-0.000 282 005 914 438 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 438(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
-0.000 282 005 914 438(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:

|-0.000 282 005 914 438| = 0.000 282 005 914 438


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: 0.000 282 005 914 438.

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) 0.000 282 005 914 438 × 2 = 0 + 0.000 564 011 828 876;
  • 2) 0.000 564 011 828 876 × 2 = 0 + 0.001 128 023 657 752;
  • 3) 0.001 128 023 657 752 × 2 = 0 + 0.002 256 047 315 504;
  • 4) 0.002 256 047 315 504 × 2 = 0 + 0.004 512 094 631 008;
  • 5) 0.004 512 094 631 008 × 2 = 0 + 0.009 024 189 262 016;
  • 6) 0.009 024 189 262 016 × 2 = 0 + 0.018 048 378 524 032;
  • 7) 0.018 048 378 524 032 × 2 = 0 + 0.036 096 757 048 064;
  • 8) 0.036 096 757 048 064 × 2 = 0 + 0.072 193 514 096 128;
  • 9) 0.072 193 514 096 128 × 2 = 0 + 0.144 387 028 192 256;
  • 10) 0.144 387 028 192 256 × 2 = 0 + 0.288 774 056 384 512;
  • 11) 0.288 774 056 384 512 × 2 = 0 + 0.577 548 112 769 024;
  • 12) 0.577 548 112 769 024 × 2 = 1 + 0.155 096 225 538 048;
  • 13) 0.155 096 225 538 048 × 2 = 0 + 0.310 192 451 076 096;
  • 14) 0.310 192 451 076 096 × 2 = 0 + 0.620 384 902 152 192;
  • 15) 0.620 384 902 152 192 × 2 = 1 + 0.240 769 804 304 384;
  • 16) 0.240 769 804 304 384 × 2 = 0 + 0.481 539 608 608 768;
  • 17) 0.481 539 608 608 768 × 2 = 0 + 0.963 079 217 217 536;
  • 18) 0.963 079 217 217 536 × 2 = 1 + 0.926 158 434 435 072;
  • 19) 0.926 158 434 435 072 × 2 = 1 + 0.852 316 868 870 144;
  • 20) 0.852 316 868 870 144 × 2 = 1 + 0.704 633 737 740 288;
  • 21) 0.704 633 737 740 288 × 2 = 1 + 0.409 267 475 480 576;
  • 22) 0.409 267 475 480 576 × 2 = 0 + 0.818 534 950 961 152;
  • 23) 0.818 534 950 961 152 × 2 = 1 + 0.637 069 901 922 304;
  • 24) 0.637 069 901 922 304 × 2 = 1 + 0.274 139 803 844 608;
  • 25) 0.274 139 803 844 608 × 2 = 0 + 0.548 279 607 689 216;
  • 26) 0.548 279 607 689 216 × 2 = 1 + 0.096 559 215 378 432;
  • 27) 0.096 559 215 378 432 × 2 = 0 + 0.193 118 430 756 864;
  • 28) 0.193 118 430 756 864 × 2 = 0 + 0.386 236 861 513 728;
  • 29) 0.386 236 861 513 728 × 2 = 0 + 0.772 473 723 027 456;
  • 30) 0.772 473 723 027 456 × 2 = 1 + 0.544 947 446 054 912;
  • 31) 0.544 947 446 054 912 × 2 = 1 + 0.089 894 892 109 824;
  • 32) 0.089 894 892 109 824 × 2 = 0 + 0.179 789 784 219 648;
  • 33) 0.179 789 784 219 648 × 2 = 0 + 0.359 579 568 439 296;
  • 34) 0.359 579 568 439 296 × 2 = 0 + 0.719 159 136 878 592;
  • 35) 0.719 159 136 878 592 × 2 = 1 + 0.438 318 273 757 184;
  • 36) 0.438 318 273 757 184 × 2 = 0 + 0.876 636 547 514 368;
  • 37) 0.876 636 547 514 368 × 2 = 1 + 0.753 273 095 028 736;
  • 38) 0.753 273 095 028 736 × 2 = 1 + 0.506 546 190 057 472;
  • 39) 0.506 546 190 057 472 × 2 = 1 + 0.013 092 380 114 944;
  • 40) 0.013 092 380 114 944 × 2 = 0 + 0.026 184 760 229 888;
  • 41) 0.026 184 760 229 888 × 2 = 0 + 0.052 369 520 459 776;
  • 42) 0.052 369 520 459 776 × 2 = 0 + 0.104 739 040 919 552;
  • 43) 0.104 739 040 919 552 × 2 = 0 + 0.209 478 081 839 104;
  • 44) 0.209 478 081 839 104 × 2 = 0 + 0.418 956 163 678 208;
  • 45) 0.418 956 163 678 208 × 2 = 0 + 0.837 912 327 356 416;
  • 46) 0.837 912 327 356 416 × 2 = 1 + 0.675 824 654 712 832;
  • 47) 0.675 824 654 712 832 × 2 = 1 + 0.351 649 309 425 664;
  • 48) 0.351 649 309 425 664 × 2 = 0 + 0.703 298 618 851 328;
  • 49) 0.703 298 618 851 328 × 2 = 1 + 0.406 597 237 702 656;
  • 50) 0.406 597 237 702 656 × 2 = 0 + 0.813 194 475 405 312;
  • 51) 0.813 194 475 405 312 × 2 = 1 + 0.626 388 950 810 624;
  • 52) 0.626 388 950 810 624 × 2 = 1 + 0.252 777 901 621 248;
  • 53) 0.252 777 901 621 248 × 2 = 0 + 0.505 555 803 242 496;
  • 54) 0.505 555 803 242 496 × 2 = 1 + 0.011 111 606 484 992;
  • 55) 0.011 111 606 484 992 × 2 = 0 + 0.022 223 212 969 984;
  • 56) 0.022 223 212 969 984 × 2 = 0 + 0.044 446 425 939 968;
  • 57) 0.044 446 425 939 968 × 2 = 0 + 0.088 892 851 879 936;
  • 58) 0.088 892 851 879 936 × 2 = 0 + 0.177 785 703 759 872;
  • 59) 0.177 785 703 759 872 × 2 = 0 + 0.355 571 407 519 744;
  • 60) 0.355 571 407 519 744 × 2 = 0 + 0.711 142 815 039 488;
  • 61) 0.711 142 815 039 488 × 2 = 1 + 0.422 285 630 078 976;
  • 62) 0.422 285 630 078 976 × 2 = 0 + 0.844 571 260 157 952;
  • 63) 0.844 571 260 157 952 × 2 = 1 + 0.689 142 520 315 904;
  • 64) 0.689 142 520 315 904 × 2 = 1 + 0.378 285 040 631 808;

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:


0.000 282 005 914 438(10) =

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

6. Positive number before normalization:

0.000 282 005 914 438(10) =

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


0.000 282 005 914 438(10) =

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

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

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


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 0110 1011 0100 0000 1011 =

0010 0111 1011 0100 0110 0010 1110 0000 0110 1011 0100 0000 1011


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 0110 1011 0100 0000 1011


Decimal number -0.000 282 005 914 438 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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