-0.000 001 334 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 334(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
-0.000 001 334(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:

|-0.000 001 334| = 0.000 001 334


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: 0.000 001 334.

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) 0.000 001 334 × 2 = 0 + 0.000 002 668;
  • 2) 0.000 002 668 × 2 = 0 + 0.000 005 336;
  • 3) 0.000 005 336 × 2 = 0 + 0.000 010 672;
  • 4) 0.000 010 672 × 2 = 0 + 0.000 021 344;
  • 5) 0.000 021 344 × 2 = 0 + 0.000 042 688;
  • 6) 0.000 042 688 × 2 = 0 + 0.000 085 376;
  • 7) 0.000 085 376 × 2 = 0 + 0.000 170 752;
  • 8) 0.000 170 752 × 2 = 0 + 0.000 341 504;
  • 9) 0.000 341 504 × 2 = 0 + 0.000 683 008;
  • 10) 0.000 683 008 × 2 = 0 + 0.001 366 016;
  • 11) 0.001 366 016 × 2 = 0 + 0.002 732 032;
  • 12) 0.002 732 032 × 2 = 0 + 0.005 464 064;
  • 13) 0.005 464 064 × 2 = 0 + 0.010 928 128;
  • 14) 0.010 928 128 × 2 = 0 + 0.021 856 256;
  • 15) 0.021 856 256 × 2 = 0 + 0.043 712 512;
  • 16) 0.043 712 512 × 2 = 0 + 0.087 425 024;
  • 17) 0.087 425 024 × 2 = 0 + 0.174 850 048;
  • 18) 0.174 850 048 × 2 = 0 + 0.349 700 096;
  • 19) 0.349 700 096 × 2 = 0 + 0.699 400 192;
  • 20) 0.699 400 192 × 2 = 1 + 0.398 800 384;
  • 21) 0.398 800 384 × 2 = 0 + 0.797 600 768;
  • 22) 0.797 600 768 × 2 = 1 + 0.595 201 536;
  • 23) 0.595 201 536 × 2 = 1 + 0.190 403 072;
  • 24) 0.190 403 072 × 2 = 0 + 0.380 806 144;
  • 25) 0.380 806 144 × 2 = 0 + 0.761 612 288;
  • 26) 0.761 612 288 × 2 = 1 + 0.523 224 576;
  • 27) 0.523 224 576 × 2 = 1 + 0.046 449 152;
  • 28) 0.046 449 152 × 2 = 0 + 0.092 898 304;
  • 29) 0.092 898 304 × 2 = 0 + 0.185 796 608;
  • 30) 0.185 796 608 × 2 = 0 + 0.371 593 216;
  • 31) 0.371 593 216 × 2 = 0 + 0.743 186 432;
  • 32) 0.743 186 432 × 2 = 1 + 0.486 372 864;
  • 33) 0.486 372 864 × 2 = 0 + 0.972 745 728;
  • 34) 0.972 745 728 × 2 = 1 + 0.945 491 456;
  • 35) 0.945 491 456 × 2 = 1 + 0.890 982 912;
  • 36) 0.890 982 912 × 2 = 1 + 0.781 965 824;
  • 37) 0.781 965 824 × 2 = 1 + 0.563 931 648;
  • 38) 0.563 931 648 × 2 = 1 + 0.127 863 296;
  • 39) 0.127 863 296 × 2 = 0 + 0.255 726 592;
  • 40) 0.255 726 592 × 2 = 0 + 0.511 453 184;
  • 41) 0.511 453 184 × 2 = 1 + 0.022 906 368;
  • 42) 0.022 906 368 × 2 = 0 + 0.045 812 736;
  • 43) 0.045 812 736 × 2 = 0 + 0.091 625 472;
  • 44) 0.091 625 472 × 2 = 0 + 0.183 250 944;
  • 45) 0.183 250 944 × 2 = 0 + 0.366 501 888;
  • 46) 0.366 501 888 × 2 = 0 + 0.733 003 776;
  • 47) 0.733 003 776 × 2 = 1 + 0.466 007 552;
  • 48) 0.466 007 552 × 2 = 0 + 0.932 015 104;
  • 49) 0.932 015 104 × 2 = 1 + 0.864 030 208;
  • 50) 0.864 030 208 × 2 = 1 + 0.728 060 416;
  • 51) 0.728 060 416 × 2 = 1 + 0.456 120 832;
  • 52) 0.456 120 832 × 2 = 0 + 0.912 241 664;
  • 53) 0.912 241 664 × 2 = 1 + 0.824 483 328;
  • 54) 0.824 483 328 × 2 = 1 + 0.648 966 656;
  • 55) 0.648 966 656 × 2 = 1 + 0.297 933 312;
  • 56) 0.297 933 312 × 2 = 0 + 0.595 866 624;
  • 57) 0.595 866 624 × 2 = 1 + 0.191 733 248;
  • 58) 0.191 733 248 × 2 = 0 + 0.383 466 496;
  • 59) 0.383 466 496 × 2 = 0 + 0.766 932 992;
  • 60) 0.766 932 992 × 2 = 1 + 0.533 865 984;
  • 61) 0.533 865 984 × 2 = 1 + 0.067 731 968;
  • 62) 0.067 731 968 × 2 = 0 + 0.135 463 936;
  • 63) 0.135 463 936 × 2 = 0 + 0.270 927 872;
  • 64) 0.270 927 872 × 2 = 0 + 0.541 855 744;
  • 65) 0.541 855 744 × 2 = 1 + 0.083 711 488;
  • 66) 0.083 711 488 × 2 = 0 + 0.167 422 976;
  • 67) 0.167 422 976 × 2 = 0 + 0.334 845 952;
  • 68) 0.334 845 952 × 2 = 0 + 0.669 691 904;
  • 69) 0.669 691 904 × 2 = 1 + 0.339 383 808;
  • 70) 0.339 383 808 × 2 = 0 + 0.678 767 616;
  • 71) 0.678 767 616 × 2 = 1 + 0.357 535 232;
  • 72) 0.357 535 232 × 2 = 0 + 0.715 070 464;

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:


0.000 001 334(10) =

0.0000 0000 0000 0000 0001 0110 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 1010(2)

6. Positive number before normalization:

0.000 001 334(10) =

0.0000 0000 0000 0000 0001 0110 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 1010(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:


0.000 001 334(10) =

0.0000 0000 0000 0000 0001 0110 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 1010(2) =

0.0000 0000 0000 0000 0001 0110 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 1010(2) × 20 =

1.0110 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 1010(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 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 1010


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 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 1010 =

0110 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 1010


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 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 1010


Decimal number -0.000 001 334 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1011 - 0110 0110 0001 0111 1100 1000 0010 1110 1110 1001 1000 1000 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