-0.000 282 005 886 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 886(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
-0.000 282 005 886(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:

|-0.000 282 005 886| = 0.000 282 005 886


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: 0.000 282 005 886.

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) 0.000 282 005 886 × 2 = 0 + 0.000 564 011 772;
  • 2) 0.000 564 011 772 × 2 = 0 + 0.001 128 023 544;
  • 3) 0.001 128 023 544 × 2 = 0 + 0.002 256 047 088;
  • 4) 0.002 256 047 088 × 2 = 0 + 0.004 512 094 176;
  • 5) 0.004 512 094 176 × 2 = 0 + 0.009 024 188 352;
  • 6) 0.009 024 188 352 × 2 = 0 + 0.018 048 376 704;
  • 7) 0.018 048 376 704 × 2 = 0 + 0.036 096 753 408;
  • 8) 0.036 096 753 408 × 2 = 0 + 0.072 193 506 816;
  • 9) 0.072 193 506 816 × 2 = 0 + 0.144 387 013 632;
  • 10) 0.144 387 013 632 × 2 = 0 + 0.288 774 027 264;
  • 11) 0.288 774 027 264 × 2 = 0 + 0.577 548 054 528;
  • 12) 0.577 548 054 528 × 2 = 1 + 0.155 096 109 056;
  • 13) 0.155 096 109 056 × 2 = 0 + 0.310 192 218 112;
  • 14) 0.310 192 218 112 × 2 = 0 + 0.620 384 436 224;
  • 15) 0.620 384 436 224 × 2 = 1 + 0.240 768 872 448;
  • 16) 0.240 768 872 448 × 2 = 0 + 0.481 537 744 896;
  • 17) 0.481 537 744 896 × 2 = 0 + 0.963 075 489 792;
  • 18) 0.963 075 489 792 × 2 = 1 + 0.926 150 979 584;
  • 19) 0.926 150 979 584 × 2 = 1 + 0.852 301 959 168;
  • 20) 0.852 301 959 168 × 2 = 1 + 0.704 603 918 336;
  • 21) 0.704 603 918 336 × 2 = 1 + 0.409 207 836 672;
  • 22) 0.409 207 836 672 × 2 = 0 + 0.818 415 673 344;
  • 23) 0.818 415 673 344 × 2 = 1 + 0.636 831 346 688;
  • 24) 0.636 831 346 688 × 2 = 1 + 0.273 662 693 376;
  • 25) 0.273 662 693 376 × 2 = 0 + 0.547 325 386 752;
  • 26) 0.547 325 386 752 × 2 = 1 + 0.094 650 773 504;
  • 27) 0.094 650 773 504 × 2 = 0 + 0.189 301 547 008;
  • 28) 0.189 301 547 008 × 2 = 0 + 0.378 603 094 016;
  • 29) 0.378 603 094 016 × 2 = 0 + 0.757 206 188 032;
  • 30) 0.757 206 188 032 × 2 = 1 + 0.514 412 376 064;
  • 31) 0.514 412 376 064 × 2 = 1 + 0.028 824 752 128;
  • 32) 0.028 824 752 128 × 2 = 0 + 0.057 649 504 256;
  • 33) 0.057 649 504 256 × 2 = 0 + 0.115 299 008 512;
  • 34) 0.115 299 008 512 × 2 = 0 + 0.230 598 017 024;
  • 35) 0.230 598 017 024 × 2 = 0 + 0.461 196 034 048;
  • 36) 0.461 196 034 048 × 2 = 0 + 0.922 392 068 096;
  • 37) 0.922 392 068 096 × 2 = 1 + 0.844 784 136 192;
  • 38) 0.844 784 136 192 × 2 = 1 + 0.689 568 272 384;
  • 39) 0.689 568 272 384 × 2 = 1 + 0.379 136 544 768;
  • 40) 0.379 136 544 768 × 2 = 0 + 0.758 273 089 536;
  • 41) 0.758 273 089 536 × 2 = 1 + 0.516 546 179 072;
  • 42) 0.516 546 179 072 × 2 = 1 + 0.033 092 358 144;
  • 43) 0.033 092 358 144 × 2 = 0 + 0.066 184 716 288;
  • 44) 0.066 184 716 288 × 2 = 0 + 0.132 369 432 576;
  • 45) 0.132 369 432 576 × 2 = 0 + 0.264 738 865 152;
  • 46) 0.264 738 865 152 × 2 = 0 + 0.529 477 730 304;
  • 47) 0.529 477 730 304 × 2 = 1 + 0.058 955 460 608;
  • 48) 0.058 955 460 608 × 2 = 0 + 0.117 910 921 216;
  • 49) 0.117 910 921 216 × 2 = 0 + 0.235 821 842 432;
  • 50) 0.235 821 842 432 × 2 = 0 + 0.471 643 684 864;
  • 51) 0.471 643 684 864 × 2 = 0 + 0.943 287 369 728;
  • 52) 0.943 287 369 728 × 2 = 1 + 0.886 574 739 456;
  • 53) 0.886 574 739 456 × 2 = 1 + 0.773 149 478 912;
  • 54) 0.773 149 478 912 × 2 = 1 + 0.546 298 957 824;
  • 55) 0.546 298 957 824 × 2 = 1 + 0.092 597 915 648;
  • 56) 0.092 597 915 648 × 2 = 0 + 0.185 195 831 296;
  • 57) 0.185 195 831 296 × 2 = 0 + 0.370 391 662 592;
  • 58) 0.370 391 662 592 × 2 = 0 + 0.740 783 325 184;
  • 59) 0.740 783 325 184 × 2 = 1 + 0.481 566 650 368;
  • 60) 0.481 566 650 368 × 2 = 0 + 0.963 133 300 736;
  • 61) 0.963 133 300 736 × 2 = 1 + 0.926 266 601 472;
  • 62) 0.926 266 601 472 × 2 = 1 + 0.852 533 202 944;
  • 63) 0.852 533 202 944 × 2 = 1 + 0.705 066 405 888;
  • 64) 0.705 066 405 888 × 2 = 1 + 0.410 132 811 776;

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:


0.000 282 005 886(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1110 1100 0010 0001 1110 0010 1111(2)

6. Positive number before normalization:

0.000 282 005 886(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1110 1100 0010 0001 1110 0010 1111(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:


0.000 282 005 886(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1110 1100 0010 0001 1110 0010 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1110 1100 0010 0001 1110 0010 1111(2) × 20 =

1.0010 0111 1011 0100 0110 0000 1110 1100 0010 0001 1110 0010 1111(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 0000 1110 1100 0010 0001 1110 0010 1111


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 0000 1110 1100 0010 0001 1110 0010 1111 =

0010 0111 1011 0100 0110 0000 1110 1100 0010 0001 1110 0010 1111


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 0000 1110 1100 0010 0001 1110 0010 1111


Decimal number -0.000 282 005 886 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0000 1110 1100 0010 0001 1110 0010 1111

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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