-0.000 282 006 223 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 223(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
-0.000 282 006 223(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:

|-0.000 282 006 223| = 0.000 282 006 223


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: 0.000 282 006 223.

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) 0.000 282 006 223 × 2 = 0 + 0.000 564 012 446;
  • 2) 0.000 564 012 446 × 2 = 0 + 0.001 128 024 892;
  • 3) 0.001 128 024 892 × 2 = 0 + 0.002 256 049 784;
  • 4) 0.002 256 049 784 × 2 = 0 + 0.004 512 099 568;
  • 5) 0.004 512 099 568 × 2 = 0 + 0.009 024 199 136;
  • 6) 0.009 024 199 136 × 2 = 0 + 0.018 048 398 272;
  • 7) 0.018 048 398 272 × 2 = 0 + 0.036 096 796 544;
  • 8) 0.036 096 796 544 × 2 = 0 + 0.072 193 593 088;
  • 9) 0.072 193 593 088 × 2 = 0 + 0.144 387 186 176;
  • 10) 0.144 387 186 176 × 2 = 0 + 0.288 774 372 352;
  • 11) 0.288 774 372 352 × 2 = 0 + 0.577 548 744 704;
  • 12) 0.577 548 744 704 × 2 = 1 + 0.155 097 489 408;
  • 13) 0.155 097 489 408 × 2 = 0 + 0.310 194 978 816;
  • 14) 0.310 194 978 816 × 2 = 0 + 0.620 389 957 632;
  • 15) 0.620 389 957 632 × 2 = 1 + 0.240 779 915 264;
  • 16) 0.240 779 915 264 × 2 = 0 + 0.481 559 830 528;
  • 17) 0.481 559 830 528 × 2 = 0 + 0.963 119 661 056;
  • 18) 0.963 119 661 056 × 2 = 1 + 0.926 239 322 112;
  • 19) 0.926 239 322 112 × 2 = 1 + 0.852 478 644 224;
  • 20) 0.852 478 644 224 × 2 = 1 + 0.704 957 288 448;
  • 21) 0.704 957 288 448 × 2 = 1 + 0.409 914 576 896;
  • 22) 0.409 914 576 896 × 2 = 0 + 0.819 829 153 792;
  • 23) 0.819 829 153 792 × 2 = 1 + 0.639 658 307 584;
  • 24) 0.639 658 307 584 × 2 = 1 + 0.279 316 615 168;
  • 25) 0.279 316 615 168 × 2 = 0 + 0.558 633 230 336;
  • 26) 0.558 633 230 336 × 2 = 1 + 0.117 266 460 672;
  • 27) 0.117 266 460 672 × 2 = 0 + 0.234 532 921 344;
  • 28) 0.234 532 921 344 × 2 = 0 + 0.469 065 842 688;
  • 29) 0.469 065 842 688 × 2 = 0 + 0.938 131 685 376;
  • 30) 0.938 131 685 376 × 2 = 1 + 0.876 263 370 752;
  • 31) 0.876 263 370 752 × 2 = 1 + 0.752 526 741 504;
  • 32) 0.752 526 741 504 × 2 = 1 + 0.505 053 483 008;
  • 33) 0.505 053 483 008 × 2 = 1 + 0.010 106 966 016;
  • 34) 0.010 106 966 016 × 2 = 0 + 0.020 213 932 032;
  • 35) 0.020 213 932 032 × 2 = 0 + 0.040 427 864 064;
  • 36) 0.040 427 864 064 × 2 = 0 + 0.080 855 728 128;
  • 37) 0.080 855 728 128 × 2 = 0 + 0.161 711 456 256;
  • 38) 0.161 711 456 256 × 2 = 0 + 0.323 422 912 512;
  • 39) 0.323 422 912 512 × 2 = 0 + 0.646 845 825 024;
  • 40) 0.646 845 825 024 × 2 = 1 + 0.293 691 650 048;
  • 41) 0.293 691 650 048 × 2 = 0 + 0.587 383 300 096;
  • 42) 0.587 383 300 096 × 2 = 1 + 0.174 766 600 192;
  • 43) 0.174 766 600 192 × 2 = 0 + 0.349 533 200 384;
  • 44) 0.349 533 200 384 × 2 = 0 + 0.699 066 400 768;
  • 45) 0.699 066 400 768 × 2 = 1 + 0.398 132 801 536;
  • 46) 0.398 132 801 536 × 2 = 0 + 0.796 265 603 072;
  • 47) 0.796 265 603 072 × 2 = 1 + 0.592 531 206 144;
  • 48) 0.592 531 206 144 × 2 = 1 + 0.185 062 412 288;
  • 49) 0.185 062 412 288 × 2 = 0 + 0.370 124 824 576;
  • 50) 0.370 124 824 576 × 2 = 0 + 0.740 249 649 152;
  • 51) 0.740 249 649 152 × 2 = 1 + 0.480 499 298 304;
  • 52) 0.480 499 298 304 × 2 = 0 + 0.960 998 596 608;
  • 53) 0.960 998 596 608 × 2 = 1 + 0.921 997 193 216;
  • 54) 0.921 997 193 216 × 2 = 1 + 0.843 994 386 432;
  • 55) 0.843 994 386 432 × 2 = 1 + 0.687 988 772 864;
  • 56) 0.687 988 772 864 × 2 = 1 + 0.375 977 545 728;
  • 57) 0.375 977 545 728 × 2 = 0 + 0.751 955 091 456;
  • 58) 0.751 955 091 456 × 2 = 1 + 0.503 910 182 912;
  • 59) 0.503 910 182 912 × 2 = 1 + 0.007 820 365 824;
  • 60) 0.007 820 365 824 × 2 = 0 + 0.015 640 731 648;
  • 61) 0.015 640 731 648 × 2 = 0 + 0.031 281 463 296;
  • 62) 0.031 281 463 296 × 2 = 0 + 0.062 562 926 592;
  • 63) 0.062 562 926 592 × 2 = 0 + 0.125 125 853 184;
  • 64) 0.125 125 853 184 × 2 = 0 + 0.250 251 706 368;

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:


0.000 282 006 223(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1000 0001 0100 1011 0010 1111 0110 0000(2)

6. Positive number before normalization:

0.000 282 006 223(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1000 0001 0100 1011 0010 1111 0110 0000(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:


0.000 282 006 223(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1000 0001 0100 1011 0010 1111 0110 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1000 0001 0100 1011 0010 1111 0110 0000(2) × 20 =

1.0010 0111 1011 0100 0111 1000 0001 0100 1011 0010 1111 0110 0000(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 0111 1000 0001 0100 1011 0010 1111 0110 0000


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 0111 1000 0001 0100 1011 0010 1111 0110 0000 =

0010 0111 1011 0100 0111 1000 0001 0100 1011 0010 1111 0110 0000


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 0111 1000 0001 0100 1011 0010 1111 0110 0000


Decimal number -0.000 282 006 223 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1000 0001 0100 1011 0010 1111 0110 0000

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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