 96;
  • 7) 0.626 437 677 060 621 091 272 574 696 96 × 2 = 1 + 0.252 875 354 121 242 182 545 149 393 92;
  • 8) 0.252 875 354 121 242 182 545 149 393 92 × 2 = 0 + 0.505 750 708 242 484 365 090 298 787 84;
  • 9) 0.505 750 708 242 484 365 090 298 787 84 × 2 = 1 + 0.011 501 416 484 968 730 180 597 575 68;
  • 10) 0.011 501 416 484 968 730 180 597 575 68 × 2 = 0 + 0.023 002 832 969 937 460 361 195 151 36;
  • 11) 0.023 002 832 969 937 460 361 195 151 36 × 2 = 0 + 0.046 005 665 939 874 920 722 390 302 72;
  • 12) 0.046 005 665 939 874 920 722 390 302 72 × 2 = 0 + 0.092 011 331 879 749 841 444 780 605 44;
  • 13) 0.092 011 331 879 749 841 444 780 605 44 × 2 = 0 + 0.184 022 663 759 499 682 889 561 210 88;
  • 14) 0.184 022 663 759 499 682 889 561 210 88 × 2 = 0 + 0.368 045 327 518 999 365 779 122 421 76;
  • 15) 0.368 045 327 518 999 365 779 122 421 76 × 2 = 0 + 0.736 090 655 037 998 731 558 244 843 52;
  • 16) 0.736 090 655 037 998 731 558 244 843 52 × 2 = 1 + 0.472 181 310 075 997 463 116 489 687 04;
  • 17) 0.472 181 310 075 997 463 116 489 687 04 × 2 = 0 + 0.944 362 620 151 994 926 232 979 374 08;
  • 18) 0.944 362 620 151 994 926 232 979 374 08 × 2 = 1 + 0.888 725 240 303 989 852 465 958 748 16;
  • 19) 0.888 725 240 303 989 852 465 958 748 16 × 2 = 1 + 0.777 450 480 607 979 704 931 917 496 32;
  • 20) 0.777 450 480 607 979 704 931 917 496 32 × 2 = 1 + 0.554 900 961 215 959 409 863 834 992 64;
  • 21) 0.554 900 961 215 959 409 863 834 992 64 × 2 = 1 + 0.109 801 922 431 918 819 727 669 985 28;
  • 22) 0.109 801 922 431 918 819 727 669 985 28 × 2 = 0 + 0.219 603 844 863 837 639 455 339 970 56;
  • 23) 0.219 603 844 863 837 639 455 339 970 56 × 2 = 0 + 0.439 207 689 727 675 278 910 679 941 12;
  • 24) 0.439 207 689 727 675 278 910 679 941 12 × 2 = 0 + 0.878 415 379 455 350 557 821 359 882 24;
  • 25) 0.878 415 379 455 350 557 821 359 882 24 × 2 = 1 + 0.756 830 758 910 701 115 642 719 764 48;
  • 26) 0.756 830 758 910 701 115 642 719 764 48 × 2 = 1 + 0.513 661 517 821 402 231 285 439 528 96;
  • 27) 0.513 661 517 821 402 231 285 439 528 96 × 2 = 1 + 0.027 323 035 642 804 462 570 879 057 92;
  • 28) 0.027 323 035 642 804 462 570 879 057 92 × 2 = 0 + 0.054 646 071 285 608 925 141 758 115 84;
  • 29) 0.054 646 071 285 608 925 141 758 115 84 × 2 = 0 + 0.109 292 142 571 217 850 283 516 231 68;
  • 30) 0.109 292 142 571 217 850 283 516 231 68 × 2 = 0 + 0.218 584 285 142 435 700 567 032 463 36;
  • 31) 0.218 584 285 142 435 700 567 032 463 36 × 2 = 0 + 0.437 168 570 284 871 401 134 064 926 72;
  • 32) 0.437 168 570 284 871 401 134 064 926 72 × 2 = 0 + 0.874 337 140 569 742 802 268 129 853 44;
  • 33) 0.874 337 140 569 742 802 268 129 853 44 × 2 = 1 + 0.748 674 281 139 485 604 536 259 706 88;
  • 34) 0.748 674 281 139 485 604 536 259 706 88 × 2 = 1 + 0.497 348 562 278 971 209 072 519 413 76;
  • 35) 0.497 348 562 278 971 209 072 519 413 76 × 2 = 0 + 0.994 697 124 557 942 418 145 038 827 52;
  • 36) 0.994 697 124 557 942 418 145 038 827 52 × 2 = 1 + 0.989 394 249 115 884 836 290 077 655 04;
  • 37) 0.989 394 249 115 884 836 290 077 655 04 × 2 = 1 + 0.978 788 498 231 769 672 580 155 310 08;
  • 38) 0.978 788 498 231 769 672 580 155 310 08 × 2 = 1 + 0.957 576 996 463 539 345 160 310 620 16;
  • 39) 0.957 576 996 463 539 345 160 310 620 16 × 2 = 1 + 0.915 153 992 927 078 690 320 621 240 32;
  • 40) 0.915 153 992 927 078 690 320 621 240 32 × 2 = 1 + 0.830 307 985 854 157 380 641 242 480 64;
  • 41) 0.830 307 985 854 157 380 641 242 480 64 × 2 = 1 + 0.660 615 971 708 314 761 282 484 961 28;
  • 42) 0.660 615 971 708 314 761 282 484 961 28 × 2 = 1 + 0.321 231 943 416 629 522 564 969 922 56;
  • 43) 0.321 231 943 416 629 522 564 969 922 56 × 2 = 0 + 0.642 463 886 833 259 045 129 939 845 12;
  • 44) 0.642 463 886 833 259 045 129 939 845 12 × 2 = 1 + 0.284 927 773 666 518 090 259 879 690 24;
  • 45) 0.284 927 773 666 518 090 259 879 690 24 × 2 = 0 + 0.569 855 547 333 036 180 519 759 380 48;
  • 46) 0.569 855 547 333 036 180 519 759 380 48 × 2 = 1 + 0.139 711 094 666 072 361 039 518 760 96;
  • 47) 0.139 711 094 666 072 361 039 518 760 96 × 2 = 0 + 0.279 422 189 332 144 722 079 037 521 92;
  • 48) 0.279 422 189 332 144 722 079 037 521 92 × 2 = 0 + 0.558 844 378 664 289 444 158 075 043 84;
  • 49) 0.558 844 378 664 289 444 158 075 043 84 × 2 = 1 + 0.117 688 757 328 578 888 316 150 087 68;
  • 50) 0.117 688 757 328 578 888 316 150 087 68 × 2 = 0 + 0.235 377 514 657 157 776 632 300 175 36;
  • 51) 0.235 377 514 657 157 776 632 300 175 36 × 2 = 0 + 0.470 755 029 314 315 553 264 600 350 72;
  • 52) 0.470 755 029 314 315 553 264 600 350 72 × 2 = 0 + 0.941 510 058 628 631 106 529 200 701 44;
  • 53) 0.941 510 058 628 631 106 529 200 701 44 × 2 = 1 + 0.883 020 117 257 262 213 058 401 402 88;

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:


0.931 663 088 704 072 204 551 133 979 64(10) =

0.1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100 1000 1(2)

6. Positive number before normalization:

29.931 663 088 704 072 204 551 133 979 64(10) =

1 1101.1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100 1000 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


29.931 663 088 704 072 204 551 133 979 64(10) =

1 1101.1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100 1000 1(2) =

1 1101.1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100 1000 1(2) × 20 =

1.1101 1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100 1000 1(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):
1.1101 1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100 1000 1


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

4 + 2(11-1) - 1 =

(4 + 1 023)(10) =

1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1027(10) =

100 0000 0011(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1101 1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100 1 0001 =

1101 1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100


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) =
100 0000 0011

Mantissa (52 bits) =
1101 1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100


Decimal number -29.931 663 088 704 072 204 551 133 979 64 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0011 - 1101 1110 1110 1000 0001 0111 1000 1110 0000 1101 1111 1101 0100

» Convert decimal -29.931 663 088 704 072 204 551 133 978 98 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