-0.000 375 658 489 03 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 375 658 489 03(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
-0.000 375 658 489 03(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:

|-0.000 375 658 489 03| = 0.000 375 658 489 03


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: 0.000 375 658 489 03.

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) 0.000 375 658 489 03 × 2 = 0 + 0.000 751 316 978 06;
  • 2) 0.000 751 316 978 06 × 2 = 0 + 0.001 502 633 956 12;
  • 3) 0.001 502 633 956 12 × 2 = 0 + 0.003 005 267 912 24;
  • 4) 0.003 005 267 912 24 × 2 = 0 + 0.006 010 535 824 48;
  • 5) 0.006 010 535 824 48 × 2 = 0 + 0.012 021 071 648 96;
  • 6) 0.012 021 071 648 96 × 2 = 0 + 0.024 042 143 297 92;
  • 7) 0.024 042 143 297 92 × 2 = 0 + 0.048 084 286 595 84;
  • 8) 0.048 084 286 595 84 × 2 = 0 + 0.096 168 573 191 68;
  • 9) 0.096 168 573 191 68 × 2 = 0 + 0.192 337 146 383 36;
  • 10) 0.192 337 146 383 36 × 2 = 0 + 0.384 674 292 766 72;
  • 11) 0.384 674 292 766 72 × 2 = 0 + 0.769 348 585 533 44;
  • 12) 0.769 348 585 533 44 × 2 = 1 + 0.538 697 171 066 88;
  • 13) 0.538 697 171 066 88 × 2 = 1 + 0.077 394 342 133 76;
  • 14) 0.077 394 342 133 76 × 2 = 0 + 0.154 788 684 267 52;
  • 15) 0.154 788 684 267 52 × 2 = 0 + 0.309 577 368 535 04;
  • 16) 0.309 577 368 535 04 × 2 = 0 + 0.619 154 737 070 08;
  • 17) 0.619 154 737 070 08 × 2 = 1 + 0.238 309 474 140 16;
  • 18) 0.238 309 474 140 16 × 2 = 0 + 0.476 618 948 280 32;
  • 19) 0.476 618 948 280 32 × 2 = 0 + 0.953 237 896 560 64;
  • 20) 0.953 237 896 560 64 × 2 = 1 + 0.906 475 793 121 28;
  • 21) 0.906 475 793 121 28 × 2 = 1 + 0.812 951 586 242 56;
  • 22) 0.812 951 586 242 56 × 2 = 1 + 0.625 903 172 485 12;
  • 23) 0.625 903 172 485 12 × 2 = 1 + 0.251 806 344 970 24;
  • 24) 0.251 806 344 970 24 × 2 = 0 + 0.503 612 689 940 48;
  • 25) 0.503 612 689 940 48 × 2 = 1 + 0.007 225 379 880 96;
  • 26) 0.007 225 379 880 96 × 2 = 0 + 0.014 450 759 761 92;
  • 27) 0.014 450 759 761 92 × 2 = 0 + 0.028 901 519 523 84;
  • 28) 0.028 901 519 523 84 × 2 = 0 + 0.057 803 039 047 68;
  • 29) 0.057 803 039 047 68 × 2 = 0 + 0.115 606 078 095 36;
  • 30) 0.115 606 078 095 36 × 2 = 0 + 0.231 212 156 190 72;
  • 31) 0.231 212 156 190 72 × 2 = 0 + 0.462 424 312 381 44;
  • 32) 0.462 424 312 381 44 × 2 = 0 + 0.924 848 624 762 88;
  • 33) 0.924 848 624 762 88 × 2 = 1 + 0.849 697 249 525 76;
  • 34) 0.849 697 249 525 76 × 2 = 1 + 0.699 394 499 051 52;
  • 35) 0.699 394 499 051 52 × 2 = 1 + 0.398 788 998 103 04;
  • 36) 0.398 788 998 103 04 × 2 = 0 + 0.797 577 996 206 08;
  • 37) 0.797 577 996 206 08 × 2 = 1 + 0.595 155 992 412 16;
  • 38) 0.595 155 992 412 16 × 2 = 1 + 0.190 311 984 824 32;
  • 39) 0.190 311 984 824 32 × 2 = 0 + 0.380 623 969 648 64;
  • 40) 0.380 623 969 648 64 × 2 = 0 + 0.761 247 939 297 28;
  • 41) 0.761 247 939 297 28 × 2 = 1 + 0.522 495 878 594 56;
  • 42) 0.522 495 878 594 56 × 2 = 1 + 0.044 991 757 189 12;
  • 43) 0.044 991 757 189 12 × 2 = 0 + 0.089 983 514 378 24;
  • 44) 0.089 983 514 378 24 × 2 = 0 + 0.179 967 028 756 48;
  • 45) 0.179 967 028 756 48 × 2 = 0 + 0.359 934 057 512 96;
  • 46) 0.359 934 057 512 96 × 2 = 0 + 0.719 868 115 025 92;
  • 47) 0.719 868 115 025 92 × 2 = 1 + 0.439 736 230 051 84;
  • 48) 0.439 736 230 051 84 × 2 = 0 + 0.879 472 460 103 68;
  • 49) 0.879 472 460 103 68 × 2 = 1 + 0.758 944 920 207 36;
  • 50) 0.758 944 920 207 36 × 2 = 1 + 0.517 889 840 414 72;
  • 51) 0.517 889 840 414 72 × 2 = 1 + 0.035 779 680 829 44;
  • 52) 0.035 779 680 829 44 × 2 = 0 + 0.071 559 361 658 88;
  • 53) 0.071 559 361 658 88 × 2 = 0 + 0.143 118 723 317 76;
  • 54) 0.143 118 723 317 76 × 2 = 0 + 0.286 237 446 635 52;
  • 55) 0.286 237 446 635 52 × 2 = 0 + 0.572 474 893 271 04;
  • 56) 0.572 474 893 271 04 × 2 = 1 + 0.144 949 786 542 08;
  • 57) 0.144 949 786 542 08 × 2 = 0 + 0.289 899 573 084 16;
  • 58) 0.289 899 573 084 16 × 2 = 0 + 0.579 799 146 168 32;
  • 59) 0.579 799 146 168 32 × 2 = 1 + 0.159 598 292 336 64;
  • 60) 0.159 598 292 336 64 × 2 = 0 + 0.319 196 584 673 28;
  • 61) 0.319 196 584 673 28 × 2 = 0 + 0.638 393 169 346 56;
  • 62) 0.638 393 169 346 56 × 2 = 1 + 0.276 786 338 693 12;
  • 63) 0.276 786 338 693 12 × 2 = 0 + 0.553 572 677 386 24;
  • 64) 0.553 572 677 386 24 × 2 = 1 + 0.107 145 354 772 48;

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:


0.000 375 658 489 03(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101(2)

6. Positive number before normalization:

0.000 375 658 489 03(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 375 658 489 03(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101(2) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101(2) × 20 =

1.1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101(2) × 2-12


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): -12

Mantissa (not normalized):
1.1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101 =

1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101


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 0011

Mantissa (52 bits) =
1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101


Decimal number -0.000 375 658 489 03 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 1000 1001 1110 1000 0000 1110 1100 1100 0010 1110 0001 0010 0101

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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