-0.000 282 487 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 487(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
-0.000 282 487(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:

|-0.000 282 487| = 0.000 282 487


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: 0.000 282 487.

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) 0.000 282 487 × 2 = 0 + 0.000 564 974;
  • 2) 0.000 564 974 × 2 = 0 + 0.001 129 948;
  • 3) 0.001 129 948 × 2 = 0 + 0.002 259 896;
  • 4) 0.002 259 896 × 2 = 0 + 0.004 519 792;
  • 5) 0.004 519 792 × 2 = 0 + 0.009 039 584;
  • 6) 0.009 039 584 × 2 = 0 + 0.018 079 168;
  • 7) 0.018 079 168 × 2 = 0 + 0.036 158 336;
  • 8) 0.036 158 336 × 2 = 0 + 0.072 316 672;
  • 9) 0.072 316 672 × 2 = 0 + 0.144 633 344;
  • 10) 0.144 633 344 × 2 = 0 + 0.289 266 688;
  • 11) 0.289 266 688 × 2 = 0 + 0.578 533 376;
  • 12) 0.578 533 376 × 2 = 1 + 0.157 066 752;
  • 13) 0.157 066 752 × 2 = 0 + 0.314 133 504;
  • 14) 0.314 133 504 × 2 = 0 + 0.628 267 008;
  • 15) 0.628 267 008 × 2 = 1 + 0.256 534 016;
  • 16) 0.256 534 016 × 2 = 0 + 0.513 068 032;
  • 17) 0.513 068 032 × 2 = 1 + 0.026 136 064;
  • 18) 0.026 136 064 × 2 = 0 + 0.052 272 128;
  • 19) 0.052 272 128 × 2 = 0 + 0.104 544 256;
  • 20) 0.104 544 256 × 2 = 0 + 0.209 088 512;
  • 21) 0.209 088 512 × 2 = 0 + 0.418 177 024;
  • 22) 0.418 177 024 × 2 = 0 + 0.836 354 048;
  • 23) 0.836 354 048 × 2 = 1 + 0.672 708 096;
  • 24) 0.672 708 096 × 2 = 1 + 0.345 416 192;
  • 25) 0.345 416 192 × 2 = 0 + 0.690 832 384;
  • 26) 0.690 832 384 × 2 = 1 + 0.381 664 768;
  • 27) 0.381 664 768 × 2 = 0 + 0.763 329 536;
  • 28) 0.763 329 536 × 2 = 1 + 0.526 659 072;
  • 29) 0.526 659 072 × 2 = 1 + 0.053 318 144;
  • 30) 0.053 318 144 × 2 = 0 + 0.106 636 288;
  • 31) 0.106 636 288 × 2 = 0 + 0.213 272 576;
  • 32) 0.213 272 576 × 2 = 0 + 0.426 545 152;
  • 33) 0.426 545 152 × 2 = 0 + 0.853 090 304;
  • 34) 0.853 090 304 × 2 = 1 + 0.706 180 608;
  • 35) 0.706 180 608 × 2 = 1 + 0.412 361 216;
  • 36) 0.412 361 216 × 2 = 0 + 0.824 722 432;
  • 37) 0.824 722 432 × 2 = 1 + 0.649 444 864;
  • 38) 0.649 444 864 × 2 = 1 + 0.298 889 728;
  • 39) 0.298 889 728 × 2 = 0 + 0.597 779 456;
  • 40) 0.597 779 456 × 2 = 1 + 0.195 558 912;
  • 41) 0.195 558 912 × 2 = 0 + 0.391 117 824;
  • 42) 0.391 117 824 × 2 = 0 + 0.782 235 648;
  • 43) 0.782 235 648 × 2 = 1 + 0.564 471 296;
  • 44) 0.564 471 296 × 2 = 1 + 0.128 942 592;
  • 45) 0.128 942 592 × 2 = 0 + 0.257 885 184;
  • 46) 0.257 885 184 × 2 = 0 + 0.515 770 368;
  • 47) 0.515 770 368 × 2 = 1 + 0.031 540 736;
  • 48) 0.031 540 736 × 2 = 0 + 0.063 081 472;
  • 49) 0.063 081 472 × 2 = 0 + 0.126 162 944;
  • 50) 0.126 162 944 × 2 = 0 + 0.252 325 888;
  • 51) 0.252 325 888 × 2 = 0 + 0.504 651 776;
  • 52) 0.504 651 776 × 2 = 1 + 0.009 303 552;
  • 53) 0.009 303 552 × 2 = 0 + 0.018 607 104;
  • 54) 0.018 607 104 × 2 = 0 + 0.037 214 208;
  • 55) 0.037 214 208 × 2 = 0 + 0.074 428 416;
  • 56) 0.074 428 416 × 2 = 0 + 0.148 856 832;
  • 57) 0.148 856 832 × 2 = 0 + 0.297 713 664;
  • 58) 0.297 713 664 × 2 = 0 + 0.595 427 328;
  • 59) 0.595 427 328 × 2 = 1 + 0.190 854 656;
  • 60) 0.190 854 656 × 2 = 0 + 0.381 709 312;
  • 61) 0.381 709 312 × 2 = 0 + 0.763 418 624;
  • 62) 0.763 418 624 × 2 = 1 + 0.526 837 248;
  • 63) 0.526 837 248 × 2 = 1 + 0.053 674 496;
  • 64) 0.053 674 496 × 2 = 0 + 0.107 348 992;

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:


0.000 282 487(10) =

0.0000 0000 0001 0010 1000 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110(2)

6. Positive number before normalization:

0.000 282 487(10) =

0.0000 0000 0001 0010 1000 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110(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:


0.000 282 487(10) =

0.0000 0000 0001 0010 1000 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110(2) =

0.0000 0000 0001 0010 1000 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110(2) × 20 =

1.0010 1000 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110(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 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110


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 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110 =

0010 1000 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110


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 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110


Decimal number -0.000 282 487 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0011 0101 1000 0110 1101 0011 0010 0001 0000 0010 0110

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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