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

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

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


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: 0.000 282 006 282.

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) 0.000 282 006 282 × 2 = 0 + 0.000 564 012 564;
  • 2) 0.000 564 012 564 × 2 = 0 + 0.001 128 025 128;
  • 3) 0.001 128 025 128 × 2 = 0 + 0.002 256 050 256;
  • 4) 0.002 256 050 256 × 2 = 0 + 0.004 512 100 512;
  • 5) 0.004 512 100 512 × 2 = 0 + 0.009 024 201 024;
  • 6) 0.009 024 201 024 × 2 = 0 + 0.018 048 402 048;
  • 7) 0.018 048 402 048 × 2 = 0 + 0.036 096 804 096;
  • 8) 0.036 096 804 096 × 2 = 0 + 0.072 193 608 192;
  • 9) 0.072 193 608 192 × 2 = 0 + 0.144 387 216 384;
  • 10) 0.144 387 216 384 × 2 = 0 + 0.288 774 432 768;
  • 11) 0.288 774 432 768 × 2 = 0 + 0.577 548 865 536;
  • 12) 0.577 548 865 536 × 2 = 1 + 0.155 097 731 072;
  • 13) 0.155 097 731 072 × 2 = 0 + 0.310 195 462 144;
  • 14) 0.310 195 462 144 × 2 = 0 + 0.620 390 924 288;
  • 15) 0.620 390 924 288 × 2 = 1 + 0.240 781 848 576;
  • 16) 0.240 781 848 576 × 2 = 0 + 0.481 563 697 152;
  • 17) 0.481 563 697 152 × 2 = 0 + 0.963 127 394 304;
  • 18) 0.963 127 394 304 × 2 = 1 + 0.926 254 788 608;
  • 19) 0.926 254 788 608 × 2 = 1 + 0.852 509 577 216;
  • 20) 0.852 509 577 216 × 2 = 1 + 0.705 019 154 432;
  • 21) 0.705 019 154 432 × 2 = 1 + 0.410 038 308 864;
  • 22) 0.410 038 308 864 × 2 = 0 + 0.820 076 617 728;
  • 23) 0.820 076 617 728 × 2 = 1 + 0.640 153 235 456;
  • 24) 0.640 153 235 456 × 2 = 1 + 0.280 306 470 912;
  • 25) 0.280 306 470 912 × 2 = 0 + 0.560 612 941 824;
  • 26) 0.560 612 941 824 × 2 = 1 + 0.121 225 883 648;
  • 27) 0.121 225 883 648 × 2 = 0 + 0.242 451 767 296;
  • 28) 0.242 451 767 296 × 2 = 0 + 0.484 903 534 592;
  • 29) 0.484 903 534 592 × 2 = 0 + 0.969 807 069 184;
  • 30) 0.969 807 069 184 × 2 = 1 + 0.939 614 138 368;
  • 31) 0.939 614 138 368 × 2 = 1 + 0.879 228 276 736;
  • 32) 0.879 228 276 736 × 2 = 1 + 0.758 456 553 472;
  • 33) 0.758 456 553 472 × 2 = 1 + 0.516 913 106 944;
  • 34) 0.516 913 106 944 × 2 = 1 + 0.033 826 213 888;
  • 35) 0.033 826 213 888 × 2 = 0 + 0.067 652 427 776;
  • 36) 0.067 652 427 776 × 2 = 0 + 0.135 304 855 552;
  • 37) 0.135 304 855 552 × 2 = 0 + 0.270 609 711 104;
  • 38) 0.270 609 711 104 × 2 = 0 + 0.541 219 422 208;
  • 39) 0.541 219 422 208 × 2 = 1 + 0.082 438 844 416;
  • 40) 0.082 438 844 416 × 2 = 0 + 0.164 877 688 832;
  • 41) 0.164 877 688 832 × 2 = 0 + 0.329 755 377 664;
  • 42) 0.329 755 377 664 × 2 = 0 + 0.659 510 755 328;
  • 43) 0.659 510 755 328 × 2 = 1 + 0.319 021 510 656;
  • 44) 0.319 021 510 656 × 2 = 0 + 0.638 043 021 312;
  • 45) 0.638 043 021 312 × 2 = 1 + 0.276 086 042 624;
  • 46) 0.276 086 042 624 × 2 = 0 + 0.552 172 085 248;
  • 47) 0.552 172 085 248 × 2 = 1 + 0.104 344 170 496;
  • 48) 0.104 344 170 496 × 2 = 0 + 0.208 688 340 992;
  • 49) 0.208 688 340 992 × 2 = 0 + 0.417 376 681 984;
  • 50) 0.417 376 681 984 × 2 = 0 + 0.834 753 363 968;
  • 51) 0.834 753 363 968 × 2 = 1 + 0.669 506 727 936;
  • 52) 0.669 506 727 936 × 2 = 1 + 0.339 013 455 872;
  • 53) 0.339 013 455 872 × 2 = 0 + 0.678 026 911 744;
  • 54) 0.678 026 911 744 × 2 = 1 + 0.356 053 823 488;
  • 55) 0.356 053 823 488 × 2 = 0 + 0.712 107 646 976;
  • 56) 0.712 107 646 976 × 2 = 1 + 0.424 215 293 952;
  • 57) 0.424 215 293 952 × 2 = 0 + 0.848 430 587 904;
  • 58) 0.848 430 587 904 × 2 = 1 + 0.696 861 175 808;
  • 59) 0.696 861 175 808 × 2 = 1 + 0.393 722 351 616;
  • 60) 0.393 722 351 616 × 2 = 0 + 0.787 444 703 232;
  • 61) 0.787 444 703 232 × 2 = 1 + 0.574 889 406 464;
  • 62) 0.574 889 406 464 × 2 = 1 + 0.149 778 812 928;
  • 63) 0.149 778 812 928 × 2 = 0 + 0.299 557 625 856;
  • 64) 0.299 557 625 856 × 2 = 0 + 0.599 115 251 712;

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:


0.000 282 006 282(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 0010 0010 1010 0011 0101 0110 1100(2)

6. Positive number before normalization:

0.000 282 006 282(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 0010 0010 1010 0011 0101 0110 1100(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:


0.000 282 006 282(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 0010 0010 1010 0011 0101 0110 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1100 0010 0010 1010 0011 0101 0110 1100(2) × 20 =

1.0010 0111 1011 0100 0111 1100 0010 0010 1010 0011 0101 0110 1100(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 1100 0010 0010 1010 0011 0101 0110 1100


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 1100 0010 0010 1010 0011 0101 0110 1100 =

0010 0111 1011 0100 0111 1100 0010 0010 1010 0011 0101 0110 1100


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 1100 0010 0010 1010 0011 0101 0110 1100


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

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1100 0010 0010 1010 0011 0101 0110 1100

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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