0.001 087 033 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.001 087 033(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
0.001 087 033(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: 0.001 087 033.

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) 0.001 087 033 × 2 = 0 + 0.002 174 066;
  • 2) 0.002 174 066 × 2 = 0 + 0.004 348 132;
  • 3) 0.004 348 132 × 2 = 0 + 0.008 696 264;
  • 4) 0.008 696 264 × 2 = 0 + 0.017 392 528;
  • 5) 0.017 392 528 × 2 = 0 + 0.034 785 056;
  • 6) 0.034 785 056 × 2 = 0 + 0.069 570 112;
  • 7) 0.069 570 112 × 2 = 0 + 0.139 140 224;
  • 8) 0.139 140 224 × 2 = 0 + 0.278 280 448;
  • 9) 0.278 280 448 × 2 = 0 + 0.556 560 896;
  • 10) 0.556 560 896 × 2 = 1 + 0.113 121 792;
  • 11) 0.113 121 792 × 2 = 0 + 0.226 243 584;
  • 12) 0.226 243 584 × 2 = 0 + 0.452 487 168;
  • 13) 0.452 487 168 × 2 = 0 + 0.904 974 336;
  • 14) 0.904 974 336 × 2 = 1 + 0.809 948 672;
  • 15) 0.809 948 672 × 2 = 1 + 0.619 897 344;
  • 16) 0.619 897 344 × 2 = 1 + 0.239 794 688;
  • 17) 0.239 794 688 × 2 = 0 + 0.479 589 376;
  • 18) 0.479 589 376 × 2 = 0 + 0.959 178 752;
  • 19) 0.959 178 752 × 2 = 1 + 0.918 357 504;
  • 20) 0.918 357 504 × 2 = 1 + 0.836 715 008;
  • 21) 0.836 715 008 × 2 = 1 + 0.673 430 016;
  • 22) 0.673 430 016 × 2 = 1 + 0.346 860 032;
  • 23) 0.346 860 032 × 2 = 0 + 0.693 720 064;
  • 24) 0.693 720 064 × 2 = 1 + 0.387 440 128;
  • 25) 0.387 440 128 × 2 = 0 + 0.774 880 256;
  • 26) 0.774 880 256 × 2 = 1 + 0.549 760 512;
  • 27) 0.549 760 512 × 2 = 1 + 0.099 521 024;
  • 28) 0.099 521 024 × 2 = 0 + 0.199 042 048;
  • 29) 0.199 042 048 × 2 = 0 + 0.398 084 096;
  • 30) 0.398 084 096 × 2 = 0 + 0.796 168 192;
  • 31) 0.796 168 192 × 2 = 1 + 0.592 336 384;
  • 32) 0.592 336 384 × 2 = 1 + 0.184 672 768;
  • 33) 0.184 672 768 × 2 = 0 + 0.369 345 536;
  • 34) 0.369 345 536 × 2 = 0 + 0.738 691 072;
  • 35) 0.738 691 072 × 2 = 1 + 0.477 382 144;
  • 36) 0.477 382 144 × 2 = 0 + 0.954 764 288;
  • 37) 0.954 764 288 × 2 = 1 + 0.909 528 576;
  • 38) 0.909 528 576 × 2 = 1 + 0.819 057 152;
  • 39) 0.819 057 152 × 2 = 1 + 0.638 114 304;
  • 40) 0.638 114 304 × 2 = 1 + 0.276 228 608;
  • 41) 0.276 228 608 × 2 = 0 + 0.552 457 216;
  • 42) 0.552 457 216 × 2 = 1 + 0.104 914 432;
  • 43) 0.104 914 432 × 2 = 0 + 0.209 828 864;
  • 44) 0.209 828 864 × 2 = 0 + 0.419 657 728;
  • 45) 0.419 657 728 × 2 = 0 + 0.839 315 456;
  • 46) 0.839 315 456 × 2 = 1 + 0.678 630 912;
  • 47) 0.678 630 912 × 2 = 1 + 0.357 261 824;
  • 48) 0.357 261 824 × 2 = 0 + 0.714 523 648;
  • 49) 0.714 523 648 × 2 = 1 + 0.429 047 296;
  • 50) 0.429 047 296 × 2 = 0 + 0.858 094 592;
  • 51) 0.858 094 592 × 2 = 1 + 0.716 189 184;
  • 52) 0.716 189 184 × 2 = 1 + 0.432 378 368;
  • 53) 0.432 378 368 × 2 = 0 + 0.864 756 736;
  • 54) 0.864 756 736 × 2 = 1 + 0.729 513 472;
  • 55) 0.729 513 472 × 2 = 1 + 0.459 026 944;
  • 56) 0.459 026 944 × 2 = 0 + 0.918 053 888;
  • 57) 0.918 053 888 × 2 = 1 + 0.836 107 776;
  • 58) 0.836 107 776 × 2 = 1 + 0.672 215 552;
  • 59) 0.672 215 552 × 2 = 1 + 0.344 431 104;
  • 60) 0.344 431 104 × 2 = 0 + 0.688 862 208;
  • 61) 0.688 862 208 × 2 = 1 + 0.377 724 416;
  • 62) 0.377 724 416 × 2 = 0 + 0.755 448 832;

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:


0.001 087 033(10) =

0.0000 0000 0100 0111 0011 1101 0110 0011 0010 1111 0100 0110 1011 0110 1110 10(2)

5. Positive number before normalization:

0.001 087 033(10) =

0.0000 0000 0100 0111 0011 1101 0110 0011 0010 1111 0100 0110 1011 0110 1110 10(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:


0.001 087 033(10) =

0.0000 0000 0100 0111 0011 1101 0110 0011 0010 1111 0100 0110 1011 0110 1110 10(2) =

0.0000 0000 0100 0111 0011 1101 0110 0011 0010 1111 0100 0110 1011 0110 1110 10(2) × 20 =

1.0001 1100 1111 0101 1000 1100 1011 1101 0001 1010 1101 1011 1010(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 1000 1100 1011 1101 0001 1010 1101 1011 1010


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 1000 1100 1011 1101 0001 1010 1101 1011 1010 =

0001 1100 1111 0101 1000 1100 1011 1101 0001 1010 1101 1011 1010


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 1000 1100 1011 1101 0001 1010 1101 1011 1010


Decimal number 0.001 087 033 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0101 - 0001 1100 1111 0101 1000 1100 1011 1101 0001 1010 1101 1011 1010

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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