-0.000 282 006 038 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 038(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
-0.000 282 006 038(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:

|-0.000 282 006 038| = 0.000 282 006 038


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: 0.000 282 006 038.

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) 0.000 282 006 038 × 2 = 0 + 0.000 564 012 076;
  • 2) 0.000 564 012 076 × 2 = 0 + 0.001 128 024 152;
  • 3) 0.001 128 024 152 × 2 = 0 + 0.002 256 048 304;
  • 4) 0.002 256 048 304 × 2 = 0 + 0.004 512 096 608;
  • 5) 0.004 512 096 608 × 2 = 0 + 0.009 024 193 216;
  • 6) 0.009 024 193 216 × 2 = 0 + 0.018 048 386 432;
  • 7) 0.018 048 386 432 × 2 = 0 + 0.036 096 772 864;
  • 8) 0.036 096 772 864 × 2 = 0 + 0.072 193 545 728;
  • 9) 0.072 193 545 728 × 2 = 0 + 0.144 387 091 456;
  • 10) 0.144 387 091 456 × 2 = 0 + 0.288 774 182 912;
  • 11) 0.288 774 182 912 × 2 = 0 + 0.577 548 365 824;
  • 12) 0.577 548 365 824 × 2 = 1 + 0.155 096 731 648;
  • 13) 0.155 096 731 648 × 2 = 0 + 0.310 193 463 296;
  • 14) 0.310 193 463 296 × 2 = 0 + 0.620 386 926 592;
  • 15) 0.620 386 926 592 × 2 = 1 + 0.240 773 853 184;
  • 16) 0.240 773 853 184 × 2 = 0 + 0.481 547 706 368;
  • 17) 0.481 547 706 368 × 2 = 0 + 0.963 095 412 736;
  • 18) 0.963 095 412 736 × 2 = 1 + 0.926 190 825 472;
  • 19) 0.926 190 825 472 × 2 = 1 + 0.852 381 650 944;
  • 20) 0.852 381 650 944 × 2 = 1 + 0.704 763 301 888;
  • 21) 0.704 763 301 888 × 2 = 1 + 0.409 526 603 776;
  • 22) 0.409 526 603 776 × 2 = 0 + 0.819 053 207 552;
  • 23) 0.819 053 207 552 × 2 = 1 + 0.638 106 415 104;
  • 24) 0.638 106 415 104 × 2 = 1 + 0.276 212 830 208;
  • 25) 0.276 212 830 208 × 2 = 0 + 0.552 425 660 416;
  • 26) 0.552 425 660 416 × 2 = 1 + 0.104 851 320 832;
  • 27) 0.104 851 320 832 × 2 = 0 + 0.209 702 641 664;
  • 28) 0.209 702 641 664 × 2 = 0 + 0.419 405 283 328;
  • 29) 0.419 405 283 328 × 2 = 0 + 0.838 810 566 656;
  • 30) 0.838 810 566 656 × 2 = 1 + 0.677 621 133 312;
  • 31) 0.677 621 133 312 × 2 = 1 + 0.355 242 266 624;
  • 32) 0.355 242 266 624 × 2 = 0 + 0.710 484 533 248;
  • 33) 0.710 484 533 248 × 2 = 1 + 0.420 969 066 496;
  • 34) 0.420 969 066 496 × 2 = 0 + 0.841 938 132 992;
  • 35) 0.841 938 132 992 × 2 = 1 + 0.683 876 265 984;
  • 36) 0.683 876 265 984 × 2 = 1 + 0.367 752 531 968;
  • 37) 0.367 752 531 968 × 2 = 0 + 0.735 505 063 936;
  • 38) 0.735 505 063 936 × 2 = 1 + 0.471 010 127 872;
  • 39) 0.471 010 127 872 × 2 = 0 + 0.942 020 255 744;
  • 40) 0.942 020 255 744 × 2 = 1 + 0.884 040 511 488;
  • 41) 0.884 040 511 488 × 2 = 1 + 0.768 081 022 976;
  • 42) 0.768 081 022 976 × 2 = 1 + 0.536 162 045 952;
  • 43) 0.536 162 045 952 × 2 = 1 + 0.072 324 091 904;
  • 44) 0.072 324 091 904 × 2 = 0 + 0.144 648 183 808;
  • 45) 0.144 648 183 808 × 2 = 0 + 0.289 296 367 616;
  • 46) 0.289 296 367 616 × 2 = 0 + 0.578 592 735 232;
  • 47) 0.578 592 735 232 × 2 = 1 + 0.157 185 470 464;
  • 48) 0.157 185 470 464 × 2 = 0 + 0.314 370 940 928;
  • 49) 0.314 370 940 928 × 2 = 0 + 0.628 741 881 856;
  • 50) 0.628 741 881 856 × 2 = 1 + 0.257 483 763 712;
  • 51) 0.257 483 763 712 × 2 = 0 + 0.514 967 527 424;
  • 52) 0.514 967 527 424 × 2 = 1 + 0.029 935 054 848;
  • 53) 0.029 935 054 848 × 2 = 0 + 0.059 870 109 696;
  • 54) 0.059 870 109 696 × 2 = 0 + 0.119 740 219 392;
  • 55) 0.119 740 219 392 × 2 = 0 + 0.239 480 438 784;
  • 56) 0.239 480 438 784 × 2 = 0 + 0.478 960 877 568;
  • 57) 0.478 960 877 568 × 2 = 0 + 0.957 921 755 136;
  • 58) 0.957 921 755 136 × 2 = 1 + 0.915 843 510 272;
  • 59) 0.915 843 510 272 × 2 = 1 + 0.831 687 020 544;
  • 60) 0.831 687 020 544 × 2 = 1 + 0.663 374 041 088;
  • 61) 0.663 374 041 088 × 2 = 1 + 0.326 748 082 176;
  • 62) 0.326 748 082 176 × 2 = 0 + 0.653 496 164 352;
  • 63) 0.653 496 164 352 × 2 = 1 + 0.306 992 328 704;
  • 64) 0.306 992 328 704 × 2 = 0 + 0.613 984 657 408;

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:


0.000 282 006 038(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 0101 1110 0010 0101 0000 0111 1010(2)

6. Positive number before normalization:

0.000 282 006 038(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 0101 1110 0010 0101 0000 0111 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:


0.000 282 006 038(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 0101 1110 0010 0101 0000 0111 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 0101 1110 0010 0101 0000 0111 1010(2) × 20 =

1.0010 0111 1011 0100 0110 1011 0101 1110 0010 0101 0000 0111 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 0110 1011 0101 1110 0010 0101 0000 0111 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 0110 1011 0101 1110 0010 0101 0000 0111 1010 =

0010 0111 1011 0100 0110 1011 0101 1110 0010 0101 0000 0111 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 0110 1011 0101 1110 0010 0101 0000 0111 1010


Decimal number -0.000 282 006 038 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1011 0101 1110 0010 0101 0000 0111 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