-0.000 282 005 914 366 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 366(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
-0.000 282 005 914 366(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:

|-0.000 282 005 914 366| = 0.000 282 005 914 366


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: 0.000 282 005 914 366.

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) 0.000 282 005 914 366 × 2 = 0 + 0.000 564 011 828 732;
  • 2) 0.000 564 011 828 732 × 2 = 0 + 0.001 128 023 657 464;
  • 3) 0.001 128 023 657 464 × 2 = 0 + 0.002 256 047 314 928;
  • 4) 0.002 256 047 314 928 × 2 = 0 + 0.004 512 094 629 856;
  • 5) 0.004 512 094 629 856 × 2 = 0 + 0.009 024 189 259 712;
  • 6) 0.009 024 189 259 712 × 2 = 0 + 0.018 048 378 519 424;
  • 7) 0.018 048 378 519 424 × 2 = 0 + 0.036 096 757 038 848;
  • 8) 0.036 096 757 038 848 × 2 = 0 + 0.072 193 514 077 696;
  • 9) 0.072 193 514 077 696 × 2 = 0 + 0.144 387 028 155 392;
  • 10) 0.144 387 028 155 392 × 2 = 0 + 0.288 774 056 310 784;
  • 11) 0.288 774 056 310 784 × 2 = 0 + 0.577 548 112 621 568;
  • 12) 0.577 548 112 621 568 × 2 = 1 + 0.155 096 225 243 136;
  • 13) 0.155 096 225 243 136 × 2 = 0 + 0.310 192 450 486 272;
  • 14) 0.310 192 450 486 272 × 2 = 0 + 0.620 384 900 972 544;
  • 15) 0.620 384 900 972 544 × 2 = 1 + 0.240 769 801 945 088;
  • 16) 0.240 769 801 945 088 × 2 = 0 + 0.481 539 603 890 176;
  • 17) 0.481 539 603 890 176 × 2 = 0 + 0.963 079 207 780 352;
  • 18) 0.963 079 207 780 352 × 2 = 1 + 0.926 158 415 560 704;
  • 19) 0.926 158 415 560 704 × 2 = 1 + 0.852 316 831 121 408;
  • 20) 0.852 316 831 121 408 × 2 = 1 + 0.704 633 662 242 816;
  • 21) 0.704 633 662 242 816 × 2 = 1 + 0.409 267 324 485 632;
  • 22) 0.409 267 324 485 632 × 2 = 0 + 0.818 534 648 971 264;
  • 23) 0.818 534 648 971 264 × 2 = 1 + 0.637 069 297 942 528;
  • 24) 0.637 069 297 942 528 × 2 = 1 + 0.274 138 595 885 056;
  • 25) 0.274 138 595 885 056 × 2 = 0 + 0.548 277 191 770 112;
  • 26) 0.548 277 191 770 112 × 2 = 1 + 0.096 554 383 540 224;
  • 27) 0.096 554 383 540 224 × 2 = 0 + 0.193 108 767 080 448;
  • 28) 0.193 108 767 080 448 × 2 = 0 + 0.386 217 534 160 896;
  • 29) 0.386 217 534 160 896 × 2 = 0 + 0.772 435 068 321 792;
  • 30) 0.772 435 068 321 792 × 2 = 1 + 0.544 870 136 643 584;
  • 31) 0.544 870 136 643 584 × 2 = 1 + 0.089 740 273 287 168;
  • 32) 0.089 740 273 287 168 × 2 = 0 + 0.179 480 546 574 336;
  • 33) 0.179 480 546 574 336 × 2 = 0 + 0.358 961 093 148 672;
  • 34) 0.358 961 093 148 672 × 2 = 0 + 0.717 922 186 297 344;
  • 35) 0.717 922 186 297 344 × 2 = 1 + 0.435 844 372 594 688;
  • 36) 0.435 844 372 594 688 × 2 = 0 + 0.871 688 745 189 376;
  • 37) 0.871 688 745 189 376 × 2 = 1 + 0.743 377 490 378 752;
  • 38) 0.743 377 490 378 752 × 2 = 1 + 0.486 754 980 757 504;
  • 39) 0.486 754 980 757 504 × 2 = 0 + 0.973 509 961 515 008;
  • 40) 0.973 509 961 515 008 × 2 = 1 + 0.947 019 923 030 016;
  • 41) 0.947 019 923 030 016 × 2 = 1 + 0.894 039 846 060 032;
  • 42) 0.894 039 846 060 032 × 2 = 1 + 0.788 079 692 120 064;
  • 43) 0.788 079 692 120 064 × 2 = 1 + 0.576 159 384 240 128;
  • 44) 0.576 159 384 240 128 × 2 = 1 + 0.152 318 768 480 256;
  • 45) 0.152 318 768 480 256 × 2 = 0 + 0.304 637 536 960 512;
  • 46) 0.304 637 536 960 512 × 2 = 0 + 0.609 275 073 921 024;
  • 47) 0.609 275 073 921 024 × 2 = 1 + 0.218 550 147 842 048;
  • 48) 0.218 550 147 842 048 × 2 = 0 + 0.437 100 295 684 096;
  • 49) 0.437 100 295 684 096 × 2 = 0 + 0.874 200 591 368 192;
  • 50) 0.874 200 591 368 192 × 2 = 1 + 0.748 401 182 736 384;
  • 51) 0.748 401 182 736 384 × 2 = 1 + 0.496 802 365 472 768;
  • 52) 0.496 802 365 472 768 × 2 = 0 + 0.993 604 730 945 536;
  • 53) 0.993 604 730 945 536 × 2 = 1 + 0.987 209 461 891 072;
  • 54) 0.987 209 461 891 072 × 2 = 1 + 0.974 418 923 782 144;
  • 55) 0.974 418 923 782 144 × 2 = 1 + 0.948 837 847 564 288;
  • 56) 0.948 837 847 564 288 × 2 = 1 + 0.897 675 695 128 576;
  • 57) 0.897 675 695 128 576 × 2 = 1 + 0.795 351 390 257 152;
  • 58) 0.795 351 390 257 152 × 2 = 1 + 0.590 702 780 514 304;
  • 59) 0.590 702 780 514 304 × 2 = 1 + 0.181 405 561 028 608;
  • 60) 0.181 405 561 028 608 × 2 = 0 + 0.362 811 122 057 216;
  • 61) 0.362 811 122 057 216 × 2 = 0 + 0.725 622 244 114 432;
  • 62) 0.725 622 244 114 432 × 2 = 1 + 0.451 244 488 228 864;
  • 63) 0.451 244 488 228 864 × 2 = 0 + 0.902 488 976 457 728;
  • 64) 0.902 488 976 457 728 × 2 = 1 + 0.804 977 952 915 456;

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:


0.000 282 005 914 366(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0010 0110 1111 1110 0101(2)

6. Positive number before normalization:

0.000 282 005 914 366(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0010 0110 1111 1110 0101(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:


0.000 282 005 914 366(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0010 0110 1111 1110 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0010 0110 1111 1110 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 0010 0110 1111 1110 0101(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 0110 0010 1101 1111 0010 0110 1111 1110 0101


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 0110 0010 1101 1111 0010 0110 1111 1110 0101 =

0010 0111 1011 0100 0110 0010 1101 1111 0010 0110 1111 1110 0101


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 0110 0010 1101 1111 0010 0110 1111 1110 0101


Decimal number -0.000 282 005 914 366 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 0010 0110 1111 1110 0101

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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