-0.000 282 006 256 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 256(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
-0.000 282 006 256(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:

|-0.000 282 006 256| = 0.000 282 006 256


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: 0.000 282 006 256.

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) 0.000 282 006 256 × 2 = 0 + 0.000 564 012 512;
  • 2) 0.000 564 012 512 × 2 = 0 + 0.001 128 025 024;
  • 3) 0.001 128 025 024 × 2 = 0 + 0.002 256 050 048;
  • 4) 0.002 256 050 048 × 2 = 0 + 0.004 512 100 096;
  • 5) 0.004 512 100 096 × 2 = 0 + 0.009 024 200 192;
  • 6) 0.009 024 200 192 × 2 = 0 + 0.018 048 400 384;
  • 7) 0.018 048 400 384 × 2 = 0 + 0.036 096 800 768;
  • 8) 0.036 096 800 768 × 2 = 0 + 0.072 193 601 536;
  • 9) 0.072 193 601 536 × 2 = 0 + 0.144 387 203 072;
  • 10) 0.144 387 203 072 × 2 = 0 + 0.288 774 406 144;
  • 11) 0.288 774 406 144 × 2 = 0 + 0.577 548 812 288;
  • 12) 0.577 548 812 288 × 2 = 1 + 0.155 097 624 576;
  • 13) 0.155 097 624 576 × 2 = 0 + 0.310 195 249 152;
  • 14) 0.310 195 249 152 × 2 = 0 + 0.620 390 498 304;
  • 15) 0.620 390 498 304 × 2 = 1 + 0.240 780 996 608;
  • 16) 0.240 780 996 608 × 2 = 0 + 0.481 561 993 216;
  • 17) 0.481 561 993 216 × 2 = 0 + 0.963 123 986 432;
  • 18) 0.963 123 986 432 × 2 = 1 + 0.926 247 972 864;
  • 19) 0.926 247 972 864 × 2 = 1 + 0.852 495 945 728;
  • 20) 0.852 495 945 728 × 2 = 1 + 0.704 991 891 456;
  • 21) 0.704 991 891 456 × 2 = 1 + 0.409 983 782 912;
  • 22) 0.409 983 782 912 × 2 = 0 + 0.819 967 565 824;
  • 23) 0.819 967 565 824 × 2 = 1 + 0.639 935 131 648;
  • 24) 0.639 935 131 648 × 2 = 1 + 0.279 870 263 296;
  • 25) 0.279 870 263 296 × 2 = 0 + 0.559 740 526 592;
  • 26) 0.559 740 526 592 × 2 = 1 + 0.119 481 053 184;
  • 27) 0.119 481 053 184 × 2 = 0 + 0.238 962 106 368;
  • 28) 0.238 962 106 368 × 2 = 0 + 0.477 924 212 736;
  • 29) 0.477 924 212 736 × 2 = 0 + 0.955 848 425 472;
  • 30) 0.955 848 425 472 × 2 = 1 + 0.911 696 850 944;
  • 31) 0.911 696 850 944 × 2 = 1 + 0.823 393 701 888;
  • 32) 0.823 393 701 888 × 2 = 1 + 0.646 787 403 776;
  • 33) 0.646 787 403 776 × 2 = 1 + 0.293 574 807 552;
  • 34) 0.293 574 807 552 × 2 = 0 + 0.587 149 615 104;
  • 35) 0.587 149 615 104 × 2 = 1 + 0.174 299 230 208;
  • 36) 0.174 299 230 208 × 2 = 0 + 0.348 598 460 416;
  • 37) 0.348 598 460 416 × 2 = 0 + 0.697 196 920 832;
  • 38) 0.697 196 920 832 × 2 = 1 + 0.394 393 841 664;
  • 39) 0.394 393 841 664 × 2 = 0 + 0.788 787 683 328;
  • 40) 0.788 787 683 328 × 2 = 1 + 0.577 575 366 656;
  • 41) 0.577 575 366 656 × 2 = 1 + 0.155 150 733 312;
  • 42) 0.155 150 733 312 × 2 = 0 + 0.310 301 466 624;
  • 43) 0.310 301 466 624 × 2 = 0 + 0.620 602 933 248;
  • 44) 0.620 602 933 248 × 2 = 1 + 0.241 205 866 496;
  • 45) 0.241 205 866 496 × 2 = 0 + 0.482 411 732 992;
  • 46) 0.482 411 732 992 × 2 = 0 + 0.964 823 465 984;
  • 47) 0.964 823 465 984 × 2 = 1 + 0.929 646 931 968;
  • 48) 0.929 646 931 968 × 2 = 1 + 0.859 293 863 936;
  • 49) 0.859 293 863 936 × 2 = 1 + 0.718 587 727 872;
  • 50) 0.718 587 727 872 × 2 = 1 + 0.437 175 455 744;
  • 51) 0.437 175 455 744 × 2 = 0 + 0.874 350 911 488;
  • 52) 0.874 350 911 488 × 2 = 1 + 0.748 701 822 976;
  • 53) 0.748 701 822 976 × 2 = 1 + 0.497 403 645 952;
  • 54) 0.497 403 645 952 × 2 = 0 + 0.994 807 291 904;
  • 55) 0.994 807 291 904 × 2 = 1 + 0.989 614 583 808;
  • 56) 0.989 614 583 808 × 2 = 1 + 0.979 229 167 616;
  • 57) 0.979 229 167 616 × 2 = 1 + 0.958 458 335 232;
  • 58) 0.958 458 335 232 × 2 = 1 + 0.916 916 670 464;
  • 59) 0.916 916 670 464 × 2 = 1 + 0.833 833 340 928;
  • 60) 0.833 833 340 928 × 2 = 1 + 0.667 666 681 856;
  • 61) 0.667 666 681 856 × 2 = 1 + 0.335 333 363 712;
  • 62) 0.335 333 363 712 × 2 = 0 + 0.670 666 727 424;
  • 63) 0.670 666 727 424 × 2 = 1 + 0.341 333 454 848;
  • 64) 0.341 333 454 848 × 2 = 0 + 0.682 666 909 696;

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:


0.000 282 006 256(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 0101 1001 0011 1101 1011 1111 1010(2)

6. Positive number before normalization:

0.000 282 006 256(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 0101 1001 0011 1101 1011 1111 1010(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:


0.000 282 006 256(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 0101 1001 0011 1101 1011 1111 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1010 0101 1001 0011 1101 1011 1111 1010(2) × 20 =

1.0010 0111 1011 0100 0111 1010 0101 1001 0011 1101 1011 1111 1010(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 0111 1010 0101 1001 0011 1101 1011 1111 1010


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 0111 1010 0101 1001 0011 1101 1011 1111 1010 =

0010 0111 1011 0100 0111 1010 0101 1001 0011 1101 1011 1111 1010


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 0111 1010 0101 1001 0011 1101 1011 1111 1010


Decimal number -0.000 282 006 256 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1010 0101 1001 0011 1101 1011 1111 1010

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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