0.001 087 039 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.001 087 039(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
0.001 087 039(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.001 087 039.

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) 0.001 087 039 × 2 = 0 + 0.002 174 078;
  • 2) 0.002 174 078 × 2 = 0 + 0.004 348 156;
  • 3) 0.004 348 156 × 2 = 0 + 0.008 696 312;
  • 4) 0.008 696 312 × 2 = 0 + 0.017 392 624;
  • 5) 0.017 392 624 × 2 = 0 + 0.034 785 248;
  • 6) 0.034 785 248 × 2 = 0 + 0.069 570 496;
  • 7) 0.069 570 496 × 2 = 0 + 0.139 140 992;
  • 8) 0.139 140 992 × 2 = 0 + 0.278 281 984;
  • 9) 0.278 281 984 × 2 = 0 + 0.556 563 968;
  • 10) 0.556 563 968 × 2 = 1 + 0.113 127 936;
  • 11) 0.113 127 936 × 2 = 0 + 0.226 255 872;
  • 12) 0.226 255 872 × 2 = 0 + 0.452 511 744;
  • 13) 0.452 511 744 × 2 = 0 + 0.905 023 488;
  • 14) 0.905 023 488 × 2 = 1 + 0.810 046 976;
  • 15) 0.810 046 976 × 2 = 1 + 0.620 093 952;
  • 16) 0.620 093 952 × 2 = 1 + 0.240 187 904;
  • 17) 0.240 187 904 × 2 = 0 + 0.480 375 808;
  • 18) 0.480 375 808 × 2 = 0 + 0.960 751 616;
  • 19) 0.960 751 616 × 2 = 1 + 0.921 503 232;
  • 20) 0.921 503 232 × 2 = 1 + 0.843 006 464;
  • 21) 0.843 006 464 × 2 = 1 + 0.686 012 928;
  • 22) 0.686 012 928 × 2 = 1 + 0.372 025 856;
  • 23) 0.372 025 856 × 2 = 0 + 0.744 051 712;
  • 24) 0.744 051 712 × 2 = 1 + 0.488 103 424;
  • 25) 0.488 103 424 × 2 = 0 + 0.976 206 848;
  • 26) 0.976 206 848 × 2 = 1 + 0.952 413 696;
  • 27) 0.952 413 696 × 2 = 1 + 0.904 827 392;
  • 28) 0.904 827 392 × 2 = 1 + 0.809 654 784;
  • 29) 0.809 654 784 × 2 = 1 + 0.619 309 568;
  • 30) 0.619 309 568 × 2 = 1 + 0.238 619 136;
  • 31) 0.238 619 136 × 2 = 0 + 0.477 238 272;
  • 32) 0.477 238 272 × 2 = 0 + 0.954 476 544;
  • 33) 0.954 476 544 × 2 = 1 + 0.908 953 088;
  • 34) 0.908 953 088 × 2 = 1 + 0.817 906 176;
  • 35) 0.817 906 176 × 2 = 1 + 0.635 812 352;
  • 36) 0.635 812 352 × 2 = 1 + 0.271 624 704;
  • 37) 0.271 624 704 × 2 = 0 + 0.543 249 408;
  • 38) 0.543 249 408 × 2 = 1 + 0.086 498 816;
  • 39) 0.086 498 816 × 2 = 0 + 0.172 997 632;
  • 40) 0.172 997 632 × 2 = 0 + 0.345 995 264;
  • 41) 0.345 995 264 × 2 = 0 + 0.691 990 528;
  • 42) 0.691 990 528 × 2 = 1 + 0.383 981 056;
  • 43) 0.383 981 056 × 2 = 0 + 0.767 962 112;
  • 44) 0.767 962 112 × 2 = 1 + 0.535 924 224;
  • 45) 0.535 924 224 × 2 = 1 + 0.071 848 448;
  • 46) 0.071 848 448 × 2 = 0 + 0.143 696 896;
  • 47) 0.143 696 896 × 2 = 0 + 0.287 393 792;
  • 48) 0.287 393 792 × 2 = 0 + 0.574 787 584;
  • 49) 0.574 787 584 × 2 = 1 + 0.149 575 168;
  • 50) 0.149 575 168 × 2 = 0 + 0.299 150 336;
  • 51) 0.299 150 336 × 2 = 0 + 0.598 300 672;
  • 52) 0.598 300 672 × 2 = 1 + 0.196 601 344;
  • 53) 0.196 601 344 × 2 = 0 + 0.393 202 688;
  • 54) 0.393 202 688 × 2 = 0 + 0.786 405 376;
  • 55) 0.786 405 376 × 2 = 1 + 0.572 810 752;
  • 56) 0.572 810 752 × 2 = 1 + 0.145 621 504;
  • 57) 0.145 621 504 × 2 = 0 + 0.291 243 008;
  • 58) 0.291 243 008 × 2 = 0 + 0.582 486 016;
  • 59) 0.582 486 016 × 2 = 1 + 0.164 972 032;
  • 60) 0.164 972 032 × 2 = 0 + 0.329 944 064;
  • 61) 0.329 944 064 × 2 = 0 + 0.659 888 128;
  • 62) 0.659 888 128 × 2 = 1 + 0.319 776 256;

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).


4. 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:


0.001 087 039(10) =

0.0000 0000 0100 0111 0011 1101 0111 1100 1111 0100 0101 1000 1001 0011 0010 01(2)

5. Positive number before normalization:

0.001 087 039(10) =

0.0000 0000 0100 0111 0011 1101 0111 1100 1111 0100 0101 1000 1001 0011 0010 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 087 039(10) =

0.0000 0000 0100 0111 0011 1101 0111 1100 1111 0100 0101 1000 1001 0011 0010 01(2) =

0.0000 0000 0100 0111 0011 1101 0111 1100 1111 0100 0101 1000 1001 0011 0010 01(2) × 20 =

1.0001 1100 1111 0101 1111 0011 1101 0001 0110 0010 0100 1100 1001(2) × 2-10


7. 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 0 (a positive number)

Exponent (unadjusted): -10

Mantissa (not normalized):
1.0001 1100 1111 0101 1111 0011 1101 0001 0110 0010 0100 1100 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-10 + 2(11-1) - 1 =

(-10 + 1 023)(10) =

1 013(10)


9. 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;
  • 1 013 ÷ 2 = 506 + 1;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1013(10) =

011 1111 0101(2)


11. 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 1100 1111 0101 1111 0011 1101 0001 0110 0010 0100 1100 1001 =

0001 1100 1111 0101 1111 0011 1101 0001 0110 0010 0100 1100 1001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 0101

Mantissa (52 bits) =
0001 1100 1111 0101 1111 0011 1101 0001 0110 0010 0100 1100 1001


Decimal number 0.001 087 039 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0101 - 0001 1100 1111 0101 1111 0011 1101 0001 0110 0010 0100 1100 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