-0.000 282 464 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 464(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
-0.000 282 464(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:

|-0.000 282 464| = 0.000 282 464


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: 0.000 282 464.

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) 0.000 282 464 × 2 = 0 + 0.000 564 928;
  • 2) 0.000 564 928 × 2 = 0 + 0.001 129 856;
  • 3) 0.001 129 856 × 2 = 0 + 0.002 259 712;
  • 4) 0.002 259 712 × 2 = 0 + 0.004 519 424;
  • 5) 0.004 519 424 × 2 = 0 + 0.009 038 848;
  • 6) 0.009 038 848 × 2 = 0 + 0.018 077 696;
  • 7) 0.018 077 696 × 2 = 0 + 0.036 155 392;
  • 8) 0.036 155 392 × 2 = 0 + 0.072 310 784;
  • 9) 0.072 310 784 × 2 = 0 + 0.144 621 568;
  • 10) 0.144 621 568 × 2 = 0 + 0.289 243 136;
  • 11) 0.289 243 136 × 2 = 0 + 0.578 486 272;
  • 12) 0.578 486 272 × 2 = 1 + 0.156 972 544;
  • 13) 0.156 972 544 × 2 = 0 + 0.313 945 088;
  • 14) 0.313 945 088 × 2 = 0 + 0.627 890 176;
  • 15) 0.627 890 176 × 2 = 1 + 0.255 780 352;
  • 16) 0.255 780 352 × 2 = 0 + 0.511 560 704;
  • 17) 0.511 560 704 × 2 = 1 + 0.023 121 408;
  • 18) 0.023 121 408 × 2 = 0 + 0.046 242 816;
  • 19) 0.046 242 816 × 2 = 0 + 0.092 485 632;
  • 20) 0.092 485 632 × 2 = 0 + 0.184 971 264;
  • 21) 0.184 971 264 × 2 = 0 + 0.369 942 528;
  • 22) 0.369 942 528 × 2 = 0 + 0.739 885 056;
  • 23) 0.739 885 056 × 2 = 1 + 0.479 770 112;
  • 24) 0.479 770 112 × 2 = 0 + 0.959 540 224;
  • 25) 0.959 540 224 × 2 = 1 + 0.919 080 448;
  • 26) 0.919 080 448 × 2 = 1 + 0.838 160 896;
  • 27) 0.838 160 896 × 2 = 1 + 0.676 321 792;
  • 28) 0.676 321 792 × 2 = 1 + 0.352 643 584;
  • 29) 0.352 643 584 × 2 = 0 + 0.705 287 168;
  • 30) 0.705 287 168 × 2 = 1 + 0.410 574 336;
  • 31) 0.410 574 336 × 2 = 0 + 0.821 148 672;
  • 32) 0.821 148 672 × 2 = 1 + 0.642 297 344;
  • 33) 0.642 297 344 × 2 = 1 + 0.284 594 688;
  • 34) 0.284 594 688 × 2 = 0 + 0.569 189 376;
  • 35) 0.569 189 376 × 2 = 1 + 0.138 378 752;
  • 36) 0.138 378 752 × 2 = 0 + 0.276 757 504;
  • 37) 0.276 757 504 × 2 = 0 + 0.553 515 008;
  • 38) 0.553 515 008 × 2 = 1 + 0.107 030 016;
  • 39) 0.107 030 016 × 2 = 0 + 0.214 060 032;
  • 40) 0.214 060 032 × 2 = 0 + 0.428 120 064;
  • 41) 0.428 120 064 × 2 = 0 + 0.856 240 128;
  • 42) 0.856 240 128 × 2 = 1 + 0.712 480 256;
  • 43) 0.712 480 256 × 2 = 1 + 0.424 960 512;
  • 44) 0.424 960 512 × 2 = 0 + 0.849 921 024;
  • 45) 0.849 921 024 × 2 = 1 + 0.699 842 048;
  • 46) 0.699 842 048 × 2 = 1 + 0.399 684 096;
  • 47) 0.399 684 096 × 2 = 0 + 0.799 368 192;
  • 48) 0.799 368 192 × 2 = 1 + 0.598 736 384;
  • 49) 0.598 736 384 × 2 = 1 + 0.197 472 768;
  • 50) 0.197 472 768 × 2 = 0 + 0.394 945 536;
  • 51) 0.394 945 536 × 2 = 0 + 0.789 891 072;
  • 52) 0.789 891 072 × 2 = 1 + 0.579 782 144;
  • 53) 0.579 782 144 × 2 = 1 + 0.159 564 288;
  • 54) 0.159 564 288 × 2 = 0 + 0.319 128 576;
  • 55) 0.319 128 576 × 2 = 0 + 0.638 257 152;
  • 56) 0.638 257 152 × 2 = 1 + 0.276 514 304;
  • 57) 0.276 514 304 × 2 = 0 + 0.553 028 608;
  • 58) 0.553 028 608 × 2 = 1 + 0.106 057 216;
  • 59) 0.106 057 216 × 2 = 0 + 0.212 114 432;
  • 60) 0.212 114 432 × 2 = 0 + 0.424 228 864;
  • 61) 0.424 228 864 × 2 = 0 + 0.848 457 728;
  • 62) 0.848 457 728 × 2 = 1 + 0.696 915 456;
  • 63) 0.696 915 456 × 2 = 1 + 0.393 830 912;
  • 64) 0.393 830 912 × 2 = 0 + 0.787 661 824;

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:


0.000 282 464(10) =

0.0000 0000 0001 0010 1000 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 0110(2)

6. Positive number before normalization:

0.000 282 464(10) =

0.0000 0000 0001 0010 1000 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 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:


0.000 282 464(10) =

0.0000 0000 0001 0010 1000 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 0110(2) =

0.0000 0000 0001 0010 1000 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 0110(2) × 20 =

1.0010 1000 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 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 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 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 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 0110 =

0010 1000 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 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 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 0110


Decimal number -0.000 282 464 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0010 1111 0101 1010 0100 0110 1101 1001 1001 0100 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