-0.000 282 005 914 248 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 248(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
-0.000 282 005 914 248(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:

|-0.000 282 005 914 248| = 0.000 282 005 914 248


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: 0.000 282 005 914 248.

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) 0.000 282 005 914 248 × 2 = 0 + 0.000 564 011 828 496;
  • 2) 0.000 564 011 828 496 × 2 = 0 + 0.001 128 023 656 992;
  • 3) 0.001 128 023 656 992 × 2 = 0 + 0.002 256 047 313 984;
  • 4) 0.002 256 047 313 984 × 2 = 0 + 0.004 512 094 627 968;
  • 5) 0.004 512 094 627 968 × 2 = 0 + 0.009 024 189 255 936;
  • 6) 0.009 024 189 255 936 × 2 = 0 + 0.018 048 378 511 872;
  • 7) 0.018 048 378 511 872 × 2 = 0 + 0.036 096 757 023 744;
  • 8) 0.036 096 757 023 744 × 2 = 0 + 0.072 193 514 047 488;
  • 9) 0.072 193 514 047 488 × 2 = 0 + 0.144 387 028 094 976;
  • 10) 0.144 387 028 094 976 × 2 = 0 + 0.288 774 056 189 952;
  • 11) 0.288 774 056 189 952 × 2 = 0 + 0.577 548 112 379 904;
  • 12) 0.577 548 112 379 904 × 2 = 1 + 0.155 096 224 759 808;
  • 13) 0.155 096 224 759 808 × 2 = 0 + 0.310 192 449 519 616;
  • 14) 0.310 192 449 519 616 × 2 = 0 + 0.620 384 899 039 232;
  • 15) 0.620 384 899 039 232 × 2 = 1 + 0.240 769 798 078 464;
  • 16) 0.240 769 798 078 464 × 2 = 0 + 0.481 539 596 156 928;
  • 17) 0.481 539 596 156 928 × 2 = 0 + 0.963 079 192 313 856;
  • 18) 0.963 079 192 313 856 × 2 = 1 + 0.926 158 384 627 712;
  • 19) 0.926 158 384 627 712 × 2 = 1 + 0.852 316 769 255 424;
  • 20) 0.852 316 769 255 424 × 2 = 1 + 0.704 633 538 510 848;
  • 21) 0.704 633 538 510 848 × 2 = 1 + 0.409 267 077 021 696;
  • 22) 0.409 267 077 021 696 × 2 = 0 + 0.818 534 154 043 392;
  • 23) 0.818 534 154 043 392 × 2 = 1 + 0.637 068 308 086 784;
  • 24) 0.637 068 308 086 784 × 2 = 1 + 0.274 136 616 173 568;
  • 25) 0.274 136 616 173 568 × 2 = 0 + 0.548 273 232 347 136;
  • 26) 0.548 273 232 347 136 × 2 = 1 + 0.096 546 464 694 272;
  • 27) 0.096 546 464 694 272 × 2 = 0 + 0.193 092 929 388 544;
  • 28) 0.193 092 929 388 544 × 2 = 0 + 0.386 185 858 777 088;
  • 29) 0.386 185 858 777 088 × 2 = 0 + 0.772 371 717 554 176;
  • 30) 0.772 371 717 554 176 × 2 = 1 + 0.544 743 435 108 352;
  • 31) 0.544 743 435 108 352 × 2 = 1 + 0.089 486 870 216 704;
  • 32) 0.089 486 870 216 704 × 2 = 0 + 0.178 973 740 433 408;
  • 33) 0.178 973 740 433 408 × 2 = 0 + 0.357 947 480 866 816;
  • 34) 0.357 947 480 866 816 × 2 = 0 + 0.715 894 961 733 632;
  • 35) 0.715 894 961 733 632 × 2 = 1 + 0.431 789 923 467 264;
  • 36) 0.431 789 923 467 264 × 2 = 0 + 0.863 579 846 934 528;
  • 37) 0.863 579 846 934 528 × 2 = 1 + 0.727 159 693 869 056;
  • 38) 0.727 159 693 869 056 × 2 = 1 + 0.454 319 387 738 112;
  • 39) 0.454 319 387 738 112 × 2 = 0 + 0.908 638 775 476 224;
  • 40) 0.908 638 775 476 224 × 2 = 1 + 0.817 277 550 952 448;
  • 41) 0.817 277 550 952 448 × 2 = 1 + 0.634 555 101 904 896;
  • 42) 0.634 555 101 904 896 × 2 = 1 + 0.269 110 203 809 792;
  • 43) 0.269 110 203 809 792 × 2 = 0 + 0.538 220 407 619 584;
  • 44) 0.538 220 407 619 584 × 2 = 1 + 0.076 440 815 239 168;
  • 45) 0.076 440 815 239 168 × 2 = 0 + 0.152 881 630 478 336;
  • 46) 0.152 881 630 478 336 × 2 = 0 + 0.305 763 260 956 672;
  • 47) 0.305 763 260 956 672 × 2 = 0 + 0.611 526 521 913 344;
  • 48) 0.611 526 521 913 344 × 2 = 1 + 0.223 053 043 826 688;
  • 49) 0.223 053 043 826 688 × 2 = 0 + 0.446 106 087 653 376;
  • 50) 0.446 106 087 653 376 × 2 = 0 + 0.892 212 175 306 752;
  • 51) 0.892 212 175 306 752 × 2 = 1 + 0.784 424 350 613 504;
  • 52) 0.784 424 350 613 504 × 2 = 1 + 0.568 848 701 227 008;
  • 53) 0.568 848 701 227 008 × 2 = 1 + 0.137 697 402 454 016;
  • 54) 0.137 697 402 454 016 × 2 = 0 + 0.275 394 804 908 032;
  • 55) 0.275 394 804 908 032 × 2 = 0 + 0.550 789 609 816 064;
  • 56) 0.550 789 609 816 064 × 2 = 1 + 0.101 579 219 632 128;
  • 57) 0.101 579 219 632 128 × 2 = 0 + 0.203 158 439 264 256;
  • 58) 0.203 158 439 264 256 × 2 = 0 + 0.406 316 878 528 512;
  • 59) 0.406 316 878 528 512 × 2 = 0 + 0.812 633 757 057 024;
  • 60) 0.812 633 757 057 024 × 2 = 1 + 0.625 267 514 114 048;
  • 61) 0.625 267 514 114 048 × 2 = 1 + 0.250 535 028 228 096;
  • 62) 0.250 535 028 228 096 × 2 = 0 + 0.501 070 056 456 192;
  • 63) 0.501 070 056 456 192 × 2 = 1 + 0.002 140 112 912 384;
  • 64) 0.002 140 112 912 384 × 2 = 0 + 0.004 280 225 824 768;

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:


0.000 282 005 914 248(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0001 0011 1001 0001 1010(2)

6. Positive number before normalization:

0.000 282 005 914 248(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0001 0011 1001 0001 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:


0.000 282 005 914 248(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0001 0011 1001 0001 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0001 0011 1001 0001 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1101 0001 0011 1001 0001 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 1101 0001 0011 1001 0001 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 1101 0001 0011 1001 0001 1010 =

0010 0111 1011 0100 0110 0010 1101 1101 0001 0011 1001 0001 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 1101 0001 0011 1001 0001 1010


Decimal number -0.000 282 005 914 248 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1101 0001 0011 1001 0001 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