-0.000 282 005 752 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 752(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
-0.000 282 005 752(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:

|-0.000 282 005 752| = 0.000 282 005 752


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: 0.000 282 005 752.

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) 0.000 282 005 752 × 2 = 0 + 0.000 564 011 504;
  • 2) 0.000 564 011 504 × 2 = 0 + 0.001 128 023 008;
  • 3) 0.001 128 023 008 × 2 = 0 + 0.002 256 046 016;
  • 4) 0.002 256 046 016 × 2 = 0 + 0.004 512 092 032;
  • 5) 0.004 512 092 032 × 2 = 0 + 0.009 024 184 064;
  • 6) 0.009 024 184 064 × 2 = 0 + 0.018 048 368 128;
  • 7) 0.018 048 368 128 × 2 = 0 + 0.036 096 736 256;
  • 8) 0.036 096 736 256 × 2 = 0 + 0.072 193 472 512;
  • 9) 0.072 193 472 512 × 2 = 0 + 0.144 386 945 024;
  • 10) 0.144 386 945 024 × 2 = 0 + 0.288 773 890 048;
  • 11) 0.288 773 890 048 × 2 = 0 + 0.577 547 780 096;
  • 12) 0.577 547 780 096 × 2 = 1 + 0.155 095 560 192;
  • 13) 0.155 095 560 192 × 2 = 0 + 0.310 191 120 384;
  • 14) 0.310 191 120 384 × 2 = 0 + 0.620 382 240 768;
  • 15) 0.620 382 240 768 × 2 = 1 + 0.240 764 481 536;
  • 16) 0.240 764 481 536 × 2 = 0 + 0.481 528 963 072;
  • 17) 0.481 528 963 072 × 2 = 0 + 0.963 057 926 144;
  • 18) 0.963 057 926 144 × 2 = 1 + 0.926 115 852 288;
  • 19) 0.926 115 852 288 × 2 = 1 + 0.852 231 704 576;
  • 20) 0.852 231 704 576 × 2 = 1 + 0.704 463 409 152;
  • 21) 0.704 463 409 152 × 2 = 1 + 0.408 926 818 304;
  • 22) 0.408 926 818 304 × 2 = 0 + 0.817 853 636 608;
  • 23) 0.817 853 636 608 × 2 = 1 + 0.635 707 273 216;
  • 24) 0.635 707 273 216 × 2 = 1 + 0.271 414 546 432;
  • 25) 0.271 414 546 432 × 2 = 0 + 0.542 829 092 864;
  • 26) 0.542 829 092 864 × 2 = 1 + 0.085 658 185 728;
  • 27) 0.085 658 185 728 × 2 = 0 + 0.171 316 371 456;
  • 28) 0.171 316 371 456 × 2 = 0 + 0.342 632 742 912;
  • 29) 0.342 632 742 912 × 2 = 0 + 0.685 265 485 824;
  • 30) 0.685 265 485 824 × 2 = 1 + 0.370 530 971 648;
  • 31) 0.370 530 971 648 × 2 = 0 + 0.741 061 943 296;
  • 32) 0.741 061 943 296 × 2 = 1 + 0.482 123 886 592;
  • 33) 0.482 123 886 592 × 2 = 0 + 0.964 247 773 184;
  • 34) 0.964 247 773 184 × 2 = 1 + 0.928 495 546 368;
  • 35) 0.928 495 546 368 × 2 = 1 + 0.856 991 092 736;
  • 36) 0.856 991 092 736 × 2 = 1 + 0.713 982 185 472;
  • 37) 0.713 982 185 472 × 2 = 1 + 0.427 964 370 944;
  • 38) 0.427 964 370 944 × 2 = 0 + 0.855 928 741 888;
  • 39) 0.855 928 741 888 × 2 = 1 + 0.711 857 483 776;
  • 40) 0.711 857 483 776 × 2 = 1 + 0.423 714 967 552;
  • 41) 0.423 714 967 552 × 2 = 0 + 0.847 429 935 104;
  • 42) 0.847 429 935 104 × 2 = 1 + 0.694 859 870 208;
  • 43) 0.694 859 870 208 × 2 = 1 + 0.389 719 740 416;
  • 44) 0.389 719 740 416 × 2 = 0 + 0.779 439 480 832;
  • 45) 0.779 439 480 832 × 2 = 1 + 0.558 878 961 664;
  • 46) 0.558 878 961 664 × 2 = 1 + 0.117 757 923 328;
  • 47) 0.117 757 923 328 × 2 = 0 + 0.235 515 846 656;
  • 48) 0.235 515 846 656 × 2 = 0 + 0.471 031 693 312;
  • 49) 0.471 031 693 312 × 2 = 0 + 0.942 063 386 624;
  • 50) 0.942 063 386 624 × 2 = 1 + 0.884 126 773 248;
  • 51) 0.884 126 773 248 × 2 = 1 + 0.768 253 546 496;
  • 52) 0.768 253 546 496 × 2 = 1 + 0.536 507 092 992;
  • 53) 0.536 507 092 992 × 2 = 1 + 0.073 014 185 984;
  • 54) 0.073 014 185 984 × 2 = 0 + 0.146 028 371 968;
  • 55) 0.146 028 371 968 × 2 = 0 + 0.292 056 743 936;
  • 56) 0.292 056 743 936 × 2 = 0 + 0.584 113 487 872;
  • 57) 0.584 113 487 872 × 2 = 1 + 0.168 226 975 744;
  • 58) 0.168 226 975 744 × 2 = 0 + 0.336 453 951 488;
  • 59) 0.336 453 951 488 × 2 = 0 + 0.672 907 902 976;
  • 60) 0.672 907 902 976 × 2 = 1 + 0.345 815 805 952;
  • 61) 0.345 815 805 952 × 2 = 0 + 0.691 631 611 904;
  • 62) 0.691 631 611 904 × 2 = 1 + 0.383 263 223 808;
  • 63) 0.383 263 223 808 × 2 = 0 + 0.766 526 447 616;
  • 64) 0.766 526 447 616 × 2 = 1 + 0.533 052 895 232;

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:


0.000 282 005 752(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101(2)

6. Positive number before normalization:

0.000 282 005 752(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 752(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101(2) × 20 =

1.0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101 =

0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101


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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101


Decimal number -0.000 282 005 752 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0111 1011 0110 1100 0111 1000 1001 0101

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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