-0.000 282 005 909 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 909 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
-0.000 282 005 909 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:

|-0.000 282 005 909 7| = 0.000 282 005 909 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: 0.000 282 005 909 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) 0.000 282 005 909 7 × 2 = 0 + 0.000 564 011 819 4;
  • 2) 0.000 564 011 819 4 × 2 = 0 + 0.001 128 023 638 8;
  • 3) 0.001 128 023 638 8 × 2 = 0 + 0.002 256 047 277 6;
  • 4) 0.002 256 047 277 6 × 2 = 0 + 0.004 512 094 555 2;
  • 5) 0.004 512 094 555 2 × 2 = 0 + 0.009 024 189 110 4;
  • 6) 0.009 024 189 110 4 × 2 = 0 + 0.018 048 378 220 8;
  • 7) 0.018 048 378 220 8 × 2 = 0 + 0.036 096 756 441 6;
  • 8) 0.036 096 756 441 6 × 2 = 0 + 0.072 193 512 883 2;
  • 9) 0.072 193 512 883 2 × 2 = 0 + 0.144 387 025 766 4;
  • 10) 0.144 387 025 766 4 × 2 = 0 + 0.288 774 051 532 8;
  • 11) 0.288 774 051 532 8 × 2 = 0 + 0.577 548 103 065 6;
  • 12) 0.577 548 103 065 6 × 2 = 1 + 0.155 096 206 131 2;
  • 13) 0.155 096 206 131 2 × 2 = 0 + 0.310 192 412 262 4;
  • 14) 0.310 192 412 262 4 × 2 = 0 + 0.620 384 824 524 8;
  • 15) 0.620 384 824 524 8 × 2 = 1 + 0.240 769 649 049 6;
  • 16) 0.240 769 649 049 6 × 2 = 0 + 0.481 539 298 099 2;
  • 17) 0.481 539 298 099 2 × 2 = 0 + 0.963 078 596 198 4;
  • 18) 0.963 078 596 198 4 × 2 = 1 + 0.926 157 192 396 8;
  • 19) 0.926 157 192 396 8 × 2 = 1 + 0.852 314 384 793 6;
  • 20) 0.852 314 384 793 6 × 2 = 1 + 0.704 628 769 587 2;
  • 21) 0.704 628 769 587 2 × 2 = 1 + 0.409 257 539 174 4;
  • 22) 0.409 257 539 174 4 × 2 = 0 + 0.818 515 078 348 8;
  • 23) 0.818 515 078 348 8 × 2 = 1 + 0.637 030 156 697 6;
  • 24) 0.637 030 156 697 6 × 2 = 1 + 0.274 060 313 395 2;
  • 25) 0.274 060 313 395 2 × 2 = 0 + 0.548 120 626 790 4;
  • 26) 0.548 120 626 790 4 × 2 = 1 + 0.096 241 253 580 8;
  • 27) 0.096 241 253 580 8 × 2 = 0 + 0.192 482 507 161 6;
  • 28) 0.192 482 507 161 6 × 2 = 0 + 0.384 965 014 323 2;
  • 29) 0.384 965 014 323 2 × 2 = 0 + 0.769 930 028 646 4;
  • 30) 0.769 930 028 646 4 × 2 = 1 + 0.539 860 057 292 8;
  • 31) 0.539 860 057 292 8 × 2 = 1 + 0.079 720 114 585 6;
  • 32) 0.079 720 114 585 6 × 2 = 0 + 0.159 440 229 171 2;
  • 33) 0.159 440 229 171 2 × 2 = 0 + 0.318 880 458 342 4;
  • 34) 0.318 880 458 342 4 × 2 = 0 + 0.637 760 916 684 8;
  • 35) 0.637 760 916 684 8 × 2 = 1 + 0.275 521 833 369 6;
  • 36) 0.275 521 833 369 6 × 2 = 0 + 0.551 043 666 739 2;
  • 37) 0.551 043 666 739 2 × 2 = 1 + 0.102 087 333 478 4;
  • 38) 0.102 087 333 478 4 × 2 = 0 + 0.204 174 666 956 8;
  • 39) 0.204 174 666 956 8 × 2 = 0 + 0.408 349 333 913 6;
  • 40) 0.408 349 333 913 6 × 2 = 0 + 0.816 698 667 827 2;
  • 41) 0.816 698 667 827 2 × 2 = 1 + 0.633 397 335 654 4;
  • 42) 0.633 397 335 654 4 × 2 = 1 + 0.266 794 671 308 8;
  • 43) 0.266 794 671 308 8 × 2 = 0 + 0.533 589 342 617 6;
  • 44) 0.533 589 342 617 6 × 2 = 1 + 0.067 178 685 235 2;
  • 45) 0.067 178 685 235 2 × 2 = 0 + 0.134 357 370 470 4;
  • 46) 0.134 357 370 470 4 × 2 = 0 + 0.268 714 740 940 8;
  • 47) 0.268 714 740 940 8 × 2 = 0 + 0.537 429 481 881 6;
  • 48) 0.537 429 481 881 6 × 2 = 1 + 0.074 858 963 763 2;
  • 49) 0.074 858 963 763 2 × 2 = 0 + 0.149 717 927 526 4;
  • 50) 0.149 717 927 526 4 × 2 = 0 + 0.299 435 855 052 8;
  • 51) 0.299 435 855 052 8 × 2 = 0 + 0.598 871 710 105 6;
  • 52) 0.598 871 710 105 6 × 2 = 1 + 0.197 743 420 211 2;
  • 53) 0.197 743 420 211 2 × 2 = 0 + 0.395 486 840 422 4;
  • 54) 0.395 486 840 422 4 × 2 = 0 + 0.790 973 680 844 8;
  • 55) 0.790 973 680 844 8 × 2 = 1 + 0.581 947 361 689 6;
  • 56) 0.581 947 361 689 6 × 2 = 1 + 0.163 894 723 379 2;
  • 57) 0.163 894 723 379 2 × 2 = 0 + 0.327 789 446 758 4;
  • 58) 0.327 789 446 758 4 × 2 = 0 + 0.655 578 893 516 8;
  • 59) 0.655 578 893 516 8 × 2 = 1 + 0.311 157 787 033 6;
  • 60) 0.311 157 787 033 6 × 2 = 0 + 0.622 315 574 067 2;
  • 61) 0.622 315 574 067 2 × 2 = 1 + 0.244 631 148 134 4;
  • 62) 0.244 631 148 134 4 × 2 = 0 + 0.489 262 296 268 8;
  • 63) 0.489 262 296 268 8 × 2 = 0 + 0.978 524 592 537 6;
  • 64) 0.978 524 592 537 6 × 2 = 1 + 0.957 049 185 075 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:


0.000 282 005 909 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1000 1101 0001 0001 0011 0010 1001(2)

6. Positive number before normalization:

0.000 282 005 909 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1000 1101 0001 0001 0011 0010 1001(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:


0.000 282 005 909 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1000 1101 0001 0001 0011 0010 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1000 1101 0001 0001 0011 0010 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1000 1101 0001 0001 0011 0010 1001(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 1000 1101 0001 0001 0011 0010 1001


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 1000 1101 0001 0001 0011 0010 1001 =

0010 0111 1011 0100 0110 0010 1000 1101 0001 0001 0011 0010 1001


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 1000 1101 0001 0001 0011 0010 1001


Decimal number -0.000 282 005 909 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1000 1101 0001 0001 0011 0010 1001

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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