-0.000 282 005 914 222 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 222(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
-0.000 282 005 914 222(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:

|-0.000 282 005 914 222| = 0.000 282 005 914 222


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: 0.000 282 005 914 222.

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) 0.000 282 005 914 222 × 2 = 0 + 0.000 564 011 828 444;
  • 2) 0.000 564 011 828 444 × 2 = 0 + 0.001 128 023 656 888;
  • 3) 0.001 128 023 656 888 × 2 = 0 + 0.002 256 047 313 776;
  • 4) 0.002 256 047 313 776 × 2 = 0 + 0.004 512 094 627 552;
  • 5) 0.004 512 094 627 552 × 2 = 0 + 0.009 024 189 255 104;
  • 6) 0.009 024 189 255 104 × 2 = 0 + 0.018 048 378 510 208;
  • 7) 0.018 048 378 510 208 × 2 = 0 + 0.036 096 757 020 416;
  • 8) 0.036 096 757 020 416 × 2 = 0 + 0.072 193 514 040 832;
  • 9) 0.072 193 514 040 832 × 2 = 0 + 0.144 387 028 081 664;
  • 10) 0.144 387 028 081 664 × 2 = 0 + 0.288 774 056 163 328;
  • 11) 0.288 774 056 163 328 × 2 = 0 + 0.577 548 112 326 656;
  • 12) 0.577 548 112 326 656 × 2 = 1 + 0.155 096 224 653 312;
  • 13) 0.155 096 224 653 312 × 2 = 0 + 0.310 192 449 306 624;
  • 14) 0.310 192 449 306 624 × 2 = 0 + 0.620 384 898 613 248;
  • 15) 0.620 384 898 613 248 × 2 = 1 + 0.240 769 797 226 496;
  • 16) 0.240 769 797 226 496 × 2 = 0 + 0.481 539 594 452 992;
  • 17) 0.481 539 594 452 992 × 2 = 0 + 0.963 079 188 905 984;
  • 18) 0.963 079 188 905 984 × 2 = 1 + 0.926 158 377 811 968;
  • 19) 0.926 158 377 811 968 × 2 = 1 + 0.852 316 755 623 936;
  • 20) 0.852 316 755 623 936 × 2 = 1 + 0.704 633 511 247 872;
  • 21) 0.704 633 511 247 872 × 2 = 1 + 0.409 267 022 495 744;
  • 22) 0.409 267 022 495 744 × 2 = 0 + 0.818 534 044 991 488;
  • 23) 0.818 534 044 991 488 × 2 = 1 + 0.637 068 089 982 976;
  • 24) 0.637 068 089 982 976 × 2 = 1 + 0.274 136 179 965 952;
  • 25) 0.274 136 179 965 952 × 2 = 0 + 0.548 272 359 931 904;
  • 26) 0.548 272 359 931 904 × 2 = 1 + 0.096 544 719 863 808;
  • 27) 0.096 544 719 863 808 × 2 = 0 + 0.193 089 439 727 616;
  • 28) 0.193 089 439 727 616 × 2 = 0 + 0.386 178 879 455 232;
  • 29) 0.386 178 879 455 232 × 2 = 0 + 0.772 357 758 910 464;
  • 30) 0.772 357 758 910 464 × 2 = 1 + 0.544 715 517 820 928;
  • 31) 0.544 715 517 820 928 × 2 = 1 + 0.089 431 035 641 856;
  • 32) 0.089 431 035 641 856 × 2 = 0 + 0.178 862 071 283 712;
  • 33) 0.178 862 071 283 712 × 2 = 0 + 0.357 724 142 567 424;
  • 34) 0.357 724 142 567 424 × 2 = 0 + 0.715 448 285 134 848;
  • 35) 0.715 448 285 134 848 × 2 = 1 + 0.430 896 570 269 696;
  • 36) 0.430 896 570 269 696 × 2 = 0 + 0.861 793 140 539 392;
  • 37) 0.861 793 140 539 392 × 2 = 1 + 0.723 586 281 078 784;
  • 38) 0.723 586 281 078 784 × 2 = 1 + 0.447 172 562 157 568;
  • 39) 0.447 172 562 157 568 × 2 = 0 + 0.894 345 124 315 136;
  • 40) 0.894 345 124 315 136 × 2 = 1 + 0.788 690 248 630 272;
  • 41) 0.788 690 248 630 272 × 2 = 1 + 0.577 380 497 260 544;
  • 42) 0.577 380 497 260 544 × 2 = 1 + 0.154 760 994 521 088;
  • 43) 0.154 760 994 521 088 × 2 = 0 + 0.309 521 989 042 176;
  • 44) 0.309 521 989 042 176 × 2 = 0 + 0.619 043 978 084 352;
  • 45) 0.619 043 978 084 352 × 2 = 1 + 0.238 087 956 168 704;
  • 46) 0.238 087 956 168 704 × 2 = 0 + 0.476 175 912 337 408;
  • 47) 0.476 175 912 337 408 × 2 = 0 + 0.952 351 824 674 816;
  • 48) 0.952 351 824 674 816 × 2 = 1 + 0.904 703 649 349 632;
  • 49) 0.904 703 649 349 632 × 2 = 1 + 0.809 407 298 699 264;
  • 50) 0.809 407 298 699 264 × 2 = 1 + 0.618 814 597 398 528;
  • 51) 0.618 814 597 398 528 × 2 = 1 + 0.237 629 194 797 056;
  • 52) 0.237 629 194 797 056 × 2 = 0 + 0.475 258 389 594 112;
  • 53) 0.475 258 389 594 112 × 2 = 0 + 0.950 516 779 188 224;
  • 54) 0.950 516 779 188 224 × 2 = 1 + 0.901 033 558 376 448;
  • 55) 0.901 033 558 376 448 × 2 = 1 + 0.802 067 116 752 896;
  • 56) 0.802 067 116 752 896 × 2 = 1 + 0.604 134 233 505 792;
  • 57) 0.604 134 233 505 792 × 2 = 1 + 0.208 268 467 011 584;
  • 58) 0.208 268 467 011 584 × 2 = 0 + 0.416 536 934 023 168;
  • 59) 0.416 536 934 023 168 × 2 = 0 + 0.833 073 868 046 336;
  • 60) 0.833 073 868 046 336 × 2 = 1 + 0.666 147 736 092 672;
  • 61) 0.666 147 736 092 672 × 2 = 1 + 0.332 295 472 185 344;
  • 62) 0.332 295 472 185 344 × 2 = 0 + 0.664 590 944 370 688;
  • 63) 0.664 590 944 370 688 × 2 = 1 + 0.329 181 888 741 376;
  • 64) 0.329 181 888 741 376 × 2 = 0 + 0.658 363 777 482 752;

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:


0.000 282 005 914 222(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1001 1110 0111 1001 1010(2)

6. Positive number before normalization:

0.000 282 005 914 222(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1001 1110 0111 1001 1010(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:


0.000 282 005 914 222(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1001 1110 0111 1001 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1001 1110 0111 1001 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1100 1001 1110 0111 1001 1010(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 1100 1001 1110 0111 1001 1010


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 1100 1001 1110 0111 1001 1010 =

0010 0111 1011 0100 0110 0010 1101 1100 1001 1110 0111 1001 1010


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 1100 1001 1110 0111 1001 1010


Decimal number -0.000 282 005 914 222 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1100 1001 1110 0111 1001 1010

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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