-0.000 282 005 888 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 888(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
-0.000 282 005 888(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:

|-0.000 282 005 888| = 0.000 282 005 888


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: 0.000 282 005 888.

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) 0.000 282 005 888 × 2 = 0 + 0.000 564 011 776;
  • 2) 0.000 564 011 776 × 2 = 0 + 0.001 128 023 552;
  • 3) 0.001 128 023 552 × 2 = 0 + 0.002 256 047 104;
  • 4) 0.002 256 047 104 × 2 = 0 + 0.004 512 094 208;
  • 5) 0.004 512 094 208 × 2 = 0 + 0.009 024 188 416;
  • 6) 0.009 024 188 416 × 2 = 0 + 0.018 048 376 832;
  • 7) 0.018 048 376 832 × 2 = 0 + 0.036 096 753 664;
  • 8) 0.036 096 753 664 × 2 = 0 + 0.072 193 507 328;
  • 9) 0.072 193 507 328 × 2 = 0 + 0.144 387 014 656;
  • 10) 0.144 387 014 656 × 2 = 0 + 0.288 774 029 312;
  • 11) 0.288 774 029 312 × 2 = 0 + 0.577 548 058 624;
  • 12) 0.577 548 058 624 × 2 = 1 + 0.155 096 117 248;
  • 13) 0.155 096 117 248 × 2 = 0 + 0.310 192 234 496;
  • 14) 0.310 192 234 496 × 2 = 0 + 0.620 384 468 992;
  • 15) 0.620 384 468 992 × 2 = 1 + 0.240 768 937 984;
  • 16) 0.240 768 937 984 × 2 = 0 + 0.481 537 875 968;
  • 17) 0.481 537 875 968 × 2 = 0 + 0.963 075 751 936;
  • 18) 0.963 075 751 936 × 2 = 1 + 0.926 151 503 872;
  • 19) 0.926 151 503 872 × 2 = 1 + 0.852 303 007 744;
  • 20) 0.852 303 007 744 × 2 = 1 + 0.704 606 015 488;
  • 21) 0.704 606 015 488 × 2 = 1 + 0.409 212 030 976;
  • 22) 0.409 212 030 976 × 2 = 0 + 0.818 424 061 952;
  • 23) 0.818 424 061 952 × 2 = 1 + 0.636 848 123 904;
  • 24) 0.636 848 123 904 × 2 = 1 + 0.273 696 247 808;
  • 25) 0.273 696 247 808 × 2 = 0 + 0.547 392 495 616;
  • 26) 0.547 392 495 616 × 2 = 1 + 0.094 784 991 232;
  • 27) 0.094 784 991 232 × 2 = 0 + 0.189 569 982 464;
  • 28) 0.189 569 982 464 × 2 = 0 + 0.379 139 964 928;
  • 29) 0.379 139 964 928 × 2 = 0 + 0.758 279 929 856;
  • 30) 0.758 279 929 856 × 2 = 1 + 0.516 559 859 712;
  • 31) 0.516 559 859 712 × 2 = 1 + 0.033 119 719 424;
  • 32) 0.033 119 719 424 × 2 = 0 + 0.066 239 438 848;
  • 33) 0.066 239 438 848 × 2 = 0 + 0.132 478 877 696;
  • 34) 0.132 478 877 696 × 2 = 0 + 0.264 957 755 392;
  • 35) 0.264 957 755 392 × 2 = 0 + 0.529 915 510 784;
  • 36) 0.529 915 510 784 × 2 = 1 + 0.059 831 021 568;
  • 37) 0.059 831 021 568 × 2 = 0 + 0.119 662 043 136;
  • 38) 0.119 662 043 136 × 2 = 0 + 0.239 324 086 272;
  • 39) 0.239 324 086 272 × 2 = 0 + 0.478 648 172 544;
  • 40) 0.478 648 172 544 × 2 = 0 + 0.957 296 345 088;
  • 41) 0.957 296 345 088 × 2 = 1 + 0.914 592 690 176;
  • 42) 0.914 592 690 176 × 2 = 1 + 0.829 185 380 352;
  • 43) 0.829 185 380 352 × 2 = 1 + 0.658 370 760 704;
  • 44) 0.658 370 760 704 × 2 = 1 + 0.316 741 521 408;
  • 45) 0.316 741 521 408 × 2 = 0 + 0.633 483 042 816;
  • 46) 0.633 483 042 816 × 2 = 1 + 0.266 966 085 632;
  • 47) 0.266 966 085 632 × 2 = 0 + 0.533 932 171 264;
  • 48) 0.533 932 171 264 × 2 = 1 + 0.067 864 342 528;
  • 49) 0.067 864 342 528 × 2 = 0 + 0.135 728 685 056;
  • 50) 0.135 728 685 056 × 2 = 0 + 0.271 457 370 112;
  • 51) 0.271 457 370 112 × 2 = 0 + 0.542 914 740 224;
  • 52) 0.542 914 740 224 × 2 = 1 + 0.085 829 480 448;
  • 53) 0.085 829 480 448 × 2 = 0 + 0.171 658 960 896;
  • 54) 0.171 658 960 896 × 2 = 0 + 0.343 317 921 792;
  • 55) 0.343 317 921 792 × 2 = 0 + 0.686 635 843 584;
  • 56) 0.686 635 843 584 × 2 = 1 + 0.373 271 687 168;
  • 57) 0.373 271 687 168 × 2 = 0 + 0.746 543 374 336;
  • 58) 0.746 543 374 336 × 2 = 1 + 0.493 086 748 672;
  • 59) 0.493 086 748 672 × 2 = 0 + 0.986 173 497 344;
  • 60) 0.986 173 497 344 × 2 = 1 + 0.972 346 994 688;
  • 61) 0.972 346 994 688 × 2 = 1 + 0.944 693 989 376;
  • 62) 0.944 693 989 376 × 2 = 1 + 0.889 387 978 752;
  • 63) 0.889 387 978 752 × 2 = 1 + 0.778 775 957 504;
  • 64) 0.778 775 957 504 × 2 = 1 + 0.557 551 915 008;

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:


0.000 282 005 888(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 1111 0101 0001 0001 0101 1111(2)

6. Positive number before normalization:

0.000 282 005 888(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 1111 0101 0001 0001 0101 1111(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:


0.000 282 005 888(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 1111 0101 0001 0001 0101 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 1111 0101 0001 0001 0101 1111(2) × 20 =

1.0010 0111 1011 0100 0110 0001 0000 1111 0101 0001 0001 0101 1111(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 0110 0001 0000 1111 0101 0001 0001 0101 1111


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 0110 0001 0000 1111 0101 0001 0001 0101 1111 =

0010 0111 1011 0100 0110 0001 0000 1111 0101 0001 0001 0101 1111


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 0110 0001 0000 1111 0101 0001 0001 0101 1111


Decimal number -0.000 282 005 888 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0000 1111 0101 0001 0001 0101 1111

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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