-0.000 282 005 869 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 869(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
-0.000 282 005 869(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:

|-0.000 282 005 869| = 0.000 282 005 869


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: 0.000 282 005 869.

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) 0.000 282 005 869 × 2 = 0 + 0.000 564 011 738;
  • 2) 0.000 564 011 738 × 2 = 0 + 0.001 128 023 476;
  • 3) 0.001 128 023 476 × 2 = 0 + 0.002 256 046 952;
  • 4) 0.002 256 046 952 × 2 = 0 + 0.004 512 093 904;
  • 5) 0.004 512 093 904 × 2 = 0 + 0.009 024 187 808;
  • 6) 0.009 024 187 808 × 2 = 0 + 0.018 048 375 616;
  • 7) 0.018 048 375 616 × 2 = 0 + 0.036 096 751 232;
  • 8) 0.036 096 751 232 × 2 = 0 + 0.072 193 502 464;
  • 9) 0.072 193 502 464 × 2 = 0 + 0.144 387 004 928;
  • 10) 0.144 387 004 928 × 2 = 0 + 0.288 774 009 856;
  • 11) 0.288 774 009 856 × 2 = 0 + 0.577 548 019 712;
  • 12) 0.577 548 019 712 × 2 = 1 + 0.155 096 039 424;
  • 13) 0.155 096 039 424 × 2 = 0 + 0.310 192 078 848;
  • 14) 0.310 192 078 848 × 2 = 0 + 0.620 384 157 696;
  • 15) 0.620 384 157 696 × 2 = 1 + 0.240 768 315 392;
  • 16) 0.240 768 315 392 × 2 = 0 + 0.481 536 630 784;
  • 17) 0.481 536 630 784 × 2 = 0 + 0.963 073 261 568;
  • 18) 0.963 073 261 568 × 2 = 1 + 0.926 146 523 136;
  • 19) 0.926 146 523 136 × 2 = 1 + 0.852 293 046 272;
  • 20) 0.852 293 046 272 × 2 = 1 + 0.704 586 092 544;
  • 21) 0.704 586 092 544 × 2 = 1 + 0.409 172 185 088;
  • 22) 0.409 172 185 088 × 2 = 0 + 0.818 344 370 176;
  • 23) 0.818 344 370 176 × 2 = 1 + 0.636 688 740 352;
  • 24) 0.636 688 740 352 × 2 = 1 + 0.273 377 480 704;
  • 25) 0.273 377 480 704 × 2 = 0 + 0.546 754 961 408;
  • 26) 0.546 754 961 408 × 2 = 1 + 0.093 509 922 816;
  • 27) 0.093 509 922 816 × 2 = 0 + 0.187 019 845 632;
  • 28) 0.187 019 845 632 × 2 = 0 + 0.374 039 691 264;
  • 29) 0.374 039 691 264 × 2 = 0 + 0.748 079 382 528;
  • 30) 0.748 079 382 528 × 2 = 1 + 0.496 158 765 056;
  • 31) 0.496 158 765 056 × 2 = 0 + 0.992 317 530 112;
  • 32) 0.992 317 530 112 × 2 = 1 + 0.984 635 060 224;
  • 33) 0.984 635 060 224 × 2 = 1 + 0.969 270 120 448;
  • 34) 0.969 270 120 448 × 2 = 1 + 0.938 540 240 896;
  • 35) 0.938 540 240 896 × 2 = 1 + 0.877 080 481 792;
  • 36) 0.877 080 481 792 × 2 = 1 + 0.754 160 963 584;
  • 37) 0.754 160 963 584 × 2 = 1 + 0.508 321 927 168;
  • 38) 0.508 321 927 168 × 2 = 1 + 0.016 643 854 336;
  • 39) 0.016 643 854 336 × 2 = 0 + 0.033 287 708 672;
  • 40) 0.033 287 708 672 × 2 = 0 + 0.066 575 417 344;
  • 41) 0.066 575 417 344 × 2 = 0 + 0.133 150 834 688;
  • 42) 0.133 150 834 688 × 2 = 0 + 0.266 301 669 376;
  • 43) 0.266 301 669 376 × 2 = 0 + 0.532 603 338 752;
  • 44) 0.532 603 338 752 × 2 = 1 + 0.065 206 677 504;
  • 45) 0.065 206 677 504 × 2 = 0 + 0.130 413 355 008;
  • 46) 0.130 413 355 008 × 2 = 0 + 0.260 826 710 016;
  • 47) 0.260 826 710 016 × 2 = 0 + 0.521 653 420 032;
  • 48) 0.521 653 420 032 × 2 = 1 + 0.043 306 840 064;
  • 49) 0.043 306 840 064 × 2 = 0 + 0.086 613 680 128;
  • 50) 0.086 613 680 128 × 2 = 0 + 0.173 227 360 256;
  • 51) 0.173 227 360 256 × 2 = 0 + 0.346 454 720 512;
  • 52) 0.346 454 720 512 × 2 = 0 + 0.692 909 441 024;
  • 53) 0.692 909 441 024 × 2 = 1 + 0.385 818 882 048;
  • 54) 0.385 818 882 048 × 2 = 0 + 0.771 637 764 096;
  • 55) 0.771 637 764 096 × 2 = 1 + 0.543 275 528 192;
  • 56) 0.543 275 528 192 × 2 = 1 + 0.086 551 056 384;
  • 57) 0.086 551 056 384 × 2 = 0 + 0.173 102 112 768;
  • 58) 0.173 102 112 768 × 2 = 0 + 0.346 204 225 536;
  • 59) 0.346 204 225 536 × 2 = 0 + 0.692 408 451 072;
  • 60) 0.692 408 451 072 × 2 = 1 + 0.384 816 902 144;
  • 61) 0.384 816 902 144 × 2 = 0 + 0.769 633 804 288;
  • 62) 0.769 633 804 288 × 2 = 1 + 0.539 267 608 576;
  • 63) 0.539 267 608 576 × 2 = 1 + 0.078 535 217 152;
  • 64) 0.078 535 217 152 × 2 = 0 + 0.157 070 434 304;

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:


0.000 282 005 869(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1100 0001 0001 0000 1011 0001 0110(2)

6. Positive number before normalization:

0.000 282 005 869(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1100 0001 0001 0000 1011 0001 0110(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:


0.000 282 005 869(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1100 0001 0001 0000 1011 0001 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1100 0001 0001 0000 1011 0001 0110(2) × 20 =

1.0010 0111 1011 0100 0101 1111 1100 0001 0001 0000 1011 0001 0110(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 1111 1100 0001 0001 0000 1011 0001 0110


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 1111 1100 0001 0001 0000 1011 0001 0110 =

0010 0111 1011 0100 0101 1111 1100 0001 0001 0000 1011 0001 0110


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 1111 1100 0001 0001 0000 1011 0001 0110


Decimal number -0.000 282 005 869 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 1100 0001 0001 0000 1011 0001 0110

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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