-0.000 282 678 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 678(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
-0.000 282 678(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:

|-0.000 282 678| = 0.000 282 678


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: 0.000 282 678.

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) 0.000 282 678 × 2 = 0 + 0.000 565 356;
  • 2) 0.000 565 356 × 2 = 0 + 0.001 130 712;
  • 3) 0.001 130 712 × 2 = 0 + 0.002 261 424;
  • 4) 0.002 261 424 × 2 = 0 + 0.004 522 848;
  • 5) 0.004 522 848 × 2 = 0 + 0.009 045 696;
  • 6) 0.009 045 696 × 2 = 0 + 0.018 091 392;
  • 7) 0.018 091 392 × 2 = 0 + 0.036 182 784;
  • 8) 0.036 182 784 × 2 = 0 + 0.072 365 568;
  • 9) 0.072 365 568 × 2 = 0 + 0.144 731 136;
  • 10) 0.144 731 136 × 2 = 0 + 0.289 462 272;
  • 11) 0.289 462 272 × 2 = 0 + 0.578 924 544;
  • 12) 0.578 924 544 × 2 = 1 + 0.157 849 088;
  • 13) 0.157 849 088 × 2 = 0 + 0.315 698 176;
  • 14) 0.315 698 176 × 2 = 0 + 0.631 396 352;
  • 15) 0.631 396 352 × 2 = 1 + 0.262 792 704;
  • 16) 0.262 792 704 × 2 = 0 + 0.525 585 408;
  • 17) 0.525 585 408 × 2 = 1 + 0.051 170 816;
  • 18) 0.051 170 816 × 2 = 0 + 0.102 341 632;
  • 19) 0.102 341 632 × 2 = 0 + 0.204 683 264;
  • 20) 0.204 683 264 × 2 = 0 + 0.409 366 528;
  • 21) 0.409 366 528 × 2 = 0 + 0.818 733 056;
  • 22) 0.818 733 056 × 2 = 1 + 0.637 466 112;
  • 23) 0.637 466 112 × 2 = 1 + 0.274 932 224;
  • 24) 0.274 932 224 × 2 = 0 + 0.549 864 448;
  • 25) 0.549 864 448 × 2 = 1 + 0.099 728 896;
  • 26) 0.099 728 896 × 2 = 0 + 0.199 457 792;
  • 27) 0.199 457 792 × 2 = 0 + 0.398 915 584;
  • 28) 0.398 915 584 × 2 = 0 + 0.797 831 168;
  • 29) 0.797 831 168 × 2 = 1 + 0.595 662 336;
  • 30) 0.595 662 336 × 2 = 1 + 0.191 324 672;
  • 31) 0.191 324 672 × 2 = 0 + 0.382 649 344;
  • 32) 0.382 649 344 × 2 = 0 + 0.765 298 688;
  • 33) 0.765 298 688 × 2 = 1 + 0.530 597 376;
  • 34) 0.530 597 376 × 2 = 1 + 0.061 194 752;
  • 35) 0.061 194 752 × 2 = 0 + 0.122 389 504;
  • 36) 0.122 389 504 × 2 = 0 + 0.244 779 008;
  • 37) 0.244 779 008 × 2 = 0 + 0.489 558 016;
  • 38) 0.489 558 016 × 2 = 0 + 0.979 116 032;
  • 39) 0.979 116 032 × 2 = 1 + 0.958 232 064;
  • 40) 0.958 232 064 × 2 = 1 + 0.916 464 128;
  • 41) 0.916 464 128 × 2 = 1 + 0.832 928 256;
  • 42) 0.832 928 256 × 2 = 1 + 0.665 856 512;
  • 43) 0.665 856 512 × 2 = 1 + 0.331 713 024;
  • 44) 0.331 713 024 × 2 = 0 + 0.663 426 048;
  • 45) 0.663 426 048 × 2 = 1 + 0.326 852 096;
  • 46) 0.326 852 096 × 2 = 0 + 0.653 704 192;
  • 47) 0.653 704 192 × 2 = 1 + 0.307 408 384;
  • 48) 0.307 408 384 × 2 = 0 + 0.614 816 768;
  • 49) 0.614 816 768 × 2 = 1 + 0.229 633 536;
  • 50) 0.229 633 536 × 2 = 0 + 0.459 267 072;
  • 51) 0.459 267 072 × 2 = 0 + 0.918 534 144;
  • 52) 0.918 534 144 × 2 = 1 + 0.837 068 288;
  • 53) 0.837 068 288 × 2 = 1 + 0.674 136 576;
  • 54) 0.674 136 576 × 2 = 1 + 0.348 273 152;
  • 55) 0.348 273 152 × 2 = 0 + 0.696 546 304;
  • 56) 0.696 546 304 × 2 = 1 + 0.393 092 608;
  • 57) 0.393 092 608 × 2 = 0 + 0.786 185 216;
  • 58) 0.786 185 216 × 2 = 1 + 0.572 370 432;
  • 59) 0.572 370 432 × 2 = 1 + 0.144 740 864;
  • 60) 0.144 740 864 × 2 = 0 + 0.289 481 728;
  • 61) 0.289 481 728 × 2 = 0 + 0.578 963 456;
  • 62) 0.578 963 456 × 2 = 1 + 0.157 926 912;
  • 63) 0.157 926 912 × 2 = 0 + 0.315 853 824;
  • 64) 0.315 853 824 × 2 = 0 + 0.631 707 648;

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:


0.000 282 678(10) =

0.0000 0000 0001 0010 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100(2)

6. Positive number before normalization:

0.000 282 678(10) =

0.0000 0000 0001 0010 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100(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:


0.000 282 678(10) =

0.0000 0000 0001 0010 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100(2) =

0.0000 0000 0001 0010 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100(2) × 20 =

1.0010 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100(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 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100


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 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100 =

0010 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100


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 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100


Decimal number -0.000 282 678 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0110 1000 1100 1100 0011 1110 1010 1001 1101 0110 0100

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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