-0.000 282 005 823 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 823(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
-0.000 282 005 823(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:

|-0.000 282 005 823| = 0.000 282 005 823


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: 0.000 282 005 823.

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) 0.000 282 005 823 × 2 = 0 + 0.000 564 011 646;
  • 2) 0.000 564 011 646 × 2 = 0 + 0.001 128 023 292;
  • 3) 0.001 128 023 292 × 2 = 0 + 0.002 256 046 584;
  • 4) 0.002 256 046 584 × 2 = 0 + 0.004 512 093 168;
  • 5) 0.004 512 093 168 × 2 = 0 + 0.009 024 186 336;
  • 6) 0.009 024 186 336 × 2 = 0 + 0.018 048 372 672;
  • 7) 0.018 048 372 672 × 2 = 0 + 0.036 096 745 344;
  • 8) 0.036 096 745 344 × 2 = 0 + 0.072 193 490 688;
  • 9) 0.072 193 490 688 × 2 = 0 + 0.144 386 981 376;
  • 10) 0.144 386 981 376 × 2 = 0 + 0.288 773 962 752;
  • 11) 0.288 773 962 752 × 2 = 0 + 0.577 547 925 504;
  • 12) 0.577 547 925 504 × 2 = 1 + 0.155 095 851 008;
  • 13) 0.155 095 851 008 × 2 = 0 + 0.310 191 702 016;
  • 14) 0.310 191 702 016 × 2 = 0 + 0.620 383 404 032;
  • 15) 0.620 383 404 032 × 2 = 1 + 0.240 766 808 064;
  • 16) 0.240 766 808 064 × 2 = 0 + 0.481 533 616 128;
  • 17) 0.481 533 616 128 × 2 = 0 + 0.963 067 232 256;
  • 18) 0.963 067 232 256 × 2 = 1 + 0.926 134 464 512;
  • 19) 0.926 134 464 512 × 2 = 1 + 0.852 268 929 024;
  • 20) 0.852 268 929 024 × 2 = 1 + 0.704 537 858 048;
  • 21) 0.704 537 858 048 × 2 = 1 + 0.409 075 716 096;
  • 22) 0.409 075 716 096 × 2 = 0 + 0.818 151 432 192;
  • 23) 0.818 151 432 192 × 2 = 1 + 0.636 302 864 384;
  • 24) 0.636 302 864 384 × 2 = 1 + 0.272 605 728 768;
  • 25) 0.272 605 728 768 × 2 = 0 + 0.545 211 457 536;
  • 26) 0.545 211 457 536 × 2 = 1 + 0.090 422 915 072;
  • 27) 0.090 422 915 072 × 2 = 0 + 0.180 845 830 144;
  • 28) 0.180 845 830 144 × 2 = 0 + 0.361 691 660 288;
  • 29) 0.361 691 660 288 × 2 = 0 + 0.723 383 320 576;
  • 30) 0.723 383 320 576 × 2 = 1 + 0.446 766 641 152;
  • 31) 0.446 766 641 152 × 2 = 0 + 0.893 533 282 304;
  • 32) 0.893 533 282 304 × 2 = 1 + 0.787 066 564 608;
  • 33) 0.787 066 564 608 × 2 = 1 + 0.574 133 129 216;
  • 34) 0.574 133 129 216 × 2 = 1 + 0.148 266 258 432;
  • 35) 0.148 266 258 432 × 2 = 0 + 0.296 532 516 864;
  • 36) 0.296 532 516 864 × 2 = 0 + 0.593 065 033 728;
  • 37) 0.593 065 033 728 × 2 = 1 + 0.186 130 067 456;
  • 38) 0.186 130 067 456 × 2 = 0 + 0.372 260 134 912;
  • 39) 0.372 260 134 912 × 2 = 0 + 0.744 520 269 824;
  • 40) 0.744 520 269 824 × 2 = 1 + 0.489 040 539 648;
  • 41) 0.489 040 539 648 × 2 = 0 + 0.978 081 079 296;
  • 42) 0.978 081 079 296 × 2 = 1 + 0.956 162 158 592;
  • 43) 0.956 162 158 592 × 2 = 1 + 0.912 324 317 184;
  • 44) 0.912 324 317 184 × 2 = 1 + 0.824 648 634 368;
  • 45) 0.824 648 634 368 × 2 = 1 + 0.649 297 268 736;
  • 46) 0.649 297 268 736 × 2 = 1 + 0.298 594 537 472;
  • 47) 0.298 594 537 472 × 2 = 0 + 0.597 189 074 944;
  • 48) 0.597 189 074 944 × 2 = 1 + 0.194 378 149 888;
  • 49) 0.194 378 149 888 × 2 = 0 + 0.388 756 299 776;
  • 50) 0.388 756 299 776 × 2 = 0 + 0.777 512 599 552;
  • 51) 0.777 512 599 552 × 2 = 1 + 0.555 025 199 104;
  • 52) 0.555 025 199 104 × 2 = 1 + 0.110 050 398 208;
  • 53) 0.110 050 398 208 × 2 = 0 + 0.220 100 796 416;
  • 54) 0.220 100 796 416 × 2 = 0 + 0.440 201 592 832;
  • 55) 0.440 201 592 832 × 2 = 0 + 0.880 403 185 664;
  • 56) 0.880 403 185 664 × 2 = 1 + 0.760 806 371 328;
  • 57) 0.760 806 371 328 × 2 = 1 + 0.521 612 742 656;
  • 58) 0.521 612 742 656 × 2 = 1 + 0.043 225 485 312;
  • 59) 0.043 225 485 312 × 2 = 0 + 0.086 450 970 624;
  • 60) 0.086 450 970 624 × 2 = 0 + 0.172 901 941 248;
  • 61) 0.172 901 941 248 × 2 = 0 + 0.345 803 882 496;
  • 62) 0.345 803 882 496 × 2 = 0 + 0.691 607 764 992;
  • 63) 0.691 607 764 992 × 2 = 1 + 0.383 215 529 984;
  • 64) 0.383 215 529 984 × 2 = 0 + 0.766 431 059 968;

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:


0.000 282 005 823(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1100 1001 0111 1101 0011 0001 1100 0010(2)

6. Positive number before normalization:

0.000 282 005 823(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1100 1001 0111 1101 0011 0001 1100 0010(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:


0.000 282 005 823(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1100 1001 0111 1101 0011 0001 1100 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1100 1001 0111 1101 0011 0001 1100 0010(2) × 20 =

1.0010 0111 1011 0100 0101 1100 1001 0111 1101 0011 0001 1100 0010(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 1100 1001 0111 1101 0011 0001 1100 0010


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 1100 1001 0111 1101 0011 0001 1100 0010 =

0010 0111 1011 0100 0101 1100 1001 0111 1101 0011 0001 1100 0010


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 1100 1001 0111 1101 0011 0001 1100 0010


Decimal number -0.000 282 005 823 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1100 1001 0111 1101 0011 0001 1100 0010

Share

» Convert decimal -0.000 282 005 875 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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