-0.000 006 491 636 613 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 006 491 636 613(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
-0.000 006 491 636 613(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:

|-0.000 006 491 636 613| = 0.000 006 491 636 613


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: 0.000 006 491 636 613.

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) 0.000 006 491 636 613 × 2 = 0 + 0.000 012 983 273 226;
  • 2) 0.000 012 983 273 226 × 2 = 0 + 0.000 025 966 546 452;
  • 3) 0.000 025 966 546 452 × 2 = 0 + 0.000 051 933 092 904;
  • 4) 0.000 051 933 092 904 × 2 = 0 + 0.000 103 866 185 808;
  • 5) 0.000 103 866 185 808 × 2 = 0 + 0.000 207 732 371 616;
  • 6) 0.000 207 732 371 616 × 2 = 0 + 0.000 415 464 743 232;
  • 7) 0.000 415 464 743 232 × 2 = 0 + 0.000 830 929 486 464;
  • 8) 0.000 830 929 486 464 × 2 = 0 + 0.001 661 858 972 928;
  • 9) 0.001 661 858 972 928 × 2 = 0 + 0.003 323 717 945 856;
  • 10) 0.003 323 717 945 856 × 2 = 0 + 0.006 647 435 891 712;
  • 11) 0.006 647 435 891 712 × 2 = 0 + 0.013 294 871 783 424;
  • 12) 0.013 294 871 783 424 × 2 = 0 + 0.026 589 743 566 848;
  • 13) 0.026 589 743 566 848 × 2 = 0 + 0.053 179 487 133 696;
  • 14) 0.053 179 487 133 696 × 2 = 0 + 0.106 358 974 267 392;
  • 15) 0.106 358 974 267 392 × 2 = 0 + 0.212 717 948 534 784;
  • 16) 0.212 717 948 534 784 × 2 = 0 + 0.425 435 897 069 568;
  • 17) 0.425 435 897 069 568 × 2 = 0 + 0.850 871 794 139 136;
  • 18) 0.850 871 794 139 136 × 2 = 1 + 0.701 743 588 278 272;
  • 19) 0.701 743 588 278 272 × 2 = 1 + 0.403 487 176 556 544;
  • 20) 0.403 487 176 556 544 × 2 = 0 + 0.806 974 353 113 088;
  • 21) 0.806 974 353 113 088 × 2 = 1 + 0.613 948 706 226 176;
  • 22) 0.613 948 706 226 176 × 2 = 1 + 0.227 897 412 452 352;
  • 23) 0.227 897 412 452 352 × 2 = 0 + 0.455 794 824 904 704;
  • 24) 0.455 794 824 904 704 × 2 = 0 + 0.911 589 649 809 408;
  • 25) 0.911 589 649 809 408 × 2 = 1 + 0.823 179 299 618 816;
  • 26) 0.823 179 299 618 816 × 2 = 1 + 0.646 358 599 237 632;
  • 27) 0.646 358 599 237 632 × 2 = 1 + 0.292 717 198 475 264;
  • 28) 0.292 717 198 475 264 × 2 = 0 + 0.585 434 396 950 528;
  • 29) 0.585 434 396 950 528 × 2 = 1 + 0.170 868 793 901 056;
  • 30) 0.170 868 793 901 056 × 2 = 0 + 0.341 737 587 802 112;
  • 31) 0.341 737 587 802 112 × 2 = 0 + 0.683 475 175 604 224;
  • 32) 0.683 475 175 604 224 × 2 = 1 + 0.366 950 351 208 448;
  • 33) 0.366 950 351 208 448 × 2 = 0 + 0.733 900 702 416 896;
  • 34) 0.733 900 702 416 896 × 2 = 1 + 0.467 801 404 833 792;
  • 35) 0.467 801 404 833 792 × 2 = 0 + 0.935 602 809 667 584;
  • 36) 0.935 602 809 667 584 × 2 = 1 + 0.871 205 619 335 168;
  • 37) 0.871 205 619 335 168 × 2 = 1 + 0.742 411 238 670 336;
  • 38) 0.742 411 238 670 336 × 2 = 1 + 0.484 822 477 340 672;
  • 39) 0.484 822 477 340 672 × 2 = 0 + 0.969 644 954 681 344;
  • 40) 0.969 644 954 681 344 × 2 = 1 + 0.939 289 909 362 688;
  • 41) 0.939 289 909 362 688 × 2 = 1 + 0.878 579 818 725 376;
  • 42) 0.878 579 818 725 376 × 2 = 1 + 0.757 159 637 450 752;
  • 43) 0.757 159 637 450 752 × 2 = 1 + 0.514 319 274 901 504;
  • 44) 0.514 319 274 901 504 × 2 = 1 + 0.028 638 549 803 008;
  • 45) 0.028 638 549 803 008 × 2 = 0 + 0.057 277 099 606 016;
  • 46) 0.057 277 099 606 016 × 2 = 0 + 0.114 554 199 212 032;
  • 47) 0.114 554 199 212 032 × 2 = 0 + 0.229 108 398 424 064;
  • 48) 0.229 108 398 424 064 × 2 = 0 + 0.458 216 796 848 128;
  • 49) 0.458 216 796 848 128 × 2 = 0 + 0.916 433 593 696 256;
  • 50) 0.916 433 593 696 256 × 2 = 1 + 0.832 867 187 392 512;
  • 51) 0.832 867 187 392 512 × 2 = 1 + 0.665 734 374 785 024;
  • 52) 0.665 734 374 785 024 × 2 = 1 + 0.331 468 749 570 048;
  • 53) 0.331 468 749 570 048 × 2 = 0 + 0.662 937 499 140 096;
  • 54) 0.662 937 499 140 096 × 2 = 1 + 0.325 874 998 280 192;
  • 55) 0.325 874 998 280 192 × 2 = 0 + 0.651 749 996 560 384;
  • 56) 0.651 749 996 560 384 × 2 = 1 + 0.303 499 993 120 768;
  • 57) 0.303 499 993 120 768 × 2 = 0 + 0.606 999 986 241 536;
  • 58) 0.606 999 986 241 536 × 2 = 1 + 0.213 999 972 483 072;
  • 59) 0.213 999 972 483 072 × 2 = 0 + 0.427 999 944 966 144;
  • 60) 0.427 999 944 966 144 × 2 = 0 + 0.855 999 889 932 288;
  • 61) 0.855 999 889 932 288 × 2 = 1 + 0.711 999 779 864 576;
  • 62) 0.711 999 779 864 576 × 2 = 1 + 0.423 999 559 729 152;
  • 63) 0.423 999 559 729 152 × 2 = 0 + 0.847 999 119 458 304;
  • 64) 0.847 999 119 458 304 × 2 = 1 + 0.695 998 238 916 608;
  • 65) 0.695 998 238 916 608 × 2 = 1 + 0.391 996 477 833 216;
  • 66) 0.391 996 477 833 216 × 2 = 0 + 0.783 992 955 666 432;
  • 67) 0.783 992 955 666 432 × 2 = 1 + 0.567 985 911 332 864;
  • 68) 0.567 985 911 332 864 × 2 = 1 + 0.135 971 822 665 728;
  • 69) 0.135 971 822 665 728 × 2 = 0 + 0.271 943 645 331 456;
  • 70) 0.271 943 645 331 456 × 2 = 0 + 0.543 887 290 662 912;

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:


0.000 006 491 636 613(10) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1111 0000 0111 0101 0100 1101 1011 00(2)

6. Positive number before normalization:

0.000 006 491 636 613(10) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1111 0000 0111 0101 0100 1101 1011 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the right, so that only one non zero digit remains to the left of it:


0.000 006 491 636 613(10) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1111 0000 0111 0101 0100 1101 1011 00(2) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1111 0000 0111 0101 0100 1101 1011 00(2) × 20 =

1.1011 0011 1010 0101 0111 0111 1100 0001 1101 0101 0011 0110 1100(2) × 2-18


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): -18

Mantissa (not normalized):
1.1011 0011 1010 0101 0111 0111 1100 0001 1101 0101 0011 0110 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-18 + 2(11-1) - 1 =

(-18 + 1 023)(10) =

1 005(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;
  • 1 005 ÷ 2 = 502 + 1;
  • 502 ÷ 2 = 251 + 0;
  • 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;

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) =

1005(10) =

011 1110 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. 1011 0011 1010 0101 0111 0111 1100 0001 1101 0101 0011 0110 1100 =

1011 0011 1010 0101 0111 0111 1100 0001 1101 0101 0011 0110 1100


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 1110 1101

Mantissa (52 bits) =
1011 0011 1010 0101 0111 0111 1100 0001 1101 0101 0011 0110 1100


Decimal number -0.000 006 491 636 613 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1101 - 1011 0011 1010 0101 0111 0111 1100 0001 1101 0101 0011 0110 1100

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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