-0.000 282 005 519 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 519(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
-0.000 282 005 519(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:

|-0.000 282 005 519| = 0.000 282 005 519


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: 0.000 282 005 519.

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) 0.000 282 005 519 × 2 = 0 + 0.000 564 011 038;
  • 2) 0.000 564 011 038 × 2 = 0 + 0.001 128 022 076;
  • 3) 0.001 128 022 076 × 2 = 0 + 0.002 256 044 152;
  • 4) 0.002 256 044 152 × 2 = 0 + 0.004 512 088 304;
  • 5) 0.004 512 088 304 × 2 = 0 + 0.009 024 176 608;
  • 6) 0.009 024 176 608 × 2 = 0 + 0.018 048 353 216;
  • 7) 0.018 048 353 216 × 2 = 0 + 0.036 096 706 432;
  • 8) 0.036 096 706 432 × 2 = 0 + 0.072 193 412 864;
  • 9) 0.072 193 412 864 × 2 = 0 + 0.144 386 825 728;
  • 10) 0.144 386 825 728 × 2 = 0 + 0.288 773 651 456;
  • 11) 0.288 773 651 456 × 2 = 0 + 0.577 547 302 912;
  • 12) 0.577 547 302 912 × 2 = 1 + 0.155 094 605 824;
  • 13) 0.155 094 605 824 × 2 = 0 + 0.310 189 211 648;
  • 14) 0.310 189 211 648 × 2 = 0 + 0.620 378 423 296;
  • 15) 0.620 378 423 296 × 2 = 1 + 0.240 756 846 592;
  • 16) 0.240 756 846 592 × 2 = 0 + 0.481 513 693 184;
  • 17) 0.481 513 693 184 × 2 = 0 + 0.963 027 386 368;
  • 18) 0.963 027 386 368 × 2 = 1 + 0.926 054 772 736;
  • 19) 0.926 054 772 736 × 2 = 1 + 0.852 109 545 472;
  • 20) 0.852 109 545 472 × 2 = 1 + 0.704 219 090 944;
  • 21) 0.704 219 090 944 × 2 = 1 + 0.408 438 181 888;
  • 22) 0.408 438 181 888 × 2 = 0 + 0.816 876 363 776;
  • 23) 0.816 876 363 776 × 2 = 1 + 0.633 752 727 552;
  • 24) 0.633 752 727 552 × 2 = 1 + 0.267 505 455 104;
  • 25) 0.267 505 455 104 × 2 = 0 + 0.535 010 910 208;
  • 26) 0.535 010 910 208 × 2 = 1 + 0.070 021 820 416;
  • 27) 0.070 021 820 416 × 2 = 0 + 0.140 043 640 832;
  • 28) 0.140 043 640 832 × 2 = 0 + 0.280 087 281 664;
  • 29) 0.280 087 281 664 × 2 = 0 + 0.560 174 563 328;
  • 30) 0.560 174 563 328 × 2 = 1 + 0.120 349 126 656;
  • 31) 0.120 349 126 656 × 2 = 0 + 0.240 698 253 312;
  • 32) 0.240 698 253 312 × 2 = 0 + 0.481 396 506 624;
  • 33) 0.481 396 506 624 × 2 = 0 + 0.962 793 013 248;
  • 34) 0.962 793 013 248 × 2 = 1 + 0.925 586 026 496;
  • 35) 0.925 586 026 496 × 2 = 1 + 0.851 172 052 992;
  • 36) 0.851 172 052 992 × 2 = 1 + 0.702 344 105 984;
  • 37) 0.702 344 105 984 × 2 = 1 + 0.404 688 211 968;
  • 38) 0.404 688 211 968 × 2 = 0 + 0.809 376 423 936;
  • 39) 0.809 376 423 936 × 2 = 1 + 0.618 752 847 872;
  • 40) 0.618 752 847 872 × 2 = 1 + 0.237 505 695 744;
  • 41) 0.237 505 695 744 × 2 = 0 + 0.475 011 391 488;
  • 42) 0.475 011 391 488 × 2 = 0 + 0.950 022 782 976;
  • 43) 0.950 022 782 976 × 2 = 1 + 0.900 045 565 952;
  • 44) 0.900 045 565 952 × 2 = 1 + 0.800 091 131 904;
  • 45) 0.800 091 131 904 × 2 = 1 + 0.600 182 263 808;
  • 46) 0.600 182 263 808 × 2 = 1 + 0.200 364 527 616;
  • 47) 0.200 364 527 616 × 2 = 0 + 0.400 729 055 232;
  • 48) 0.400 729 055 232 × 2 = 0 + 0.801 458 110 464;
  • 49) 0.801 458 110 464 × 2 = 1 + 0.602 916 220 928;
  • 50) 0.602 916 220 928 × 2 = 1 + 0.205 832 441 856;
  • 51) 0.205 832 441 856 × 2 = 0 + 0.411 664 883 712;
  • 52) 0.411 664 883 712 × 2 = 0 + 0.823 329 767 424;
  • 53) 0.823 329 767 424 × 2 = 1 + 0.646 659 534 848;
  • 54) 0.646 659 534 848 × 2 = 1 + 0.293 319 069 696;
  • 55) 0.293 319 069 696 × 2 = 0 + 0.586 638 139 392;
  • 56) 0.586 638 139 392 × 2 = 1 + 0.173 276 278 784;
  • 57) 0.173 276 278 784 × 2 = 0 + 0.346 552 557 568;
  • 58) 0.346 552 557 568 × 2 = 0 + 0.693 105 115 136;
  • 59) 0.693 105 115 136 × 2 = 1 + 0.386 210 230 272;
  • 60) 0.386 210 230 272 × 2 = 0 + 0.772 420 460 544;
  • 61) 0.772 420 460 544 × 2 = 1 + 0.544 840 921 088;
  • 62) 0.544 840 921 088 × 2 = 1 + 0.089 681 842 176;
  • 63) 0.089 681 842 176 × 2 = 0 + 0.179 363 684 352;
  • 64) 0.179 363 684 352 × 2 = 0 + 0.358 727 368 704;

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:


0.000 282 005 519(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0111 1011 0011 1100 1100 1101 0010 1100(2)

6. Positive number before normalization:

0.000 282 005 519(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0111 1011 0011 1100 1100 1101 0010 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:


0.000 282 005 519(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0111 1011 0011 1100 1100 1101 0010 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0111 1011 0011 1100 1100 1101 0010 1100(2) × 20 =

1.0010 0111 1011 0100 0100 0111 1011 0011 1100 1100 1101 0010 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 0100 0111 1011 0011 1100 1100 1101 0010 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 0100 0111 1011 0011 1100 1100 1101 0010 1100 =

0010 0111 1011 0100 0100 0111 1011 0011 1100 1100 1101 0010 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 0100 0111 1011 0011 1100 1100 1101 0010 1100


Decimal number -0.000 282 005 519 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 0111 1011 0011 1100 1100 1101 0010 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