-0.000 282 005 469 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 469(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
-0.000 282 005 469(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:

|-0.000 282 005 469| = 0.000 282 005 469


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: 0.000 282 005 469.

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) 0.000 282 005 469 × 2 = 0 + 0.000 564 010 938;
  • 2) 0.000 564 010 938 × 2 = 0 + 0.001 128 021 876;
  • 3) 0.001 128 021 876 × 2 = 0 + 0.002 256 043 752;
  • 4) 0.002 256 043 752 × 2 = 0 + 0.004 512 087 504;
  • 5) 0.004 512 087 504 × 2 = 0 + 0.009 024 175 008;
  • 6) 0.009 024 175 008 × 2 = 0 + 0.018 048 350 016;
  • 7) 0.018 048 350 016 × 2 = 0 + 0.036 096 700 032;
  • 8) 0.036 096 700 032 × 2 = 0 + 0.072 193 400 064;
  • 9) 0.072 193 400 064 × 2 = 0 + 0.144 386 800 128;
  • 10) 0.144 386 800 128 × 2 = 0 + 0.288 773 600 256;
  • 11) 0.288 773 600 256 × 2 = 0 + 0.577 547 200 512;
  • 12) 0.577 547 200 512 × 2 = 1 + 0.155 094 401 024;
  • 13) 0.155 094 401 024 × 2 = 0 + 0.310 188 802 048;
  • 14) 0.310 188 802 048 × 2 = 0 + 0.620 377 604 096;
  • 15) 0.620 377 604 096 × 2 = 1 + 0.240 755 208 192;
  • 16) 0.240 755 208 192 × 2 = 0 + 0.481 510 416 384;
  • 17) 0.481 510 416 384 × 2 = 0 + 0.963 020 832 768;
  • 18) 0.963 020 832 768 × 2 = 1 + 0.926 041 665 536;
  • 19) 0.926 041 665 536 × 2 = 1 + 0.852 083 331 072;
  • 20) 0.852 083 331 072 × 2 = 1 + 0.704 166 662 144;
  • 21) 0.704 166 662 144 × 2 = 1 + 0.408 333 324 288;
  • 22) 0.408 333 324 288 × 2 = 0 + 0.816 666 648 576;
  • 23) 0.816 666 648 576 × 2 = 1 + 0.633 333 297 152;
  • 24) 0.633 333 297 152 × 2 = 1 + 0.266 666 594 304;
  • 25) 0.266 666 594 304 × 2 = 0 + 0.533 333 188 608;
  • 26) 0.533 333 188 608 × 2 = 1 + 0.066 666 377 216;
  • 27) 0.066 666 377 216 × 2 = 0 + 0.133 332 754 432;
  • 28) 0.133 332 754 432 × 2 = 0 + 0.266 665 508 864;
  • 29) 0.266 665 508 864 × 2 = 0 + 0.533 331 017 728;
  • 30) 0.533 331 017 728 × 2 = 1 + 0.066 662 035 456;
  • 31) 0.066 662 035 456 × 2 = 0 + 0.133 324 070 912;
  • 32) 0.133 324 070 912 × 2 = 0 + 0.266 648 141 824;
  • 33) 0.266 648 141 824 × 2 = 0 + 0.533 296 283 648;
  • 34) 0.533 296 283 648 × 2 = 1 + 0.066 592 567 296;
  • 35) 0.066 592 567 296 × 2 = 0 + 0.133 185 134 592;
  • 36) 0.133 185 134 592 × 2 = 0 + 0.266 370 269 184;
  • 37) 0.266 370 269 184 × 2 = 0 + 0.532 740 538 368;
  • 38) 0.532 740 538 368 × 2 = 1 + 0.065 481 076 736;
  • 39) 0.065 481 076 736 × 2 = 0 + 0.130 962 153 472;
  • 40) 0.130 962 153 472 × 2 = 0 + 0.261 924 306 944;
  • 41) 0.261 924 306 944 × 2 = 0 + 0.523 848 613 888;
  • 42) 0.523 848 613 888 × 2 = 1 + 0.047 697 227 776;
  • 43) 0.047 697 227 776 × 2 = 0 + 0.095 394 455 552;
  • 44) 0.095 394 455 552 × 2 = 0 + 0.190 788 911 104;
  • 45) 0.190 788 911 104 × 2 = 0 + 0.381 577 822 208;
  • 46) 0.381 577 822 208 × 2 = 0 + 0.763 155 644 416;
  • 47) 0.763 155 644 416 × 2 = 1 + 0.526 311 288 832;
  • 48) 0.526 311 288 832 × 2 = 1 + 0.052 622 577 664;
  • 49) 0.052 622 577 664 × 2 = 0 + 0.105 245 155 328;
  • 50) 0.105 245 155 328 × 2 = 0 + 0.210 490 310 656;
  • 51) 0.210 490 310 656 × 2 = 0 + 0.420 980 621 312;
  • 52) 0.420 980 621 312 × 2 = 0 + 0.841 961 242 624;
  • 53) 0.841 961 242 624 × 2 = 1 + 0.683 922 485 248;
  • 54) 0.683 922 485 248 × 2 = 1 + 0.367 844 970 496;
  • 55) 0.367 844 970 496 × 2 = 0 + 0.735 689 940 992;
  • 56) 0.735 689 940 992 × 2 = 1 + 0.471 379 881 984;
  • 57) 0.471 379 881 984 × 2 = 0 + 0.942 759 763 968;
  • 58) 0.942 759 763 968 × 2 = 1 + 0.885 519 527 936;
  • 59) 0.885 519 527 936 × 2 = 1 + 0.771 039 055 872;
  • 60) 0.771 039 055 872 × 2 = 1 + 0.542 078 111 744;
  • 61) 0.542 078 111 744 × 2 = 1 + 0.084 156 223 488;
  • 62) 0.084 156 223 488 × 2 = 0 + 0.168 312 446 976;
  • 63) 0.168 312 446 976 × 2 = 0 + 0.336 624 893 952;
  • 64) 0.336 624 893 952 × 2 = 0 + 0.673 249 787 904;

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:


0.000 282 005 469(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0100 0100 0100 0011 0000 1101 0111 1000(2)

6. Positive number before normalization:

0.000 282 005 469(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0100 0100 0100 0011 0000 1101 0111 1000(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:


0.000 282 005 469(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0100 0100 0100 0011 0000 1101 0111 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 0100 0100 0100 0011 0000 1101 0111 1000(2) × 20 =

1.0010 0111 1011 0100 0100 0100 0100 0100 0011 0000 1101 0111 1000(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 0100 0100 0100 0100 0011 0000 1101 0111 1000


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 0100 0100 0100 0100 0011 0000 1101 0111 1000 =

0010 0111 1011 0100 0100 0100 0100 0100 0011 0000 1101 0111 1000


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 0100 0100 0100 0100 0011 0000 1101 0111 1000


Decimal number -0.000 282 005 469 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 0100 0100 0100 0011 0000 1101 0111 1000

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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