-0.000 282 005 703 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 703(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
-0.000 282 005 703(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:

|-0.000 282 005 703| = 0.000 282 005 703


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: 0.000 282 005 703.

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) 0.000 282 005 703 × 2 = 0 + 0.000 564 011 406;
  • 2) 0.000 564 011 406 × 2 = 0 + 0.001 128 022 812;
  • 3) 0.001 128 022 812 × 2 = 0 + 0.002 256 045 624;
  • 4) 0.002 256 045 624 × 2 = 0 + 0.004 512 091 248;
  • 5) 0.004 512 091 248 × 2 = 0 + 0.009 024 182 496;
  • 6) 0.009 024 182 496 × 2 = 0 + 0.018 048 364 992;
  • 7) 0.018 048 364 992 × 2 = 0 + 0.036 096 729 984;
  • 8) 0.036 096 729 984 × 2 = 0 + 0.072 193 459 968;
  • 9) 0.072 193 459 968 × 2 = 0 + 0.144 386 919 936;
  • 10) 0.144 386 919 936 × 2 = 0 + 0.288 773 839 872;
  • 11) 0.288 773 839 872 × 2 = 0 + 0.577 547 679 744;
  • 12) 0.577 547 679 744 × 2 = 1 + 0.155 095 359 488;
  • 13) 0.155 095 359 488 × 2 = 0 + 0.310 190 718 976;
  • 14) 0.310 190 718 976 × 2 = 0 + 0.620 381 437 952;
  • 15) 0.620 381 437 952 × 2 = 1 + 0.240 762 875 904;
  • 16) 0.240 762 875 904 × 2 = 0 + 0.481 525 751 808;
  • 17) 0.481 525 751 808 × 2 = 0 + 0.963 051 503 616;
  • 18) 0.963 051 503 616 × 2 = 1 + 0.926 103 007 232;
  • 19) 0.926 103 007 232 × 2 = 1 + 0.852 206 014 464;
  • 20) 0.852 206 014 464 × 2 = 1 + 0.704 412 028 928;
  • 21) 0.704 412 028 928 × 2 = 1 + 0.408 824 057 856;
  • 22) 0.408 824 057 856 × 2 = 0 + 0.817 648 115 712;
  • 23) 0.817 648 115 712 × 2 = 1 + 0.635 296 231 424;
  • 24) 0.635 296 231 424 × 2 = 1 + 0.270 592 462 848;
  • 25) 0.270 592 462 848 × 2 = 0 + 0.541 184 925 696;
  • 26) 0.541 184 925 696 × 2 = 1 + 0.082 369 851 392;
  • 27) 0.082 369 851 392 × 2 = 0 + 0.164 739 702 784;
  • 28) 0.164 739 702 784 × 2 = 0 + 0.329 479 405 568;
  • 29) 0.329 479 405 568 × 2 = 0 + 0.658 958 811 136;
  • 30) 0.658 958 811 136 × 2 = 1 + 0.317 917 622 272;
  • 31) 0.317 917 622 272 × 2 = 0 + 0.635 835 244 544;
  • 32) 0.635 835 244 544 × 2 = 1 + 0.271 670 489 088;
  • 33) 0.271 670 489 088 × 2 = 0 + 0.543 340 978 176;
  • 34) 0.543 340 978 176 × 2 = 1 + 0.086 681 956 352;
  • 35) 0.086 681 956 352 × 2 = 0 + 0.173 363 912 704;
  • 36) 0.173 363 912 704 × 2 = 0 + 0.346 727 825 408;
  • 37) 0.346 727 825 408 × 2 = 0 + 0.693 455 650 816;
  • 38) 0.693 455 650 816 × 2 = 1 + 0.386 911 301 632;
  • 39) 0.386 911 301 632 × 2 = 0 + 0.773 822 603 264;
  • 40) 0.773 822 603 264 × 2 = 1 + 0.547 645 206 528;
  • 41) 0.547 645 206 528 × 2 = 1 + 0.095 290 413 056;
  • 42) 0.095 290 413 056 × 2 = 0 + 0.190 580 826 112;
  • 43) 0.190 580 826 112 × 2 = 0 + 0.381 161 652 224;
  • 44) 0.381 161 652 224 × 2 = 0 + 0.762 323 304 448;
  • 45) 0.762 323 304 448 × 2 = 1 + 0.524 646 608 896;
  • 46) 0.524 646 608 896 × 2 = 1 + 0.049 293 217 792;
  • 47) 0.049 293 217 792 × 2 = 0 + 0.098 586 435 584;
  • 48) 0.098 586 435 584 × 2 = 0 + 0.197 172 871 168;
  • 49) 0.197 172 871 168 × 2 = 0 + 0.394 345 742 336;
  • 50) 0.394 345 742 336 × 2 = 0 + 0.788 691 484 672;
  • 51) 0.788 691 484 672 × 2 = 1 + 0.577 382 969 344;
  • 52) 0.577 382 969 344 × 2 = 1 + 0.154 765 938 688;
  • 53) 0.154 765 938 688 × 2 = 0 + 0.309 531 877 376;
  • 54) 0.309 531 877 376 × 2 = 0 + 0.619 063 754 752;
  • 55) 0.619 063 754 752 × 2 = 1 + 0.238 127 509 504;
  • 56) 0.238 127 509 504 × 2 = 0 + 0.476 255 019 008;
  • 57) 0.476 255 019 008 × 2 = 0 + 0.952 510 038 016;
  • 58) 0.952 510 038 016 × 2 = 1 + 0.905 020 076 032;
  • 59) 0.905 020 076 032 × 2 = 1 + 0.810 040 152 064;
  • 60) 0.810 040 152 064 × 2 = 1 + 0.620 080 304 128;
  • 61) 0.620 080 304 128 × 2 = 1 + 0.240 160 608 256;
  • 62) 0.240 160 608 256 × 2 = 0 + 0.480 321 216 512;
  • 63) 0.480 321 216 512 × 2 = 0 + 0.960 642 433 024;
  • 64) 0.960 642 433 024 × 2 = 1 + 0.921 284 866 048;

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:


0.000 282 005 703(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0100 0101 1000 1100 0011 0010 0111 1001(2)

6. Positive number before normalization:

0.000 282 005 703(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0100 0101 1000 1100 0011 0010 0111 1001(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:


0.000 282 005 703(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0100 0101 1000 1100 0011 0010 0111 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0100 0101 1000 1100 0011 0010 0111 1001(2) × 20 =

1.0010 0111 1011 0100 0101 0100 0101 1000 1100 0011 0010 0111 1001(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 0101 0100 0101 1000 1100 0011 0010 0111 1001


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 0101 0100 0101 1000 1100 0011 0010 0111 1001 =

0010 0111 1011 0100 0101 0100 0101 1000 1100 0011 0010 0111 1001


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 0101 0100 0101 1000 1100 0011 0010 0111 1001


Decimal number -0.000 282 005 703 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0100 0101 1000 1100 0011 0010 0111 1001

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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