-0.000 282 005 892 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 892(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
-0.000 282 005 892(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:

|-0.000 282 005 892| = 0.000 282 005 892


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: 0.000 282 005 892.

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) 0.000 282 005 892 × 2 = 0 + 0.000 564 011 784;
  • 2) 0.000 564 011 784 × 2 = 0 + 0.001 128 023 568;
  • 3) 0.001 128 023 568 × 2 = 0 + 0.002 256 047 136;
  • 4) 0.002 256 047 136 × 2 = 0 + 0.004 512 094 272;
  • 5) 0.004 512 094 272 × 2 = 0 + 0.009 024 188 544;
  • 6) 0.009 024 188 544 × 2 = 0 + 0.018 048 377 088;
  • 7) 0.018 048 377 088 × 2 = 0 + 0.036 096 754 176;
  • 8) 0.036 096 754 176 × 2 = 0 + 0.072 193 508 352;
  • 9) 0.072 193 508 352 × 2 = 0 + 0.144 387 016 704;
  • 10) 0.144 387 016 704 × 2 = 0 + 0.288 774 033 408;
  • 11) 0.288 774 033 408 × 2 = 0 + 0.577 548 066 816;
  • 12) 0.577 548 066 816 × 2 = 1 + 0.155 096 133 632;
  • 13) 0.155 096 133 632 × 2 = 0 + 0.310 192 267 264;
  • 14) 0.310 192 267 264 × 2 = 0 + 0.620 384 534 528;
  • 15) 0.620 384 534 528 × 2 = 1 + 0.240 769 069 056;
  • 16) 0.240 769 069 056 × 2 = 0 + 0.481 538 138 112;
  • 17) 0.481 538 138 112 × 2 = 0 + 0.963 076 276 224;
  • 18) 0.963 076 276 224 × 2 = 1 + 0.926 152 552 448;
  • 19) 0.926 152 552 448 × 2 = 1 + 0.852 305 104 896;
  • 20) 0.852 305 104 896 × 2 = 1 + 0.704 610 209 792;
  • 21) 0.704 610 209 792 × 2 = 1 + 0.409 220 419 584;
  • 22) 0.409 220 419 584 × 2 = 0 + 0.818 440 839 168;
  • 23) 0.818 440 839 168 × 2 = 1 + 0.636 881 678 336;
  • 24) 0.636 881 678 336 × 2 = 1 + 0.273 763 356 672;
  • 25) 0.273 763 356 672 × 2 = 0 + 0.547 526 713 344;
  • 26) 0.547 526 713 344 × 2 = 1 + 0.095 053 426 688;
  • 27) 0.095 053 426 688 × 2 = 0 + 0.190 106 853 376;
  • 28) 0.190 106 853 376 × 2 = 0 + 0.380 213 706 752;
  • 29) 0.380 213 706 752 × 2 = 0 + 0.760 427 413 504;
  • 30) 0.760 427 413 504 × 2 = 1 + 0.520 854 827 008;
  • 31) 0.520 854 827 008 × 2 = 1 + 0.041 709 654 016;
  • 32) 0.041 709 654 016 × 2 = 0 + 0.083 419 308 032;
  • 33) 0.083 419 308 032 × 2 = 0 + 0.166 838 616 064;
  • 34) 0.166 838 616 064 × 2 = 0 + 0.333 677 232 128;
  • 35) 0.333 677 232 128 × 2 = 0 + 0.667 354 464 256;
  • 36) 0.667 354 464 256 × 2 = 1 + 0.334 708 928 512;
  • 37) 0.334 708 928 512 × 2 = 0 + 0.669 417 857 024;
  • 38) 0.669 417 857 024 × 2 = 1 + 0.338 835 714 048;
  • 39) 0.338 835 714 048 × 2 = 0 + 0.677 671 428 096;
  • 40) 0.677 671 428 096 × 2 = 1 + 0.355 342 856 192;
  • 41) 0.355 342 856 192 × 2 = 0 + 0.710 685 712 384;
  • 42) 0.710 685 712 384 × 2 = 1 + 0.421 371 424 768;
  • 43) 0.421 371 424 768 × 2 = 0 + 0.842 742 849 536;
  • 44) 0.842 742 849 536 × 2 = 1 + 0.685 485 699 072;
  • 45) 0.685 485 699 072 × 2 = 1 + 0.370 971 398 144;
  • 46) 0.370 971 398 144 × 2 = 0 + 0.741 942 796 288;
  • 47) 0.741 942 796 288 × 2 = 1 + 0.483 885 592 576;
  • 48) 0.483 885 592 576 × 2 = 0 + 0.967 771 185 152;
  • 49) 0.967 771 185 152 × 2 = 1 + 0.935 542 370 304;
  • 50) 0.935 542 370 304 × 2 = 1 + 0.871 084 740 608;
  • 51) 0.871 084 740 608 × 2 = 1 + 0.742 169 481 216;
  • 52) 0.742 169 481 216 × 2 = 1 + 0.484 338 962 432;
  • 53) 0.484 338 962 432 × 2 = 0 + 0.968 677 924 864;
  • 54) 0.968 677 924 864 × 2 = 1 + 0.937 355 849 728;
  • 55) 0.937 355 849 728 × 2 = 1 + 0.874 711 699 456;
  • 56) 0.874 711 699 456 × 2 = 1 + 0.749 423 398 912;
  • 57) 0.749 423 398 912 × 2 = 1 + 0.498 846 797 824;
  • 58) 0.498 846 797 824 × 2 = 0 + 0.997 693 595 648;
  • 59) 0.997 693 595 648 × 2 = 1 + 0.995 387 191 296;
  • 60) 0.995 387 191 296 × 2 = 1 + 0.990 774 382 592;
  • 61) 0.990 774 382 592 × 2 = 1 + 0.981 548 765 184;
  • 62) 0.981 548 765 184 × 2 = 1 + 0.963 097 530 368;
  • 63) 0.963 097 530 368 × 2 = 1 + 0.926 195 060 736;
  • 64) 0.926 195 060 736 × 2 = 1 + 0.852 390 121 472;

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:


0.000 282 005 892(10) =

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

6. Positive number before normalization:

0.000 282 005 892(10) =

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


0.000 282 005 892(10) =

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

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

1.0010 0111 1011 0100 0110 0001 0101 0101 1010 1111 0111 1011 1111(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 0101 0101 1010 1111 0111 1011 1111


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 0101 0101 1010 1111 0111 1011 1111 =

0010 0111 1011 0100 0110 0001 0101 0101 1010 1111 0111 1011 1111


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 0101 0101 1010 1111 0111 1011 1111


Decimal number -0.000 282 005 892 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0101 0101 1010 1111 0111 1011 1111

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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