-0.000 375 658 489 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 375 658 489 6(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 6(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 6| = 0.000 375 658 489 6


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 6.

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 6 × 2 = 0 + 0.000 751 316 979 2;
  • 2) 0.000 751 316 979 2 × 2 = 0 + 0.001 502 633 958 4;
  • 3) 0.001 502 633 958 4 × 2 = 0 + 0.003 005 267 916 8;
  • 4) 0.003 005 267 916 8 × 2 = 0 + 0.006 010 535 833 6;
  • 5) 0.006 010 535 833 6 × 2 = 0 + 0.012 021 071 667 2;
  • 6) 0.012 021 071 667 2 × 2 = 0 + 0.024 042 143 334 4;
  • 7) 0.024 042 143 334 4 × 2 = 0 + 0.048 084 286 668 8;
  • 8) 0.048 084 286 668 8 × 2 = 0 + 0.096 168 573 337 6;
  • 9) 0.096 168 573 337 6 × 2 = 0 + 0.192 337 146 675 2;
  • 10) 0.192 337 146 675 2 × 2 = 0 + 0.384 674 293 350 4;
  • 11) 0.384 674 293 350 4 × 2 = 0 + 0.769 348 586 700 8;
  • 12) 0.769 348 586 700 8 × 2 = 1 + 0.538 697 173 401 6;
  • 13) 0.538 697 173 401 6 × 2 = 1 + 0.077 394 346 803 2;
  • 14) 0.077 394 346 803 2 × 2 = 0 + 0.154 788 693 606 4;
  • 15) 0.154 788 693 606 4 × 2 = 0 + 0.309 577 387 212 8;
  • 16) 0.309 577 387 212 8 × 2 = 0 + 0.619 154 774 425 6;
  • 17) 0.619 154 774 425 6 × 2 = 1 + 0.238 309 548 851 2;
  • 18) 0.238 309 548 851 2 × 2 = 0 + 0.476 619 097 702 4;
  • 19) 0.476 619 097 702 4 × 2 = 0 + 0.953 238 195 404 8;
  • 20) 0.953 238 195 404 8 × 2 = 1 + 0.906 476 390 809 6;
  • 21) 0.906 476 390 809 6 × 2 = 1 + 0.812 952 781 619 2;
  • 22) 0.812 952 781 619 2 × 2 = 1 + 0.625 905 563 238 4;
  • 23) 0.625 905 563 238 4 × 2 = 1 + 0.251 811 126 476 8;
  • 24) 0.251 811 126 476 8 × 2 = 0 + 0.503 622 252 953 6;
  • 25) 0.503 622 252 953 6 × 2 = 1 + 0.007 244 505 907 2;
  • 26) 0.007 244 505 907 2 × 2 = 0 + 0.014 489 011 814 4;
  • 27) 0.014 489 011 814 4 × 2 = 0 + 0.028 978 023 628 8;
  • 28) 0.028 978 023 628 8 × 2 = 0 + 0.057 956 047 257 6;
  • 29) 0.057 956 047 257 6 × 2 = 0 + 0.115 912 094 515 2;
  • 30) 0.115 912 094 515 2 × 2 = 0 + 0.231 824 189 030 4;
  • 31) 0.231 824 189 030 4 × 2 = 0 + 0.463 648 378 060 8;
  • 32) 0.463 648 378 060 8 × 2 = 0 + 0.927 296 756 121 6;
  • 33) 0.927 296 756 121 6 × 2 = 1 + 0.854 593 512 243 2;
  • 34) 0.854 593 512 243 2 × 2 = 1 + 0.709 187 024 486 4;
  • 35) 0.709 187 024 486 4 × 2 = 1 + 0.418 374 048 972 8;
  • 36) 0.418 374 048 972 8 × 2 = 0 + 0.836 748 097 945 6;
  • 37) 0.836 748 097 945 6 × 2 = 1 + 0.673 496 195 891 2;
  • 38) 0.673 496 195 891 2 × 2 = 1 + 0.346 992 391 782 4;
  • 39) 0.346 992 391 782 4 × 2 = 0 + 0.693 984 783 564 8;
  • 40) 0.693 984 783 564 8 × 2 = 1 + 0.387 969 567 129 6;
  • 41) 0.387 969 567 129 6 × 2 = 0 + 0.775 939 134 259 2;
  • 42) 0.775 939 134 259 2 × 2 = 1 + 0.551 878 268 518 4;
  • 43) 0.551 878 268 518 4 × 2 = 1 + 0.103 756 537 036 8;
  • 44) 0.103 756 537 036 8 × 2 = 0 + 0.207 513 074 073 6;
  • 45) 0.207 513 074 073 6 × 2 = 0 + 0.415 026 148 147 2;
  • 46) 0.415 026 148 147 2 × 2 = 0 + 0.830 052 296 294 4;
  • 47) 0.830 052 296 294 4 × 2 = 1 + 0.660 104 592 588 8;
  • 48) 0.660 104 592 588 8 × 2 = 1 + 0.320 209 185 177 6;
  • 49) 0.320 209 185 177 6 × 2 = 0 + 0.640 418 370 355 2;
  • 50) 0.640 418 370 355 2 × 2 = 1 + 0.280 836 740 710 4;
  • 51) 0.280 836 740 710 4 × 2 = 0 + 0.561 673 481 420 8;
  • 52) 0.561 673 481 420 8 × 2 = 1 + 0.123 346 962 841 6;
  • 53) 0.123 346 962 841 6 × 2 = 0 + 0.246 693 925 683 2;
  • 54) 0.246 693 925 683 2 × 2 = 0 + 0.493 387 851 366 4;
  • 55) 0.493 387 851 366 4 × 2 = 0 + 0.986 775 702 732 8;
  • 56) 0.986 775 702 732 8 × 2 = 1 + 0.973 551 405 465 6;
  • 57) 0.973 551 405 465 6 × 2 = 1 + 0.947 102 810 931 2;
  • 58) 0.947 102 810 931 2 × 2 = 1 + 0.894 205 621 862 4;
  • 59) 0.894 205 621 862 4 × 2 = 1 + 0.788 411 243 724 8;
  • 60) 0.788 411 243 724 8 × 2 = 1 + 0.576 822 487 449 6;
  • 61) 0.576 822 487 449 6 × 2 = 1 + 0.153 644 974 899 2;
  • 62) 0.153 644 974 899 2 × 2 = 0 + 0.307 289 949 798 4;
  • 63) 0.307 289 949 798 4 × 2 = 0 + 0.614 579 899 596 8;
  • 64) 0.614 579 899 596 8 × 2 = 1 + 0.229 159 799 193 6;

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 6(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0110 0011 0101 0001 1111 1001(2)

6. Positive number before normalization:

0.000 375 658 489 6(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0110 0011 0101 0001 1111 1001(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 6(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0110 0011 0101 0001 1111 1001(2) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0110 0011 0101 0001 1111 1001(2) × 20 =

1.1000 1001 1110 1000 0000 1110 1101 0110 0011 0101 0001 1111 1001(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 1101 0110 0011 0101 0001 1111 1001


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 1101 0110 0011 0101 0001 1111 1001 =

1000 1001 1110 1000 0000 1110 1101 0110 0011 0101 0001 1111 1001


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 1101 0110 0011 0101 0001 1111 1001


Decimal number -0.000 375 658 489 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 1000 1001 1110 1000 0000 1110 1101 0110 0011 0101 0001 1111 1001

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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