-0.000 282 005 914 351 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 351(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
-0.000 282 005 914 351(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:

|-0.000 282 005 914 351| = 0.000 282 005 914 351


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: 0.000 282 005 914 351.

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) 0.000 282 005 914 351 × 2 = 0 + 0.000 564 011 828 702;
  • 2) 0.000 564 011 828 702 × 2 = 0 + 0.001 128 023 657 404;
  • 3) 0.001 128 023 657 404 × 2 = 0 + 0.002 256 047 314 808;
  • 4) 0.002 256 047 314 808 × 2 = 0 + 0.004 512 094 629 616;
  • 5) 0.004 512 094 629 616 × 2 = 0 + 0.009 024 189 259 232;
  • 6) 0.009 024 189 259 232 × 2 = 0 + 0.018 048 378 518 464;
  • 7) 0.018 048 378 518 464 × 2 = 0 + 0.036 096 757 036 928;
  • 8) 0.036 096 757 036 928 × 2 = 0 + 0.072 193 514 073 856;
  • 9) 0.072 193 514 073 856 × 2 = 0 + 0.144 387 028 147 712;
  • 10) 0.144 387 028 147 712 × 2 = 0 + 0.288 774 056 295 424;
  • 11) 0.288 774 056 295 424 × 2 = 0 + 0.577 548 112 590 848;
  • 12) 0.577 548 112 590 848 × 2 = 1 + 0.155 096 225 181 696;
  • 13) 0.155 096 225 181 696 × 2 = 0 + 0.310 192 450 363 392;
  • 14) 0.310 192 450 363 392 × 2 = 0 + 0.620 384 900 726 784;
  • 15) 0.620 384 900 726 784 × 2 = 1 + 0.240 769 801 453 568;
  • 16) 0.240 769 801 453 568 × 2 = 0 + 0.481 539 602 907 136;
  • 17) 0.481 539 602 907 136 × 2 = 0 + 0.963 079 205 814 272;
  • 18) 0.963 079 205 814 272 × 2 = 1 + 0.926 158 411 628 544;
  • 19) 0.926 158 411 628 544 × 2 = 1 + 0.852 316 823 257 088;
  • 20) 0.852 316 823 257 088 × 2 = 1 + 0.704 633 646 514 176;
  • 21) 0.704 633 646 514 176 × 2 = 1 + 0.409 267 293 028 352;
  • 22) 0.409 267 293 028 352 × 2 = 0 + 0.818 534 586 056 704;
  • 23) 0.818 534 586 056 704 × 2 = 1 + 0.637 069 172 113 408;
  • 24) 0.637 069 172 113 408 × 2 = 1 + 0.274 138 344 226 816;
  • 25) 0.274 138 344 226 816 × 2 = 0 + 0.548 276 688 453 632;
  • 26) 0.548 276 688 453 632 × 2 = 1 + 0.096 553 376 907 264;
  • 27) 0.096 553 376 907 264 × 2 = 0 + 0.193 106 753 814 528;
  • 28) 0.193 106 753 814 528 × 2 = 0 + 0.386 213 507 629 056;
  • 29) 0.386 213 507 629 056 × 2 = 0 + 0.772 427 015 258 112;
  • 30) 0.772 427 015 258 112 × 2 = 1 + 0.544 854 030 516 224;
  • 31) 0.544 854 030 516 224 × 2 = 1 + 0.089 708 061 032 448;
  • 32) 0.089 708 061 032 448 × 2 = 0 + 0.179 416 122 064 896;
  • 33) 0.179 416 122 064 896 × 2 = 0 + 0.358 832 244 129 792;
  • 34) 0.358 832 244 129 792 × 2 = 0 + 0.717 664 488 259 584;
  • 35) 0.717 664 488 259 584 × 2 = 1 + 0.435 328 976 519 168;
  • 36) 0.435 328 976 519 168 × 2 = 0 + 0.870 657 953 038 336;
  • 37) 0.870 657 953 038 336 × 2 = 1 + 0.741 315 906 076 672;
  • 38) 0.741 315 906 076 672 × 2 = 1 + 0.482 631 812 153 344;
  • 39) 0.482 631 812 153 344 × 2 = 0 + 0.965 263 624 306 688;
  • 40) 0.965 263 624 306 688 × 2 = 1 + 0.930 527 248 613 376;
  • 41) 0.930 527 248 613 376 × 2 = 1 + 0.861 054 497 226 752;
  • 42) 0.861 054 497 226 752 × 2 = 1 + 0.722 108 994 453 504;
  • 43) 0.722 108 994 453 504 × 2 = 1 + 0.444 217 988 907 008;
  • 44) 0.444 217 988 907 008 × 2 = 0 + 0.888 435 977 814 016;
  • 45) 0.888 435 977 814 016 × 2 = 1 + 0.776 871 955 628 032;
  • 46) 0.776 871 955 628 032 × 2 = 1 + 0.553 743 911 256 064;
  • 47) 0.553 743 911 256 064 × 2 = 1 + 0.107 487 822 512 128;
  • 48) 0.107 487 822 512 128 × 2 = 0 + 0.214 975 645 024 256;
  • 49) 0.214 975 645 024 256 × 2 = 0 + 0.429 951 290 048 512;
  • 50) 0.429 951 290 048 512 × 2 = 0 + 0.859 902 580 097 024;
  • 51) 0.859 902 580 097 024 × 2 = 1 + 0.719 805 160 194 048;
  • 52) 0.719 805 160 194 048 × 2 = 1 + 0.439 610 320 388 096;
  • 53) 0.439 610 320 388 096 × 2 = 0 + 0.879 220 640 776 192;
  • 54) 0.879 220 640 776 192 × 2 = 1 + 0.758 441 281 552 384;
  • 55) 0.758 441 281 552 384 × 2 = 1 + 0.516 882 563 104 768;
  • 56) 0.516 882 563 104 768 × 2 = 1 + 0.033 765 126 209 536;
  • 57) 0.033 765 126 209 536 × 2 = 0 + 0.067 530 252 419 072;
  • 58) 0.067 530 252 419 072 × 2 = 0 + 0.135 060 504 838 144;
  • 59) 0.135 060 504 838 144 × 2 = 0 + 0.270 121 009 676 288;
  • 60) 0.270 121 009 676 288 × 2 = 0 + 0.540 242 019 352 576;
  • 61) 0.540 242 019 352 576 × 2 = 1 + 0.080 484 038 705 152;
  • 62) 0.080 484 038 705 152 × 2 = 0 + 0.160 968 077 410 304;
  • 63) 0.160 968 077 410 304 × 2 = 0 + 0.321 936 154 820 608;
  • 64) 0.321 936 154 820 608 × 2 = 0 + 0.643 872 309 641 216;

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:


0.000 282 005 914 351(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1110 1110 0011 0111 0000 1000(2)

6. Positive number before normalization:

0.000 282 005 914 351(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1110 1110 0011 0111 0000 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:


0.000 282 005 914 351(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1110 1110 0011 0111 0000 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1110 1110 0011 0111 0000 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1110 1110 0011 0111 0000 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 0110 0010 1101 1110 1110 0011 0111 0000 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 0110 0010 1101 1110 1110 0011 0111 0000 1000 =

0010 0111 1011 0100 0110 0010 1101 1110 1110 0011 0111 0000 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 0110 0010 1101 1110 1110 0011 0111 0000 1000


Decimal number -0.000 282 005 914 351 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1110 1110 0011 0111 0000 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