-0.000 282 005 910 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 910 7(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
-0.000 282 005 910 7(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:

|-0.000 282 005 910 7| = 0.000 282 005 910 7


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: 0.000 282 005 910 7.

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) 0.000 282 005 910 7 × 2 = 0 + 0.000 564 011 821 4;
  • 2) 0.000 564 011 821 4 × 2 = 0 + 0.001 128 023 642 8;
  • 3) 0.001 128 023 642 8 × 2 = 0 + 0.002 256 047 285 6;
  • 4) 0.002 256 047 285 6 × 2 = 0 + 0.004 512 094 571 2;
  • 5) 0.004 512 094 571 2 × 2 = 0 + 0.009 024 189 142 4;
  • 6) 0.009 024 189 142 4 × 2 = 0 + 0.018 048 378 284 8;
  • 7) 0.018 048 378 284 8 × 2 = 0 + 0.036 096 756 569 6;
  • 8) 0.036 096 756 569 6 × 2 = 0 + 0.072 193 513 139 2;
  • 9) 0.072 193 513 139 2 × 2 = 0 + 0.144 387 026 278 4;
  • 10) 0.144 387 026 278 4 × 2 = 0 + 0.288 774 052 556 8;
  • 11) 0.288 774 052 556 8 × 2 = 0 + 0.577 548 105 113 6;
  • 12) 0.577 548 105 113 6 × 2 = 1 + 0.155 096 210 227 2;
  • 13) 0.155 096 210 227 2 × 2 = 0 + 0.310 192 420 454 4;
  • 14) 0.310 192 420 454 4 × 2 = 0 + 0.620 384 840 908 8;
  • 15) 0.620 384 840 908 8 × 2 = 1 + 0.240 769 681 817 6;
  • 16) 0.240 769 681 817 6 × 2 = 0 + 0.481 539 363 635 2;
  • 17) 0.481 539 363 635 2 × 2 = 0 + 0.963 078 727 270 4;
  • 18) 0.963 078 727 270 4 × 2 = 1 + 0.926 157 454 540 8;
  • 19) 0.926 157 454 540 8 × 2 = 1 + 0.852 314 909 081 6;
  • 20) 0.852 314 909 081 6 × 2 = 1 + 0.704 629 818 163 2;
  • 21) 0.704 629 818 163 2 × 2 = 1 + 0.409 259 636 326 4;
  • 22) 0.409 259 636 326 4 × 2 = 0 + 0.818 519 272 652 8;
  • 23) 0.818 519 272 652 8 × 2 = 1 + 0.637 038 545 305 6;
  • 24) 0.637 038 545 305 6 × 2 = 1 + 0.274 077 090 611 2;
  • 25) 0.274 077 090 611 2 × 2 = 0 + 0.548 154 181 222 4;
  • 26) 0.548 154 181 222 4 × 2 = 1 + 0.096 308 362 444 8;
  • 27) 0.096 308 362 444 8 × 2 = 0 + 0.192 616 724 889 6;
  • 28) 0.192 616 724 889 6 × 2 = 0 + 0.385 233 449 779 2;
  • 29) 0.385 233 449 779 2 × 2 = 0 + 0.770 466 899 558 4;
  • 30) 0.770 466 899 558 4 × 2 = 1 + 0.540 933 799 116 8;
  • 31) 0.540 933 799 116 8 × 2 = 1 + 0.081 867 598 233 6;
  • 32) 0.081 867 598 233 6 × 2 = 0 + 0.163 735 196 467 2;
  • 33) 0.163 735 196 467 2 × 2 = 0 + 0.327 470 392 934 4;
  • 34) 0.327 470 392 934 4 × 2 = 0 + 0.654 940 785 868 8;
  • 35) 0.654 940 785 868 8 × 2 = 1 + 0.309 881 571 737 6;
  • 36) 0.309 881 571 737 6 × 2 = 0 + 0.619 763 143 475 2;
  • 37) 0.619 763 143 475 2 × 2 = 1 + 0.239 526 286 950 4;
  • 38) 0.239 526 286 950 4 × 2 = 0 + 0.479 052 573 900 8;
  • 39) 0.479 052 573 900 8 × 2 = 0 + 0.958 105 147 801 6;
  • 40) 0.958 105 147 801 6 × 2 = 1 + 0.916 210 295 603 2;
  • 41) 0.916 210 295 603 2 × 2 = 1 + 0.832 420 591 206 4;
  • 42) 0.832 420 591 206 4 × 2 = 1 + 0.664 841 182 412 8;
  • 43) 0.664 841 182 412 8 × 2 = 1 + 0.329 682 364 825 6;
  • 44) 0.329 682 364 825 6 × 2 = 0 + 0.659 364 729 651 2;
  • 45) 0.659 364 729 651 2 × 2 = 1 + 0.318 729 459 302 4;
  • 46) 0.318 729 459 302 4 × 2 = 0 + 0.637 458 918 604 8;
  • 47) 0.637 458 918 604 8 × 2 = 1 + 0.274 917 837 209 6;
  • 48) 0.274 917 837 209 6 × 2 = 0 + 0.549 835 674 419 2;
  • 49) 0.549 835 674 419 2 × 2 = 1 + 0.099 671 348 838 4;
  • 50) 0.099 671 348 838 4 × 2 = 0 + 0.199 342 697 676 8;
  • 51) 0.199 342 697 676 8 × 2 = 0 + 0.398 685 395 353 6;
  • 52) 0.398 685 395 353 6 × 2 = 0 + 0.797 370 790 707 2;
  • 53) 0.797 370 790 707 2 × 2 = 1 + 0.594 741 581 414 4;
  • 54) 0.594 741 581 414 4 × 2 = 1 + 0.189 483 162 828 8;
  • 55) 0.189 483 162 828 8 × 2 = 0 + 0.378 966 325 657 6;
  • 56) 0.378 966 325 657 6 × 2 = 0 + 0.757 932 651 315 2;
  • 57) 0.757 932 651 315 2 × 2 = 1 + 0.515 865 302 630 4;
  • 58) 0.515 865 302 630 4 × 2 = 1 + 0.031 730 605 260 8;
  • 59) 0.031 730 605 260 8 × 2 = 0 + 0.063 461 210 521 6;
  • 60) 0.063 461 210 521 6 × 2 = 0 + 0.126 922 421 043 2;
  • 61) 0.126 922 421 043 2 × 2 = 0 + 0.253 844 842 086 4;
  • 62) 0.253 844 842 086 4 × 2 = 0 + 0.507 689 684 172 8;
  • 63) 0.507 689 684 172 8 × 2 = 1 + 0.015 379 368 345 6;
  • 64) 0.015 379 368 345 6 × 2 = 0 + 0.030 758 736 691 2;

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:


0.000 282 005 910 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1110 1010 1000 1100 1100 0010(2)

6. Positive number before normalization:

0.000 282 005 910 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1110 1010 1000 1100 1100 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:


0.000 282 005 910 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1110 1010 1000 1100 1100 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1110 1010 1000 1100 1100 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1001 1110 1010 1000 1100 1100 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 1001 1110 1010 1000 1100 1100 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 1001 1110 1010 1000 1100 1100 0010 =

0010 0111 1011 0100 0110 0010 1001 1110 1010 1000 1100 1100 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 1001 1110 1010 1000 1100 1100 0010


Decimal number -0.000 282 005 910 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1001 1110 1010 1000 1100 1100 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