-0.008 788 423 612 946 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 612 946(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
-0.008 788 423 612 946(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:

|-0.008 788 423 612 946| = 0.008 788 423 612 946


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: 0.008 788 423 612 946.

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) 0.008 788 423 612 946 × 2 = 0 + 0.017 576 847 225 892;
  • 2) 0.017 576 847 225 892 × 2 = 0 + 0.035 153 694 451 784;
  • 3) 0.035 153 694 451 784 × 2 = 0 + 0.070 307 388 903 568;
  • 4) 0.070 307 388 903 568 × 2 = 0 + 0.140 614 777 807 136;
  • 5) 0.140 614 777 807 136 × 2 = 0 + 0.281 229 555 614 272;
  • 6) 0.281 229 555 614 272 × 2 = 0 + 0.562 459 111 228 544;
  • 7) 0.562 459 111 228 544 × 2 = 1 + 0.124 918 222 457 088;
  • 8) 0.124 918 222 457 088 × 2 = 0 + 0.249 836 444 914 176;
  • 9) 0.249 836 444 914 176 × 2 = 0 + 0.499 672 889 828 352;
  • 10) 0.499 672 889 828 352 × 2 = 0 + 0.999 345 779 656 704;
  • 11) 0.999 345 779 656 704 × 2 = 1 + 0.998 691 559 313 408;
  • 12) 0.998 691 559 313 408 × 2 = 1 + 0.997 383 118 626 816;
  • 13) 0.997 383 118 626 816 × 2 = 1 + 0.994 766 237 253 632;
  • 14) 0.994 766 237 253 632 × 2 = 1 + 0.989 532 474 507 264;
  • 15) 0.989 532 474 507 264 × 2 = 1 + 0.979 064 949 014 528;
  • 16) 0.979 064 949 014 528 × 2 = 1 + 0.958 129 898 029 056;
  • 17) 0.958 129 898 029 056 × 2 = 1 + 0.916 259 796 058 112;
  • 18) 0.916 259 796 058 112 × 2 = 1 + 0.832 519 592 116 224;
  • 19) 0.832 519 592 116 224 × 2 = 1 + 0.665 039 184 232 448;
  • 20) 0.665 039 184 232 448 × 2 = 1 + 0.330 078 368 464 896;
  • 21) 0.330 078 368 464 896 × 2 = 0 + 0.660 156 736 929 792;
  • 22) 0.660 156 736 929 792 × 2 = 1 + 0.320 313 473 859 584;
  • 23) 0.320 313 473 859 584 × 2 = 0 + 0.640 626 947 719 168;
  • 24) 0.640 626 947 719 168 × 2 = 1 + 0.281 253 895 438 336;
  • 25) 0.281 253 895 438 336 × 2 = 0 + 0.562 507 790 876 672;
  • 26) 0.562 507 790 876 672 × 2 = 1 + 0.125 015 581 753 344;
  • 27) 0.125 015 581 753 344 × 2 = 0 + 0.250 031 163 506 688;
  • 28) 0.250 031 163 506 688 × 2 = 0 + 0.500 062 327 013 376;
  • 29) 0.500 062 327 013 376 × 2 = 1 + 0.000 124 654 026 752;
  • 30) 0.000 124 654 026 752 × 2 = 0 + 0.000 249 308 053 504;
  • 31) 0.000 249 308 053 504 × 2 = 0 + 0.000 498 616 107 008;
  • 32) 0.000 498 616 107 008 × 2 = 0 + 0.000 997 232 214 016;
  • 33) 0.000 997 232 214 016 × 2 = 0 + 0.001 994 464 428 032;
  • 34) 0.001 994 464 428 032 × 2 = 0 + 0.003 988 928 856 064;
  • 35) 0.003 988 928 856 064 × 2 = 0 + 0.007 977 857 712 128;
  • 36) 0.007 977 857 712 128 × 2 = 0 + 0.015 955 715 424 256;
  • 37) 0.015 955 715 424 256 × 2 = 0 + 0.031 911 430 848 512;
  • 38) 0.031 911 430 848 512 × 2 = 0 + 0.063 822 861 697 024;
  • 39) 0.063 822 861 697 024 × 2 = 0 + 0.127 645 723 394 048;
  • 40) 0.127 645 723 394 048 × 2 = 0 + 0.255 291 446 788 096;
  • 41) 0.255 291 446 788 096 × 2 = 0 + 0.510 582 893 576 192;
  • 42) 0.510 582 893 576 192 × 2 = 1 + 0.021 165 787 152 384;
  • 43) 0.021 165 787 152 384 × 2 = 0 + 0.042 331 574 304 768;
  • 44) 0.042 331 574 304 768 × 2 = 0 + 0.084 663 148 609 536;
  • 45) 0.084 663 148 609 536 × 2 = 0 + 0.169 326 297 219 072;
  • 46) 0.169 326 297 219 072 × 2 = 0 + 0.338 652 594 438 144;
  • 47) 0.338 652 594 438 144 × 2 = 0 + 0.677 305 188 876 288;
  • 48) 0.677 305 188 876 288 × 2 = 1 + 0.354 610 377 752 576;
  • 49) 0.354 610 377 752 576 × 2 = 0 + 0.709 220 755 505 152;
  • 50) 0.709 220 755 505 152 × 2 = 1 + 0.418 441 511 010 304;
  • 51) 0.418 441 511 010 304 × 2 = 0 + 0.836 883 022 020 608;
  • 52) 0.836 883 022 020 608 × 2 = 1 + 0.673 766 044 041 216;
  • 53) 0.673 766 044 041 216 × 2 = 1 + 0.347 532 088 082 432;
  • 54) 0.347 532 088 082 432 × 2 = 0 + 0.695 064 176 164 864;
  • 55) 0.695 064 176 164 864 × 2 = 1 + 0.390 128 352 329 728;
  • 56) 0.390 128 352 329 728 × 2 = 0 + 0.780 256 704 659 456;
  • 57) 0.780 256 704 659 456 × 2 = 1 + 0.560 513 409 318 912;
  • 58) 0.560 513 409 318 912 × 2 = 1 + 0.121 026 818 637 824;
  • 59) 0.121 026 818 637 824 × 2 = 0 + 0.242 053 637 275 648;

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:


0.008 788 423 612 946(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0100 0001 0101 1010 110(2)

6. Positive number before normalization:

0.008 788 423 612 946(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0100 0001 0101 1010 110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.008 788 423 612 946(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0100 0001 0101 1010 110(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0100 0001 0101 1010 110(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0010 0000 1010 1101 0110(2) × 2-7


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): -7

Mantissa (not normalized):
1.0001 1111 1111 1010 1010 0100 0000 0000 0010 0000 1010 1101 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1016(10) =

011 1111 1000(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. 0001 1111 1111 1010 1010 0100 0000 0000 0010 0000 1010 1101 0110 =

0001 1111 1111 1010 1010 0100 0000 0000 0010 0000 1010 1101 0110


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 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0100 0000 0000 0010 0000 1010 1101 0110


Decimal number -0.008 788 423 612 946 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 0010 0000 1010 1101 0110

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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