-0.000 282 005 898 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 898(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
-0.000 282 005 898(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:

|-0.000 282 005 898| = 0.000 282 005 898


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: 0.000 282 005 898.

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) 0.000 282 005 898 × 2 = 0 + 0.000 564 011 796;
  • 2) 0.000 564 011 796 × 2 = 0 + 0.001 128 023 592;
  • 3) 0.001 128 023 592 × 2 = 0 + 0.002 256 047 184;
  • 4) 0.002 256 047 184 × 2 = 0 + 0.004 512 094 368;
  • 5) 0.004 512 094 368 × 2 = 0 + 0.009 024 188 736;
  • 6) 0.009 024 188 736 × 2 = 0 + 0.018 048 377 472;
  • 7) 0.018 048 377 472 × 2 = 0 + 0.036 096 754 944;
  • 8) 0.036 096 754 944 × 2 = 0 + 0.072 193 509 888;
  • 9) 0.072 193 509 888 × 2 = 0 + 0.144 387 019 776;
  • 10) 0.144 387 019 776 × 2 = 0 + 0.288 774 039 552;
  • 11) 0.288 774 039 552 × 2 = 0 + 0.577 548 079 104;
  • 12) 0.577 548 079 104 × 2 = 1 + 0.155 096 158 208;
  • 13) 0.155 096 158 208 × 2 = 0 + 0.310 192 316 416;
  • 14) 0.310 192 316 416 × 2 = 0 + 0.620 384 632 832;
  • 15) 0.620 384 632 832 × 2 = 1 + 0.240 769 265 664;
  • 16) 0.240 769 265 664 × 2 = 0 + 0.481 538 531 328;
  • 17) 0.481 538 531 328 × 2 = 0 + 0.963 077 062 656;
  • 18) 0.963 077 062 656 × 2 = 1 + 0.926 154 125 312;
  • 19) 0.926 154 125 312 × 2 = 1 + 0.852 308 250 624;
  • 20) 0.852 308 250 624 × 2 = 1 + 0.704 616 501 248;
  • 21) 0.704 616 501 248 × 2 = 1 + 0.409 233 002 496;
  • 22) 0.409 233 002 496 × 2 = 0 + 0.818 466 004 992;
  • 23) 0.818 466 004 992 × 2 = 1 + 0.636 932 009 984;
  • 24) 0.636 932 009 984 × 2 = 1 + 0.273 864 019 968;
  • 25) 0.273 864 019 968 × 2 = 0 + 0.547 728 039 936;
  • 26) 0.547 728 039 936 × 2 = 1 + 0.095 456 079 872;
  • 27) 0.095 456 079 872 × 2 = 0 + 0.190 912 159 744;
  • 28) 0.190 912 159 744 × 2 = 0 + 0.381 824 319 488;
  • 29) 0.381 824 319 488 × 2 = 0 + 0.763 648 638 976;
  • 30) 0.763 648 638 976 × 2 = 1 + 0.527 297 277 952;
  • 31) 0.527 297 277 952 × 2 = 1 + 0.054 594 555 904;
  • 32) 0.054 594 555 904 × 2 = 0 + 0.109 189 111 808;
  • 33) 0.109 189 111 808 × 2 = 0 + 0.218 378 223 616;
  • 34) 0.218 378 223 616 × 2 = 0 + 0.436 756 447 232;
  • 35) 0.436 756 447 232 × 2 = 0 + 0.873 512 894 464;
  • 36) 0.873 512 894 464 × 2 = 1 + 0.747 025 788 928;
  • 37) 0.747 025 788 928 × 2 = 1 + 0.494 051 577 856;
  • 38) 0.494 051 577 856 × 2 = 0 + 0.988 103 155 712;
  • 39) 0.988 103 155 712 × 2 = 1 + 0.976 206 311 424;
  • 40) 0.976 206 311 424 × 2 = 1 + 0.952 412 622 848;
  • 41) 0.952 412 622 848 × 2 = 1 + 0.904 825 245 696;
  • 42) 0.904 825 245 696 × 2 = 1 + 0.809 650 491 392;
  • 43) 0.809 650 491 392 × 2 = 1 + 0.619 300 982 784;
  • 44) 0.619 300 982 784 × 2 = 1 + 0.238 601 965 568;
  • 45) 0.238 601 965 568 × 2 = 0 + 0.477 203 931 136;
  • 46) 0.477 203 931 136 × 2 = 0 + 0.954 407 862 272;
  • 47) 0.954 407 862 272 × 2 = 1 + 0.908 815 724 544;
  • 48) 0.908 815 724 544 × 2 = 1 + 0.817 631 449 088;
  • 49) 0.817 631 449 088 × 2 = 1 + 0.635 262 898 176;
  • 50) 0.635 262 898 176 × 2 = 1 + 0.270 525 796 352;
  • 51) 0.270 525 796 352 × 2 = 0 + 0.541 051 592 704;
  • 52) 0.541 051 592 704 × 2 = 1 + 0.082 103 185 408;
  • 53) 0.082 103 185 408 × 2 = 0 + 0.164 206 370 816;
  • 54) 0.164 206 370 816 × 2 = 0 + 0.328 412 741 632;
  • 55) 0.328 412 741 632 × 2 = 0 + 0.656 825 483 264;
  • 56) 0.656 825 483 264 × 2 = 1 + 0.313 650 966 528;
  • 57) 0.313 650 966 528 × 2 = 0 + 0.627 301 933 056;
  • 58) 0.627 301 933 056 × 2 = 1 + 0.254 603 866 112;
  • 59) 0.254 603 866 112 × 2 = 0 + 0.509 207 732 224;
  • 60) 0.509 207 732 224 × 2 = 1 + 0.018 415 464 448;
  • 61) 0.018 415 464 448 × 2 = 0 + 0.036 830 928 896;
  • 62) 0.036 830 928 896 × 2 = 0 + 0.073 661 857 792;
  • 63) 0.073 661 857 792 × 2 = 0 + 0.147 323 715 584;
  • 64) 0.147 323 715 584 × 2 = 0 + 0.294 647 431 168;

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:


0.000 282 005 898(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1011 1111 0011 1101 0001 0101 0000(2)

6. Positive number before normalization:

0.000 282 005 898(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1011 1111 0011 1101 0001 0101 0000(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:


0.000 282 005 898(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1011 1111 0011 1101 0001 0101 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1011 1111 0011 1101 0001 0101 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1011 1111 0011 1101 0001 0101 0000(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 1011 1111 0011 1101 0001 0101 0000


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 1011 1111 0011 1101 0001 0101 0000 =

0010 0111 1011 0100 0110 0001 1011 1111 0011 1101 0001 0101 0000


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 1011 1111 0011 1101 0001 0101 0000


Decimal number -0.000 282 005 898 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 1011 1111 0011 1101 0001 0101 0000

Share

» Convert decimal -0.000 282 005 832 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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