-0.000 282 005 914 292 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 292(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
-0.000 282 005 914 292(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:

|-0.000 282 005 914 292| = 0.000 282 005 914 292


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: 0.000 282 005 914 292.

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) 0.000 282 005 914 292 × 2 = 0 + 0.000 564 011 828 584;
  • 2) 0.000 564 011 828 584 × 2 = 0 + 0.001 128 023 657 168;
  • 3) 0.001 128 023 657 168 × 2 = 0 + 0.002 256 047 314 336;
  • 4) 0.002 256 047 314 336 × 2 = 0 + 0.004 512 094 628 672;
  • 5) 0.004 512 094 628 672 × 2 = 0 + 0.009 024 189 257 344;
  • 6) 0.009 024 189 257 344 × 2 = 0 + 0.018 048 378 514 688;
  • 7) 0.018 048 378 514 688 × 2 = 0 + 0.036 096 757 029 376;
  • 8) 0.036 096 757 029 376 × 2 = 0 + 0.072 193 514 058 752;
  • 9) 0.072 193 514 058 752 × 2 = 0 + 0.144 387 028 117 504;
  • 10) 0.144 387 028 117 504 × 2 = 0 + 0.288 774 056 235 008;
  • 11) 0.288 774 056 235 008 × 2 = 0 + 0.577 548 112 470 016;
  • 12) 0.577 548 112 470 016 × 2 = 1 + 0.155 096 224 940 032;
  • 13) 0.155 096 224 940 032 × 2 = 0 + 0.310 192 449 880 064;
  • 14) 0.310 192 449 880 064 × 2 = 0 + 0.620 384 899 760 128;
  • 15) 0.620 384 899 760 128 × 2 = 1 + 0.240 769 799 520 256;
  • 16) 0.240 769 799 520 256 × 2 = 0 + 0.481 539 599 040 512;
  • 17) 0.481 539 599 040 512 × 2 = 0 + 0.963 079 198 081 024;
  • 18) 0.963 079 198 081 024 × 2 = 1 + 0.926 158 396 162 048;
  • 19) 0.926 158 396 162 048 × 2 = 1 + 0.852 316 792 324 096;
  • 20) 0.852 316 792 324 096 × 2 = 1 + 0.704 633 584 648 192;
  • 21) 0.704 633 584 648 192 × 2 = 1 + 0.409 267 169 296 384;
  • 22) 0.409 267 169 296 384 × 2 = 0 + 0.818 534 338 592 768;
  • 23) 0.818 534 338 592 768 × 2 = 1 + 0.637 068 677 185 536;
  • 24) 0.637 068 677 185 536 × 2 = 1 + 0.274 137 354 371 072;
  • 25) 0.274 137 354 371 072 × 2 = 0 + 0.548 274 708 742 144;
  • 26) 0.548 274 708 742 144 × 2 = 1 + 0.096 549 417 484 288;
  • 27) 0.096 549 417 484 288 × 2 = 0 + 0.193 098 834 968 576;
  • 28) 0.193 098 834 968 576 × 2 = 0 + 0.386 197 669 937 152;
  • 29) 0.386 197 669 937 152 × 2 = 0 + 0.772 395 339 874 304;
  • 30) 0.772 395 339 874 304 × 2 = 1 + 0.544 790 679 748 608;
  • 31) 0.544 790 679 748 608 × 2 = 1 + 0.089 581 359 497 216;
  • 32) 0.089 581 359 497 216 × 2 = 0 + 0.179 162 718 994 432;
  • 33) 0.179 162 718 994 432 × 2 = 0 + 0.358 325 437 988 864;
  • 34) 0.358 325 437 988 864 × 2 = 0 + 0.716 650 875 977 728;
  • 35) 0.716 650 875 977 728 × 2 = 1 + 0.433 301 751 955 456;
  • 36) 0.433 301 751 955 456 × 2 = 0 + 0.866 603 503 910 912;
  • 37) 0.866 603 503 910 912 × 2 = 1 + 0.733 207 007 821 824;
  • 38) 0.733 207 007 821 824 × 2 = 1 + 0.466 414 015 643 648;
  • 39) 0.466 414 015 643 648 × 2 = 0 + 0.932 828 031 287 296;
  • 40) 0.932 828 031 287 296 × 2 = 1 + 0.865 656 062 574 592;
  • 41) 0.865 656 062 574 592 × 2 = 1 + 0.731 312 125 149 184;
  • 42) 0.731 312 125 149 184 × 2 = 1 + 0.462 624 250 298 368;
  • 43) 0.462 624 250 298 368 × 2 = 0 + 0.925 248 500 596 736;
  • 44) 0.925 248 500 596 736 × 2 = 1 + 0.850 497 001 193 472;
  • 45) 0.850 497 001 193 472 × 2 = 1 + 0.700 994 002 386 944;
  • 46) 0.700 994 002 386 944 × 2 = 1 + 0.401 988 004 773 888;
  • 47) 0.401 988 004 773 888 × 2 = 0 + 0.803 976 009 547 776;
  • 48) 0.803 976 009 547 776 × 2 = 1 + 0.607 952 019 095 552;
  • 49) 0.607 952 019 095 552 × 2 = 1 + 0.215 904 038 191 104;
  • 50) 0.215 904 038 191 104 × 2 = 0 + 0.431 808 076 382 208;
  • 51) 0.431 808 076 382 208 × 2 = 0 + 0.863 616 152 764 416;
  • 52) 0.863 616 152 764 416 × 2 = 1 + 0.727 232 305 528 832;
  • 53) 0.727 232 305 528 832 × 2 = 1 + 0.454 464 611 057 664;
  • 54) 0.454 464 611 057 664 × 2 = 0 + 0.908 929 222 115 328;
  • 55) 0.908 929 222 115 328 × 2 = 1 + 0.817 858 444 230 656;
  • 56) 0.817 858 444 230 656 × 2 = 1 + 0.635 716 888 461 312;
  • 57) 0.635 716 888 461 312 × 2 = 1 + 0.271 433 776 922 624;
  • 58) 0.271 433 776 922 624 × 2 = 0 + 0.542 867 553 845 248;
  • 59) 0.542 867 553 845 248 × 2 = 1 + 0.085 735 107 690 496;
  • 60) 0.085 735 107 690 496 × 2 = 0 + 0.171 470 215 380 992;
  • 61) 0.171 470 215 380 992 × 2 = 0 + 0.342 940 430 761 984;
  • 62) 0.342 940 430 761 984 × 2 = 0 + 0.685 880 861 523 968;
  • 63) 0.685 880 861 523 968 × 2 = 1 + 0.371 761 723 047 936;
  • 64) 0.371 761 723 047 936 × 2 = 0 + 0.743 523 446 095 872;

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:


0.000 282 005 914 292(10) =

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

6. Positive number before normalization:

0.000 282 005 914 292(10) =

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


0.000 282 005 914 292(10) =

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

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

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


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 1101 1001 1011 1010 0010 =

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


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 1101 1001 1011 1010 0010


Decimal number -0.000 282 005 914 292 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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