-0.000 282 005 864 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 864 2(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
-0.000 282 005 864 2(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:

|-0.000 282 005 864 2| = 0.000 282 005 864 2


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: 0.000 282 005 864 2.

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) 0.000 282 005 864 2 × 2 = 0 + 0.000 564 011 728 4;
  • 2) 0.000 564 011 728 4 × 2 = 0 + 0.001 128 023 456 8;
  • 3) 0.001 128 023 456 8 × 2 = 0 + 0.002 256 046 913 6;
  • 4) 0.002 256 046 913 6 × 2 = 0 + 0.004 512 093 827 2;
  • 5) 0.004 512 093 827 2 × 2 = 0 + 0.009 024 187 654 4;
  • 6) 0.009 024 187 654 4 × 2 = 0 + 0.018 048 375 308 8;
  • 7) 0.018 048 375 308 8 × 2 = 0 + 0.036 096 750 617 6;
  • 8) 0.036 096 750 617 6 × 2 = 0 + 0.072 193 501 235 2;
  • 9) 0.072 193 501 235 2 × 2 = 0 + 0.144 387 002 470 4;
  • 10) 0.144 387 002 470 4 × 2 = 0 + 0.288 774 004 940 8;
  • 11) 0.288 774 004 940 8 × 2 = 0 + 0.577 548 009 881 6;
  • 12) 0.577 548 009 881 6 × 2 = 1 + 0.155 096 019 763 2;
  • 13) 0.155 096 019 763 2 × 2 = 0 + 0.310 192 039 526 4;
  • 14) 0.310 192 039 526 4 × 2 = 0 + 0.620 384 079 052 8;
  • 15) 0.620 384 079 052 8 × 2 = 1 + 0.240 768 158 105 6;
  • 16) 0.240 768 158 105 6 × 2 = 0 + 0.481 536 316 211 2;
  • 17) 0.481 536 316 211 2 × 2 = 0 + 0.963 072 632 422 4;
  • 18) 0.963 072 632 422 4 × 2 = 1 + 0.926 145 264 844 8;
  • 19) 0.926 145 264 844 8 × 2 = 1 + 0.852 290 529 689 6;
  • 20) 0.852 290 529 689 6 × 2 = 1 + 0.704 581 059 379 2;
  • 21) 0.704 581 059 379 2 × 2 = 1 + 0.409 162 118 758 4;
  • 22) 0.409 162 118 758 4 × 2 = 0 + 0.818 324 237 516 8;
  • 23) 0.818 324 237 516 8 × 2 = 1 + 0.636 648 475 033 6;
  • 24) 0.636 648 475 033 6 × 2 = 1 + 0.273 296 950 067 2;
  • 25) 0.273 296 950 067 2 × 2 = 0 + 0.546 593 900 134 4;
  • 26) 0.546 593 900 134 4 × 2 = 1 + 0.093 187 800 268 8;
  • 27) 0.093 187 800 268 8 × 2 = 0 + 0.186 375 600 537 6;
  • 28) 0.186 375 600 537 6 × 2 = 0 + 0.372 751 201 075 2;
  • 29) 0.372 751 201 075 2 × 2 = 0 + 0.745 502 402 150 4;
  • 30) 0.745 502 402 150 4 × 2 = 1 + 0.491 004 804 300 8;
  • 31) 0.491 004 804 300 8 × 2 = 0 + 0.982 009 608 601 6;
  • 32) 0.982 009 608 601 6 × 2 = 1 + 0.964 019 217 203 2;
  • 33) 0.964 019 217 203 2 × 2 = 1 + 0.928 038 434 406 4;
  • 34) 0.928 038 434 406 4 × 2 = 1 + 0.856 076 868 812 8;
  • 35) 0.856 076 868 812 8 × 2 = 1 + 0.712 153 737 625 6;
  • 36) 0.712 153 737 625 6 × 2 = 1 + 0.424 307 475 251 2;
  • 37) 0.424 307 475 251 2 × 2 = 0 + 0.848 614 950 502 4;
  • 38) 0.848 614 950 502 4 × 2 = 1 + 0.697 229 901 004 8;
  • 39) 0.697 229 901 004 8 × 2 = 1 + 0.394 459 802 009 6;
  • 40) 0.394 459 802 009 6 × 2 = 0 + 0.788 919 604 019 2;
  • 41) 0.788 919 604 019 2 × 2 = 1 + 0.577 839 208 038 4;
  • 42) 0.577 839 208 038 4 × 2 = 1 + 0.155 678 416 076 8;
  • 43) 0.155 678 416 076 8 × 2 = 0 + 0.311 356 832 153 6;
  • 44) 0.311 356 832 153 6 × 2 = 0 + 0.622 713 664 307 2;
  • 45) 0.622 713 664 307 2 × 2 = 1 + 0.245 427 328 614 4;
  • 46) 0.245 427 328 614 4 × 2 = 0 + 0.490 854 657 228 8;
  • 47) 0.490 854 657 228 8 × 2 = 0 + 0.981 709 314 457 6;
  • 48) 0.981 709 314 457 6 × 2 = 1 + 0.963 418 628 915 2;
  • 49) 0.963 418 628 915 2 × 2 = 1 + 0.926 837 257 830 4;
  • 50) 0.926 837 257 830 4 × 2 = 1 + 0.853 674 515 660 8;
  • 51) 0.853 674 515 660 8 × 2 = 1 + 0.707 349 031 321 6;
  • 52) 0.707 349 031 321 6 × 2 = 1 + 0.414 698 062 643 2;
  • 53) 0.414 698 062 643 2 × 2 = 0 + 0.829 396 125 286 4;
  • 54) 0.829 396 125 286 4 × 2 = 1 + 0.658 792 250 572 8;
  • 55) 0.658 792 250 572 8 × 2 = 1 + 0.317 584 501 145 6;
  • 56) 0.317 584 501 145 6 × 2 = 0 + 0.635 169 002 291 2;
  • 57) 0.635 169 002 291 2 × 2 = 1 + 0.270 338 004 582 4;
  • 58) 0.270 338 004 582 4 × 2 = 0 + 0.540 676 009 164 8;
  • 59) 0.540 676 009 164 8 × 2 = 1 + 0.081 352 018 329 6;
  • 60) 0.081 352 018 329 6 × 2 = 0 + 0.162 704 036 659 2;
  • 61) 0.162 704 036 659 2 × 2 = 0 + 0.325 408 073 318 4;
  • 62) 0.325 408 073 318 4 × 2 = 0 + 0.650 816 146 636 8;
  • 63) 0.650 816 146 636 8 × 2 = 1 + 0.301 632 293 273 6;
  • 64) 0.301 632 293 273 6 × 2 = 0 + 0.603 264 586 547 2;

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:


0.000 282 005 864 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0110 1100 1001 1111 0110 1010 0010(2)

6. Positive number before normalization:

0.000 282 005 864 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0110 1100 1001 1111 0110 1010 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:


0.000 282 005 864 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0110 1100 1001 1111 0110 1010 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0110 1100 1001 1111 0110 1010 0010(2) × 20 =

1.0010 0111 1011 0100 0101 1111 0110 1100 1001 1111 0110 1010 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 1111 0110 1100 1001 1111 0110 1010 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 1111 0110 1100 1001 1111 0110 1010 0010 =

0010 0111 1011 0100 0101 1111 0110 1100 1001 1111 0110 1010 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 1111 0110 1100 1001 1111 0110 1010 0010


Decimal number -0.000 282 005 864 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 0110 1100 1001 1111 0110 1010 0010

Share

» Convert decimal -0.000 282 005 865 8 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