-0.000 282 005 366 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 366(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
-0.000 282 005 366(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:

|-0.000 282 005 366| = 0.000 282 005 366


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: 0.000 282 005 366.

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) 0.000 282 005 366 × 2 = 0 + 0.000 564 010 732;
  • 2) 0.000 564 010 732 × 2 = 0 + 0.001 128 021 464;
  • 3) 0.001 128 021 464 × 2 = 0 + 0.002 256 042 928;
  • 4) 0.002 256 042 928 × 2 = 0 + 0.004 512 085 856;
  • 5) 0.004 512 085 856 × 2 = 0 + 0.009 024 171 712;
  • 6) 0.009 024 171 712 × 2 = 0 + 0.018 048 343 424;
  • 7) 0.018 048 343 424 × 2 = 0 + 0.036 096 686 848;
  • 8) 0.036 096 686 848 × 2 = 0 + 0.072 193 373 696;
  • 9) 0.072 193 373 696 × 2 = 0 + 0.144 386 747 392;
  • 10) 0.144 386 747 392 × 2 = 0 + 0.288 773 494 784;
  • 11) 0.288 773 494 784 × 2 = 0 + 0.577 546 989 568;
  • 12) 0.577 546 989 568 × 2 = 1 + 0.155 093 979 136;
  • 13) 0.155 093 979 136 × 2 = 0 + 0.310 187 958 272;
  • 14) 0.310 187 958 272 × 2 = 0 + 0.620 375 916 544;
  • 15) 0.620 375 916 544 × 2 = 1 + 0.240 751 833 088;
  • 16) 0.240 751 833 088 × 2 = 0 + 0.481 503 666 176;
  • 17) 0.481 503 666 176 × 2 = 0 + 0.963 007 332 352;
  • 18) 0.963 007 332 352 × 2 = 1 + 0.926 014 664 704;
  • 19) 0.926 014 664 704 × 2 = 1 + 0.852 029 329 408;
  • 20) 0.852 029 329 408 × 2 = 1 + 0.704 058 658 816;
  • 21) 0.704 058 658 816 × 2 = 1 + 0.408 117 317 632;
  • 22) 0.408 117 317 632 × 2 = 0 + 0.816 234 635 264;
  • 23) 0.816 234 635 264 × 2 = 1 + 0.632 469 270 528;
  • 24) 0.632 469 270 528 × 2 = 1 + 0.264 938 541 056;
  • 25) 0.264 938 541 056 × 2 = 0 + 0.529 877 082 112;
  • 26) 0.529 877 082 112 × 2 = 1 + 0.059 754 164 224;
  • 27) 0.059 754 164 224 × 2 = 0 + 0.119 508 328 448;
  • 28) 0.119 508 328 448 × 2 = 0 + 0.239 016 656 896;
  • 29) 0.239 016 656 896 × 2 = 0 + 0.478 033 313 792;
  • 30) 0.478 033 313 792 × 2 = 0 + 0.956 066 627 584;
  • 31) 0.956 066 627 584 × 2 = 1 + 0.912 133 255 168;
  • 32) 0.912 133 255 168 × 2 = 1 + 0.824 266 510 336;
  • 33) 0.824 266 510 336 × 2 = 1 + 0.648 533 020 672;
  • 34) 0.648 533 020 672 × 2 = 1 + 0.297 066 041 344;
  • 35) 0.297 066 041 344 × 2 = 0 + 0.594 132 082 688;
  • 36) 0.594 132 082 688 × 2 = 1 + 0.188 264 165 376;
  • 37) 0.188 264 165 376 × 2 = 0 + 0.376 528 330 752;
  • 38) 0.376 528 330 752 × 2 = 0 + 0.753 056 661 504;
  • 39) 0.753 056 661 504 × 2 = 1 + 0.506 113 323 008;
  • 40) 0.506 113 323 008 × 2 = 1 + 0.012 226 646 016;
  • 41) 0.012 226 646 016 × 2 = 0 + 0.024 453 292 032;
  • 42) 0.024 453 292 032 × 2 = 0 + 0.048 906 584 064;
  • 43) 0.048 906 584 064 × 2 = 0 + 0.097 813 168 128;
  • 44) 0.097 813 168 128 × 2 = 0 + 0.195 626 336 256;
  • 45) 0.195 626 336 256 × 2 = 0 + 0.391 252 672 512;
  • 46) 0.391 252 672 512 × 2 = 0 + 0.782 505 345 024;
  • 47) 0.782 505 345 024 × 2 = 1 + 0.565 010 690 048;
  • 48) 0.565 010 690 048 × 2 = 1 + 0.130 021 380 096;
  • 49) 0.130 021 380 096 × 2 = 0 + 0.260 042 760 192;
  • 50) 0.260 042 760 192 × 2 = 0 + 0.520 085 520 384;
  • 51) 0.520 085 520 384 × 2 = 1 + 0.040 171 040 768;
  • 52) 0.040 171 040 768 × 2 = 0 + 0.080 342 081 536;
  • 53) 0.080 342 081 536 × 2 = 0 + 0.160 684 163 072;
  • 54) 0.160 684 163 072 × 2 = 0 + 0.321 368 326 144;
  • 55) 0.321 368 326 144 × 2 = 0 + 0.642 736 652 288;
  • 56) 0.642 736 652 288 × 2 = 1 + 0.285 473 304 576;
  • 57) 0.285 473 304 576 × 2 = 0 + 0.570 946 609 152;
  • 58) 0.570 946 609 152 × 2 = 1 + 0.141 893 218 304;
  • 59) 0.141 893 218 304 × 2 = 0 + 0.283 786 436 608;
  • 60) 0.283 786 436 608 × 2 = 0 + 0.567 572 873 216;
  • 61) 0.567 572 873 216 × 2 = 1 + 0.135 145 746 432;
  • 62) 0.135 145 746 432 × 2 = 0 + 0.270 291 492 864;
  • 63) 0.270 291 492 864 × 2 = 0 + 0.540 582 985 728;
  • 64) 0.540 582 985 728 × 2 = 1 + 0.081 165 971 456;

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:


0.000 282 005 366(10) =

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

6. Positive number before normalization:

0.000 282 005 366(10) =

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


0.000 282 005 366(10) =

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

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

1.0010 0111 1011 0100 0011 1101 0011 0000 0011 0010 0001 0100 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 0011 1101 0011 0000 0011 0010 0001 0100 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 0011 1101 0011 0000 0011 0010 0001 0100 1001 =

0010 0111 1011 0100 0011 1101 0011 0000 0011 0010 0001 0100 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 0011 1101 0011 0000 0011 0010 0001 0100 1001


Decimal number -0.000 282 005 366 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0011 1101 0011 0000 0011 0010 0001 0100 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