-0.000 001 327 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 327(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
-0.000 001 327(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:

|-0.000 001 327| = 0.000 001 327


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: 0.000 001 327.

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) 0.000 001 327 × 2 = 0 + 0.000 002 654;
  • 2) 0.000 002 654 × 2 = 0 + 0.000 005 308;
  • 3) 0.000 005 308 × 2 = 0 + 0.000 010 616;
  • 4) 0.000 010 616 × 2 = 0 + 0.000 021 232;
  • 5) 0.000 021 232 × 2 = 0 + 0.000 042 464;
  • 6) 0.000 042 464 × 2 = 0 + 0.000 084 928;
  • 7) 0.000 084 928 × 2 = 0 + 0.000 169 856;
  • 8) 0.000 169 856 × 2 = 0 + 0.000 339 712;
  • 9) 0.000 339 712 × 2 = 0 + 0.000 679 424;
  • 10) 0.000 679 424 × 2 = 0 + 0.001 358 848;
  • 11) 0.001 358 848 × 2 = 0 + 0.002 717 696;
  • 12) 0.002 717 696 × 2 = 0 + 0.005 435 392;
  • 13) 0.005 435 392 × 2 = 0 + 0.010 870 784;
  • 14) 0.010 870 784 × 2 = 0 + 0.021 741 568;
  • 15) 0.021 741 568 × 2 = 0 + 0.043 483 136;
  • 16) 0.043 483 136 × 2 = 0 + 0.086 966 272;
  • 17) 0.086 966 272 × 2 = 0 + 0.173 932 544;
  • 18) 0.173 932 544 × 2 = 0 + 0.347 865 088;
  • 19) 0.347 865 088 × 2 = 0 + 0.695 730 176;
  • 20) 0.695 730 176 × 2 = 1 + 0.391 460 352;
  • 21) 0.391 460 352 × 2 = 0 + 0.782 920 704;
  • 22) 0.782 920 704 × 2 = 1 + 0.565 841 408;
  • 23) 0.565 841 408 × 2 = 1 + 0.131 682 816;
  • 24) 0.131 682 816 × 2 = 0 + 0.263 365 632;
  • 25) 0.263 365 632 × 2 = 0 + 0.526 731 264;
  • 26) 0.526 731 264 × 2 = 1 + 0.053 462 528;
  • 27) 0.053 462 528 × 2 = 0 + 0.106 925 056;
  • 28) 0.106 925 056 × 2 = 0 + 0.213 850 112;
  • 29) 0.213 850 112 × 2 = 0 + 0.427 700 224;
  • 30) 0.427 700 224 × 2 = 0 + 0.855 400 448;
  • 31) 0.855 400 448 × 2 = 1 + 0.710 800 896;
  • 32) 0.710 800 896 × 2 = 1 + 0.421 601 792;
  • 33) 0.421 601 792 × 2 = 0 + 0.843 203 584;
  • 34) 0.843 203 584 × 2 = 1 + 0.686 407 168;
  • 35) 0.686 407 168 × 2 = 1 + 0.372 814 336;
  • 36) 0.372 814 336 × 2 = 0 + 0.745 628 672;
  • 37) 0.745 628 672 × 2 = 1 + 0.491 257 344;
  • 38) 0.491 257 344 × 2 = 0 + 0.982 514 688;
  • 39) 0.982 514 688 × 2 = 1 + 0.965 029 376;
  • 40) 0.965 029 376 × 2 = 1 + 0.930 058 752;
  • 41) 0.930 058 752 × 2 = 1 + 0.860 117 504;
  • 42) 0.860 117 504 × 2 = 1 + 0.720 235 008;
  • 43) 0.720 235 008 × 2 = 1 + 0.440 470 016;
  • 44) 0.440 470 016 × 2 = 0 + 0.880 940 032;
  • 45) 0.880 940 032 × 2 = 1 + 0.761 880 064;
  • 46) 0.761 880 064 × 2 = 1 + 0.523 760 128;
  • 47) 0.523 760 128 × 2 = 1 + 0.047 520 256;
  • 48) 0.047 520 256 × 2 = 0 + 0.095 040 512;
  • 49) 0.095 040 512 × 2 = 0 + 0.190 081 024;
  • 50) 0.190 081 024 × 2 = 0 + 0.380 162 048;
  • 51) 0.380 162 048 × 2 = 0 + 0.760 324 096;
  • 52) 0.760 324 096 × 2 = 1 + 0.520 648 192;
  • 53) 0.520 648 192 × 2 = 1 + 0.041 296 384;
  • 54) 0.041 296 384 × 2 = 0 + 0.082 592 768;
  • 55) 0.082 592 768 × 2 = 0 + 0.165 185 536;
  • 56) 0.165 185 536 × 2 = 0 + 0.330 371 072;
  • 57) 0.330 371 072 × 2 = 0 + 0.660 742 144;
  • 58) 0.660 742 144 × 2 = 1 + 0.321 484 288;
  • 59) 0.321 484 288 × 2 = 0 + 0.642 968 576;
  • 60) 0.642 968 576 × 2 = 1 + 0.285 937 152;
  • 61) 0.285 937 152 × 2 = 0 + 0.571 874 304;
  • 62) 0.571 874 304 × 2 = 1 + 0.143 748 608;
  • 63) 0.143 748 608 × 2 = 0 + 0.287 497 216;
  • 64) 0.287 497 216 × 2 = 0 + 0.574 994 432;
  • 65) 0.574 994 432 × 2 = 1 + 0.149 988 864;
  • 66) 0.149 988 864 × 2 = 0 + 0.299 977 728;
  • 67) 0.299 977 728 × 2 = 0 + 0.599 955 456;
  • 68) 0.599 955 456 × 2 = 1 + 0.199 910 912;
  • 69) 0.199 910 912 × 2 = 0 + 0.399 821 824;
  • 70) 0.399 821 824 × 2 = 0 + 0.799 643 648;
  • 71) 0.799 643 648 × 2 = 1 + 0.599 287 296;
  • 72) 0.599 287 296 × 2 = 1 + 0.198 574 592;

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:


0.000 001 327(10) =

0.0000 0000 0000 0000 0001 0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011(2)

6. Positive number before normalization:

0.000 001 327(10) =

0.0000 0000 0000 0000 0001 0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 327(10) =

0.0000 0000 0000 0000 0001 0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011(2) =

0.0000 0000 0000 0000 0001 0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011(2) × 20 =

1.0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011(2) × 2-20


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): -20

Mantissa (not normalized):
1.0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(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;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1003(10) =

011 1110 1011(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. 0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011 =

0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011


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 1110 1011

Mantissa (52 bits) =
0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011


Decimal number -0.000 001 327 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1011 - 0110 0100 0011 0110 1011 1110 1110 0001 1000 0101 0100 1001 0011

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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