-0.000 282 005 910 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 910 5(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
-0.000 282 005 910 5(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:

|-0.000 282 005 910 5| = 0.000 282 005 910 5


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: 0.000 282 005 910 5.

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) 0.000 282 005 910 5 × 2 = 0 + 0.000 564 011 821;
  • 2) 0.000 564 011 821 × 2 = 0 + 0.001 128 023 642;
  • 3) 0.001 128 023 642 × 2 = 0 + 0.002 256 047 284;
  • 4) 0.002 256 047 284 × 2 = 0 + 0.004 512 094 568;
  • 5) 0.004 512 094 568 × 2 = 0 + 0.009 024 189 136;
  • 6) 0.009 024 189 136 × 2 = 0 + 0.018 048 378 272;
  • 7) 0.018 048 378 272 × 2 = 0 + 0.036 096 756 544;
  • 8) 0.036 096 756 544 × 2 = 0 + 0.072 193 513 088;
  • 9) 0.072 193 513 088 × 2 = 0 + 0.144 387 026 176;
  • 10) 0.144 387 026 176 × 2 = 0 + 0.288 774 052 352;
  • 11) 0.288 774 052 352 × 2 = 0 + 0.577 548 104 704;
  • 12) 0.577 548 104 704 × 2 = 1 + 0.155 096 209 408;
  • 13) 0.155 096 209 408 × 2 = 0 + 0.310 192 418 816;
  • 14) 0.310 192 418 816 × 2 = 0 + 0.620 384 837 632;
  • 15) 0.620 384 837 632 × 2 = 1 + 0.240 769 675 264;
  • 16) 0.240 769 675 264 × 2 = 0 + 0.481 539 350 528;
  • 17) 0.481 539 350 528 × 2 = 0 + 0.963 078 701 056;
  • 18) 0.963 078 701 056 × 2 = 1 + 0.926 157 402 112;
  • 19) 0.926 157 402 112 × 2 = 1 + 0.852 314 804 224;
  • 20) 0.852 314 804 224 × 2 = 1 + 0.704 629 608 448;
  • 21) 0.704 629 608 448 × 2 = 1 + 0.409 259 216 896;
  • 22) 0.409 259 216 896 × 2 = 0 + 0.818 518 433 792;
  • 23) 0.818 518 433 792 × 2 = 1 + 0.637 036 867 584;
  • 24) 0.637 036 867 584 × 2 = 1 + 0.274 073 735 168;
  • 25) 0.274 073 735 168 × 2 = 0 + 0.548 147 470 336;
  • 26) 0.548 147 470 336 × 2 = 1 + 0.096 294 940 672;
  • 27) 0.096 294 940 672 × 2 = 0 + 0.192 589 881 344;
  • 28) 0.192 589 881 344 × 2 = 0 + 0.385 179 762 688;
  • 29) 0.385 179 762 688 × 2 = 0 + 0.770 359 525 376;
  • 30) 0.770 359 525 376 × 2 = 1 + 0.540 719 050 752;
  • 31) 0.540 719 050 752 × 2 = 1 + 0.081 438 101 504;
  • 32) 0.081 438 101 504 × 2 = 0 + 0.162 876 203 008;
  • 33) 0.162 876 203 008 × 2 = 0 + 0.325 752 406 016;
  • 34) 0.325 752 406 016 × 2 = 0 + 0.651 504 812 032;
  • 35) 0.651 504 812 032 × 2 = 1 + 0.303 009 624 064;
  • 36) 0.303 009 624 064 × 2 = 0 + 0.606 019 248 128;
  • 37) 0.606 019 248 128 × 2 = 1 + 0.212 038 496 256;
  • 38) 0.212 038 496 256 × 2 = 0 + 0.424 076 992 512;
  • 39) 0.424 076 992 512 × 2 = 0 + 0.848 153 985 024;
  • 40) 0.848 153 985 024 × 2 = 1 + 0.696 307 970 048;
  • 41) 0.696 307 970 048 × 2 = 1 + 0.392 615 940 096;
  • 42) 0.392 615 940 096 × 2 = 0 + 0.785 231 880 192;
  • 43) 0.785 231 880 192 × 2 = 1 + 0.570 463 760 384;
  • 44) 0.570 463 760 384 × 2 = 1 + 0.140 927 520 768;
  • 45) 0.140 927 520 768 × 2 = 0 + 0.281 855 041 536;
  • 46) 0.281 855 041 536 × 2 = 0 + 0.563 710 083 072;
  • 47) 0.563 710 083 072 × 2 = 1 + 0.127 420 166 144;
  • 48) 0.127 420 166 144 × 2 = 0 + 0.254 840 332 288;
  • 49) 0.254 840 332 288 × 2 = 0 + 0.509 680 664 576;
  • 50) 0.509 680 664 576 × 2 = 1 + 0.019 361 329 152;
  • 51) 0.019 361 329 152 × 2 = 0 + 0.038 722 658 304;
  • 52) 0.038 722 658 304 × 2 = 0 + 0.077 445 316 608;
  • 53) 0.077 445 316 608 × 2 = 0 + 0.154 890 633 216;
  • 54) 0.154 890 633 216 × 2 = 0 + 0.309 781 266 432;
  • 55) 0.309 781 266 432 × 2 = 0 + 0.619 562 532 864;
  • 56) 0.619 562 532 864 × 2 = 1 + 0.239 125 065 728;
  • 57) 0.239 125 065 728 × 2 = 0 + 0.478 250 131 456;
  • 58) 0.478 250 131 456 × 2 = 0 + 0.956 500 262 912;
  • 59) 0.956 500 262 912 × 2 = 1 + 0.913 000 525 824;
  • 60) 0.913 000 525 824 × 2 = 1 + 0.826 001 051 648;
  • 61) 0.826 001 051 648 × 2 = 1 + 0.652 002 103 296;
  • 62) 0.652 002 103 296 × 2 = 1 + 0.304 004 206 592;
  • 63) 0.304 004 206 592 × 2 = 0 + 0.608 008 413 184;
  • 64) 0.608 008 413 184 × 2 = 1 + 0.216 016 826 368;

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:


0.000 282 005 910 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1011 0010 0100 0001 0011 1101(2)

6. Positive number before normalization:

0.000 282 005 910 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1011 0010 0100 0001 0011 1101(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:


0.000 282 005 910 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1011 0010 0100 0001 0011 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1011 0010 0100 0001 0011 1101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1001 1011 0010 0100 0001 0011 1101(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 0010 1001 1011 0010 0100 0001 0011 1101


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 0010 1001 1011 0010 0100 0001 0011 1101 =

0010 0111 1011 0100 0110 0010 1001 1011 0010 0100 0001 0011 1101


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 0010 1001 1011 0010 0100 0001 0011 1101


Decimal number -0.000 282 005 910 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1001 1011 0010 0100 0001 0011 1101

Share

» Convert decimal -0.000 282 005 902 2 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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