-0.000 282 005 516 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 516(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
-0.000 282 005 516(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:

|-0.000 282 005 516| = 0.000 282 005 516


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: 0.000 282 005 516.

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) 0.000 282 005 516 × 2 = 0 + 0.000 564 011 032;
  • 2) 0.000 564 011 032 × 2 = 0 + 0.001 128 022 064;
  • 3) 0.001 128 022 064 × 2 = 0 + 0.002 256 044 128;
  • 4) 0.002 256 044 128 × 2 = 0 + 0.004 512 088 256;
  • 5) 0.004 512 088 256 × 2 = 0 + 0.009 024 176 512;
  • 6) 0.009 024 176 512 × 2 = 0 + 0.018 048 353 024;
  • 7) 0.018 048 353 024 × 2 = 0 + 0.036 096 706 048;
  • 8) 0.036 096 706 048 × 2 = 0 + 0.072 193 412 096;
  • 9) 0.072 193 412 096 × 2 = 0 + 0.144 386 824 192;
  • 10) 0.144 386 824 192 × 2 = 0 + 0.288 773 648 384;
  • 11) 0.288 773 648 384 × 2 = 0 + 0.577 547 296 768;
  • 12) 0.577 547 296 768 × 2 = 1 + 0.155 094 593 536;
  • 13) 0.155 094 593 536 × 2 = 0 + 0.310 189 187 072;
  • 14) 0.310 189 187 072 × 2 = 0 + 0.620 378 374 144;
  • 15) 0.620 378 374 144 × 2 = 1 + 0.240 756 748 288;
  • 16) 0.240 756 748 288 × 2 = 0 + 0.481 513 496 576;
  • 17) 0.481 513 496 576 × 2 = 0 + 0.963 026 993 152;
  • 18) 0.963 026 993 152 × 2 = 1 + 0.926 053 986 304;
  • 19) 0.926 053 986 304 × 2 = 1 + 0.852 107 972 608;
  • 20) 0.852 107 972 608 × 2 = 1 + 0.704 215 945 216;
  • 21) 0.704 215 945 216 × 2 = 1 + 0.408 431 890 432;
  • 22) 0.408 431 890 432 × 2 = 0 + 0.816 863 780 864;
  • 23) 0.816 863 780 864 × 2 = 1 + 0.633 727 561 728;
  • 24) 0.633 727 561 728 × 2 = 1 + 0.267 455 123 456;
  • 25) 0.267 455 123 456 × 2 = 0 + 0.534 910 246 912;
  • 26) 0.534 910 246 912 × 2 = 1 + 0.069 820 493 824;
  • 27) 0.069 820 493 824 × 2 = 0 + 0.139 640 987 648;
  • 28) 0.139 640 987 648 × 2 = 0 + 0.279 281 975 296;
  • 29) 0.279 281 975 296 × 2 = 0 + 0.558 563 950 592;
  • 30) 0.558 563 950 592 × 2 = 1 + 0.117 127 901 184;
  • 31) 0.117 127 901 184 × 2 = 0 + 0.234 255 802 368;
  • 32) 0.234 255 802 368 × 2 = 0 + 0.468 511 604 736;
  • 33) 0.468 511 604 736 × 2 = 0 + 0.937 023 209 472;
  • 34) 0.937 023 209 472 × 2 = 1 + 0.874 046 418 944;
  • 35) 0.874 046 418 944 × 2 = 1 + 0.748 092 837 888;
  • 36) 0.748 092 837 888 × 2 = 1 + 0.496 185 675 776;
  • 37) 0.496 185 675 776 × 2 = 0 + 0.992 371 351 552;
  • 38) 0.992 371 351 552 × 2 = 1 + 0.984 742 703 104;
  • 39) 0.984 742 703 104 × 2 = 1 + 0.969 485 406 208;
  • 40) 0.969 485 406 208 × 2 = 1 + 0.938 970 812 416;
  • 41) 0.938 970 812 416 × 2 = 1 + 0.877 941 624 832;
  • 42) 0.877 941 624 832 × 2 = 1 + 0.755 883 249 664;
  • 43) 0.755 883 249 664 × 2 = 1 + 0.511 766 499 328;
  • 44) 0.511 766 499 328 × 2 = 1 + 0.023 532 998 656;
  • 45) 0.023 532 998 656 × 2 = 0 + 0.047 065 997 312;
  • 46) 0.047 065 997 312 × 2 = 0 + 0.094 131 994 624;
  • 47) 0.094 131 994 624 × 2 = 0 + 0.188 263 989 248;
  • 48) 0.188 263 989 248 × 2 = 0 + 0.376 527 978 496;
  • 49) 0.376 527 978 496 × 2 = 0 + 0.753 055 956 992;
  • 50) 0.753 055 956 992 × 2 = 1 + 0.506 111 913 984;
  • 51) 0.506 111 913 984 × 2 = 1 + 0.012 223 827 968;
  • 52) 0.012 223 827 968 × 2 = 0 + 0.024 447 655 936;
  • 53) 0.024 447 655 936 × 2 = 0 + 0.048 895 311 872;
  • 54) 0.048 895 311 872 × 2 = 0 + 0.097 790 623 744;
  • 55) 0.097 790 623 744 × 2 = 0 + 0.195 581 247 488;
  • 56) 0.195 581 247 488 × 2 = 0 + 0.391 162 494 976;
  • 57) 0.391 162 494 976 × 2 = 0 + 0.782 324 989 952;
  • 58) 0.782 324 989 952 × 2 = 1 + 0.564 649 979 904;
  • 59) 0.564 649 979 904 × 2 = 1 + 0.129 299 959 808;
  • 60) 0.129 299 959 808 × 2 = 0 + 0.258 599 919 616;
  • 61) 0.258 599 919 616 × 2 = 0 + 0.517 199 839 232;
  • 62) 0.517 199 839 232 × 2 = 1 + 0.034 399 678 464;
  • 63) 0.034 399 678 464 × 2 = 0 + 0.068 799 356 928;
  • 64) 0.068 799 356 928 × 2 = 0 + 0.137 598 713 856;

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:


0.000 282 005 516(10) =

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

6. Positive number before normalization:

0.000 282 005 516(10) =

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


0.000 282 005 516(10) =

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

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

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


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 0100 0111 0111 1111 0000 0110 0000 0110 0100 =

0010 0111 1011 0100 0100 0111 0111 1111 0000 0110 0000 0110 0100


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 0100 0111 0111 1111 0000 0110 0000 0110 0100


Decimal number -0.000 282 005 516 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

Share

» Convert decimal -0.000 282 005 602 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