-0.000 282 005 914 416 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 416(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
-0.000 282 005 914 416(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:

|-0.000 282 005 914 416| = 0.000 282 005 914 416


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: 0.000 282 005 914 416.

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) 0.000 282 005 914 416 × 2 = 0 + 0.000 564 011 828 832;
  • 2) 0.000 564 011 828 832 × 2 = 0 + 0.001 128 023 657 664;
  • 3) 0.001 128 023 657 664 × 2 = 0 + 0.002 256 047 315 328;
  • 4) 0.002 256 047 315 328 × 2 = 0 + 0.004 512 094 630 656;
  • 5) 0.004 512 094 630 656 × 2 = 0 + 0.009 024 189 261 312;
  • 6) 0.009 024 189 261 312 × 2 = 0 + 0.018 048 378 522 624;
  • 7) 0.018 048 378 522 624 × 2 = 0 + 0.036 096 757 045 248;
  • 8) 0.036 096 757 045 248 × 2 = 0 + 0.072 193 514 090 496;
  • 9) 0.072 193 514 090 496 × 2 = 0 + 0.144 387 028 180 992;
  • 10) 0.144 387 028 180 992 × 2 = 0 + 0.288 774 056 361 984;
  • 11) 0.288 774 056 361 984 × 2 = 0 + 0.577 548 112 723 968;
  • 12) 0.577 548 112 723 968 × 2 = 1 + 0.155 096 225 447 936;
  • 13) 0.155 096 225 447 936 × 2 = 0 + 0.310 192 450 895 872;
  • 14) 0.310 192 450 895 872 × 2 = 0 + 0.620 384 901 791 744;
  • 15) 0.620 384 901 791 744 × 2 = 1 + 0.240 769 803 583 488;
  • 16) 0.240 769 803 583 488 × 2 = 0 + 0.481 539 607 166 976;
  • 17) 0.481 539 607 166 976 × 2 = 0 + 0.963 079 214 333 952;
  • 18) 0.963 079 214 333 952 × 2 = 1 + 0.926 158 428 667 904;
  • 19) 0.926 158 428 667 904 × 2 = 1 + 0.852 316 857 335 808;
  • 20) 0.852 316 857 335 808 × 2 = 1 + 0.704 633 714 671 616;
  • 21) 0.704 633 714 671 616 × 2 = 1 + 0.409 267 429 343 232;
  • 22) 0.409 267 429 343 232 × 2 = 0 + 0.818 534 858 686 464;
  • 23) 0.818 534 858 686 464 × 2 = 1 + 0.637 069 717 372 928;
  • 24) 0.637 069 717 372 928 × 2 = 1 + 0.274 139 434 745 856;
  • 25) 0.274 139 434 745 856 × 2 = 0 + 0.548 278 869 491 712;
  • 26) 0.548 278 869 491 712 × 2 = 1 + 0.096 557 738 983 424;
  • 27) 0.096 557 738 983 424 × 2 = 0 + 0.193 115 477 966 848;
  • 28) 0.193 115 477 966 848 × 2 = 0 + 0.386 230 955 933 696;
  • 29) 0.386 230 955 933 696 × 2 = 0 + 0.772 461 911 867 392;
  • 30) 0.772 461 911 867 392 × 2 = 1 + 0.544 923 823 734 784;
  • 31) 0.544 923 823 734 784 × 2 = 1 + 0.089 847 647 469 568;
  • 32) 0.089 847 647 469 568 × 2 = 0 + 0.179 695 294 939 136;
  • 33) 0.179 695 294 939 136 × 2 = 0 + 0.359 390 589 878 272;
  • 34) 0.359 390 589 878 272 × 2 = 0 + 0.718 781 179 756 544;
  • 35) 0.718 781 179 756 544 × 2 = 1 + 0.437 562 359 513 088;
  • 36) 0.437 562 359 513 088 × 2 = 0 + 0.875 124 719 026 176;
  • 37) 0.875 124 719 026 176 × 2 = 1 + 0.750 249 438 052 352;
  • 38) 0.750 249 438 052 352 × 2 = 1 + 0.500 498 876 104 704;
  • 39) 0.500 498 876 104 704 × 2 = 1 + 0.000 997 752 209 408;
  • 40) 0.000 997 752 209 408 × 2 = 0 + 0.001 995 504 418 816;
  • 41) 0.001 995 504 418 816 × 2 = 0 + 0.003 991 008 837 632;
  • 42) 0.003 991 008 837 632 × 2 = 0 + 0.007 982 017 675 264;
  • 43) 0.007 982 017 675 264 × 2 = 0 + 0.015 964 035 350 528;
  • 44) 0.015 964 035 350 528 × 2 = 0 + 0.031 928 070 701 056;
  • 45) 0.031 928 070 701 056 × 2 = 0 + 0.063 856 141 402 112;
  • 46) 0.063 856 141 402 112 × 2 = 0 + 0.127 712 282 804 224;
  • 47) 0.127 712 282 804 224 × 2 = 0 + 0.255 424 565 608 448;
  • 48) 0.255 424 565 608 448 × 2 = 0 + 0.510 849 131 216 896;
  • 49) 0.510 849 131 216 896 × 2 = 1 + 0.021 698 262 433 792;
  • 50) 0.021 698 262 433 792 × 2 = 0 + 0.043 396 524 867 584;
  • 51) 0.043 396 524 867 584 × 2 = 0 + 0.086 793 049 735 168;
  • 52) 0.086 793 049 735 168 × 2 = 0 + 0.173 586 099 470 336;
  • 53) 0.173 586 099 470 336 × 2 = 0 + 0.347 172 198 940 672;
  • 54) 0.347 172 198 940 672 × 2 = 0 + 0.694 344 397 881 344;
  • 55) 0.694 344 397 881 344 × 2 = 1 + 0.388 688 795 762 688;
  • 56) 0.388 688 795 762 688 × 2 = 0 + 0.777 377 591 525 376;
  • 57) 0.777 377 591 525 376 × 2 = 1 + 0.554 755 183 050 752;
  • 58) 0.554 755 183 050 752 × 2 = 1 + 0.109 510 366 101 504;
  • 59) 0.109 510 366 101 504 × 2 = 0 + 0.219 020 732 203 008;
  • 60) 0.219 020 732 203 008 × 2 = 0 + 0.438 041 464 406 016;
  • 61) 0.438 041 464 406 016 × 2 = 0 + 0.876 082 928 812 032;
  • 62) 0.876 082 928 812 032 × 2 = 1 + 0.752 165 857 624 064;
  • 63) 0.752 165 857 624 064 × 2 = 1 + 0.504 331 715 248 128;
  • 64) 0.504 331 715 248 128 × 2 = 1 + 0.008 663 430 496 256;

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:


0.000 282 005 914 416(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0000 1000 0010 1100 0111(2)

6. Positive number before normalization:

0.000 282 005 914 416(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0000 1000 0010 1100 0111(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:


0.000 282 005 914 416(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0000 1000 0010 1100 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0000 1000 0010 1100 0111(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0000 0000 1000 0010 1100 0111(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 0010 1110 0000 0000 1000 0010 1100 0111


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 0010 1110 0000 0000 1000 0010 1100 0111 =

0010 0111 1011 0100 0110 0010 1110 0000 0000 1000 0010 1100 0111


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 0010 1110 0000 0000 1000 0010 1100 0111


Decimal number -0.000 282 005 914 416 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 0000 1000 0010 1100 0111

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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