-0.000 792 035 636 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 792 035 636 42(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
-0.000 792 035 636 42(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:

|-0.000 792 035 636 42| = 0.000 792 035 636 42


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: 0.000 792 035 636 42.

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) 0.000 792 035 636 42 × 2 = 0 + 0.001 584 071 272 84;
  • 2) 0.001 584 071 272 84 × 2 = 0 + 0.003 168 142 545 68;
  • 3) 0.003 168 142 545 68 × 2 = 0 + 0.006 336 285 091 36;
  • 4) 0.006 336 285 091 36 × 2 = 0 + 0.012 672 570 182 72;
  • 5) 0.012 672 570 182 72 × 2 = 0 + 0.025 345 140 365 44;
  • 6) 0.025 345 140 365 44 × 2 = 0 + 0.050 690 280 730 88;
  • 7) 0.050 690 280 730 88 × 2 = 0 + 0.101 380 561 461 76;
  • 8) 0.101 380 561 461 76 × 2 = 0 + 0.202 761 122 923 52;
  • 9) 0.202 761 122 923 52 × 2 = 0 + 0.405 522 245 847 04;
  • 10) 0.405 522 245 847 04 × 2 = 0 + 0.811 044 491 694 08;
  • 11) 0.811 044 491 694 08 × 2 = 1 + 0.622 088 983 388 16;
  • 12) 0.622 088 983 388 16 × 2 = 1 + 0.244 177 966 776 32;
  • 13) 0.244 177 966 776 32 × 2 = 0 + 0.488 355 933 552 64;
  • 14) 0.488 355 933 552 64 × 2 = 0 + 0.976 711 867 105 28;
  • 15) 0.976 711 867 105 28 × 2 = 1 + 0.953 423 734 210 56;
  • 16) 0.953 423 734 210 56 × 2 = 1 + 0.906 847 468 421 12;
  • 17) 0.906 847 468 421 12 × 2 = 1 + 0.813 694 936 842 24;
  • 18) 0.813 694 936 842 24 × 2 = 1 + 0.627 389 873 684 48;
  • 19) 0.627 389 873 684 48 × 2 = 1 + 0.254 779 747 368 96;
  • 20) 0.254 779 747 368 96 × 2 = 0 + 0.509 559 494 737 92;
  • 21) 0.509 559 494 737 92 × 2 = 1 + 0.019 118 989 475 84;
  • 22) 0.019 118 989 475 84 × 2 = 0 + 0.038 237 978 951 68;
  • 23) 0.038 237 978 951 68 × 2 = 0 + 0.076 475 957 903 36;
  • 24) 0.076 475 957 903 36 × 2 = 0 + 0.152 951 915 806 72;
  • 25) 0.152 951 915 806 72 × 2 = 0 + 0.305 903 831 613 44;
  • 26) 0.305 903 831 613 44 × 2 = 0 + 0.611 807 663 226 88;
  • 27) 0.611 807 663 226 88 × 2 = 1 + 0.223 615 326 453 76;
  • 28) 0.223 615 326 453 76 × 2 = 0 + 0.447 230 652 907 52;
  • 29) 0.447 230 652 907 52 × 2 = 0 + 0.894 461 305 815 04;
  • 30) 0.894 461 305 815 04 × 2 = 1 + 0.788 922 611 630 08;
  • 31) 0.788 922 611 630 08 × 2 = 1 + 0.577 845 223 260 16;
  • 32) 0.577 845 223 260 16 × 2 = 1 + 0.155 690 446 520 32;
  • 33) 0.155 690 446 520 32 × 2 = 0 + 0.311 380 893 040 64;
  • 34) 0.311 380 893 040 64 × 2 = 0 + 0.622 761 786 081 28;
  • 35) 0.622 761 786 081 28 × 2 = 1 + 0.245 523 572 162 56;
  • 36) 0.245 523 572 162 56 × 2 = 0 + 0.491 047 144 325 12;
  • 37) 0.491 047 144 325 12 × 2 = 0 + 0.982 094 288 650 24;
  • 38) 0.982 094 288 650 24 × 2 = 1 + 0.964 188 577 300 48;
  • 39) 0.964 188 577 300 48 × 2 = 1 + 0.928 377 154 600 96;
  • 40) 0.928 377 154 600 96 × 2 = 1 + 0.856 754 309 201 92;
  • 41) 0.856 754 309 201 92 × 2 = 1 + 0.713 508 618 403 84;
  • 42) 0.713 508 618 403 84 × 2 = 1 + 0.427 017 236 807 68;
  • 43) 0.427 017 236 807 68 × 2 = 0 + 0.854 034 473 615 36;
  • 44) 0.854 034 473 615 36 × 2 = 1 + 0.708 068 947 230 72;
  • 45) 0.708 068 947 230 72 × 2 = 1 + 0.416 137 894 461 44;
  • 46) 0.416 137 894 461 44 × 2 = 0 + 0.832 275 788 922 88;
  • 47) 0.832 275 788 922 88 × 2 = 1 + 0.664 551 577 845 76;
  • 48) 0.664 551 577 845 76 × 2 = 1 + 0.329 103 155 691 52;
  • 49) 0.329 103 155 691 52 × 2 = 0 + 0.658 206 311 383 04;
  • 50) 0.658 206 311 383 04 × 2 = 1 + 0.316 412 622 766 08;
  • 51) 0.316 412 622 766 08 × 2 = 0 + 0.632 825 245 532 16;
  • 52) 0.632 825 245 532 16 × 2 = 1 + 0.265 650 491 064 32;
  • 53) 0.265 650 491 064 32 × 2 = 0 + 0.531 300 982 128 64;
  • 54) 0.531 300 982 128 64 × 2 = 1 + 0.062 601 964 257 28;
  • 55) 0.062 601 964 257 28 × 2 = 0 + 0.125 203 928 514 56;
  • 56) 0.125 203 928 514 56 × 2 = 0 + 0.250 407 857 029 12;
  • 57) 0.250 407 857 029 12 × 2 = 0 + 0.500 815 714 058 24;
  • 58) 0.500 815 714 058 24 × 2 = 1 + 0.001 631 428 116 48;
  • 59) 0.001 631 428 116 48 × 2 = 0 + 0.003 262 856 232 96;
  • 60) 0.003 262 856 232 96 × 2 = 0 + 0.006 525 712 465 92;
  • 61) 0.006 525 712 465 92 × 2 = 0 + 0.013 051 424 931 84;
  • 62) 0.013 051 424 931 84 × 2 = 0 + 0.026 102 849 863 68;
  • 63) 0.026 102 849 863 68 × 2 = 0 + 0.052 205 699 727 36;

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:


0.000 792 035 636 42(10) =

0.0000 0000 0011 0011 1110 1000 0010 0111 0010 0111 1101 1011 0101 0100 0100 000(2)

6. Positive number before normalization:

0.000 792 035 636 42(10) =

0.0000 0000 0011 0011 1110 1000 0010 0111 0010 0111 1101 1011 0101 0100 0100 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 792 035 636 42(10) =

0.0000 0000 0011 0011 1110 1000 0010 0111 0010 0111 1101 1011 0101 0100 0100 000(2) =

0.0000 0000 0011 0011 1110 1000 0010 0111 0010 0111 1101 1011 0101 0100 0100 000(2) × 20 =

1.1001 1111 0100 0001 0011 1001 0011 1110 1101 1010 1010 0010 0000(2) × 2-11


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): -11

Mantissa (not normalized):
1.1001 1111 0100 0001 0011 1001 0011 1110 1101 1010 1010 0010 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-11 + 1 023)(10) =

1 012(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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 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) =

1012(10) =

011 1111 0100(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. 1001 1111 0100 0001 0011 1001 0011 1110 1101 1010 1010 0010 0000 =

1001 1111 0100 0001 0011 1001 0011 1110 1101 1010 1010 0010 0000


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 0100

Mantissa (52 bits) =
1001 1111 0100 0001 0011 1001 0011 1110 1101 1010 1010 0010 0000


Decimal number -0.000 792 035 636 42 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0100 - 1001 1111 0100 0001 0011 1001 0011 1110 1101 1010 1010 0010 0000

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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