-0.000 282 005 913 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 15(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
-0.000 282 005 913 15(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:

|-0.000 282 005 913 15| = 0.000 282 005 913 15


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: 0.000 282 005 913 15.

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) 0.000 282 005 913 15 × 2 = 0 + 0.000 564 011 826 3;
  • 2) 0.000 564 011 826 3 × 2 = 0 + 0.001 128 023 652 6;
  • 3) 0.001 128 023 652 6 × 2 = 0 + 0.002 256 047 305 2;
  • 4) 0.002 256 047 305 2 × 2 = 0 + 0.004 512 094 610 4;
  • 5) 0.004 512 094 610 4 × 2 = 0 + 0.009 024 189 220 8;
  • 6) 0.009 024 189 220 8 × 2 = 0 + 0.018 048 378 441 6;
  • 7) 0.018 048 378 441 6 × 2 = 0 + 0.036 096 756 883 2;
  • 8) 0.036 096 756 883 2 × 2 = 0 + 0.072 193 513 766 4;
  • 9) 0.072 193 513 766 4 × 2 = 0 + 0.144 387 027 532 8;
  • 10) 0.144 387 027 532 8 × 2 = 0 + 0.288 774 055 065 6;
  • 11) 0.288 774 055 065 6 × 2 = 0 + 0.577 548 110 131 2;
  • 12) 0.577 548 110 131 2 × 2 = 1 + 0.155 096 220 262 4;
  • 13) 0.155 096 220 262 4 × 2 = 0 + 0.310 192 440 524 8;
  • 14) 0.310 192 440 524 8 × 2 = 0 + 0.620 384 881 049 6;
  • 15) 0.620 384 881 049 6 × 2 = 1 + 0.240 769 762 099 2;
  • 16) 0.240 769 762 099 2 × 2 = 0 + 0.481 539 524 198 4;
  • 17) 0.481 539 524 198 4 × 2 = 0 + 0.963 079 048 396 8;
  • 18) 0.963 079 048 396 8 × 2 = 1 + 0.926 158 096 793 6;
  • 19) 0.926 158 096 793 6 × 2 = 1 + 0.852 316 193 587 2;
  • 20) 0.852 316 193 587 2 × 2 = 1 + 0.704 632 387 174 4;
  • 21) 0.704 632 387 174 4 × 2 = 1 + 0.409 264 774 348 8;
  • 22) 0.409 264 774 348 8 × 2 = 0 + 0.818 529 548 697 6;
  • 23) 0.818 529 548 697 6 × 2 = 1 + 0.637 059 097 395 2;
  • 24) 0.637 059 097 395 2 × 2 = 1 + 0.274 118 194 790 4;
  • 25) 0.274 118 194 790 4 × 2 = 0 + 0.548 236 389 580 8;
  • 26) 0.548 236 389 580 8 × 2 = 1 + 0.096 472 779 161 6;
  • 27) 0.096 472 779 161 6 × 2 = 0 + 0.192 945 558 323 2;
  • 28) 0.192 945 558 323 2 × 2 = 0 + 0.385 891 116 646 4;
  • 29) 0.385 891 116 646 4 × 2 = 0 + 0.771 782 233 292 8;
  • 30) 0.771 782 233 292 8 × 2 = 1 + 0.543 564 466 585 6;
  • 31) 0.543 564 466 585 6 × 2 = 1 + 0.087 128 933 171 2;
  • 32) 0.087 128 933 171 2 × 2 = 0 + 0.174 257 866 342 4;
  • 33) 0.174 257 866 342 4 × 2 = 0 + 0.348 515 732 684 8;
  • 34) 0.348 515 732 684 8 × 2 = 0 + 0.697 031 465 369 6;
  • 35) 0.697 031 465 369 6 × 2 = 1 + 0.394 062 930 739 2;
  • 36) 0.394 062 930 739 2 × 2 = 0 + 0.788 125 861 478 4;
  • 37) 0.788 125 861 478 4 × 2 = 1 + 0.576 251 722 956 8;
  • 38) 0.576 251 722 956 8 × 2 = 1 + 0.152 503 445 913 6;
  • 39) 0.152 503 445 913 6 × 2 = 0 + 0.305 006 891 827 2;
  • 40) 0.305 006 891 827 2 × 2 = 0 + 0.610 013 783 654 4;
  • 41) 0.610 013 783 654 4 × 2 = 1 + 0.220 027 567 308 8;
  • 42) 0.220 027 567 308 8 × 2 = 0 + 0.440 055 134 617 6;
  • 43) 0.440 055 134 617 6 × 2 = 0 + 0.880 110 269 235 2;
  • 44) 0.880 110 269 235 2 × 2 = 1 + 0.760 220 538 470 4;
  • 45) 0.760 220 538 470 4 × 2 = 1 + 0.520 441 076 940 8;
  • 46) 0.520 441 076 940 8 × 2 = 1 + 0.040 882 153 881 6;
  • 47) 0.040 882 153 881 6 × 2 = 0 + 0.081 764 307 763 2;
  • 48) 0.081 764 307 763 2 × 2 = 0 + 0.163 528 615 526 4;
  • 49) 0.163 528 615 526 4 × 2 = 0 + 0.327 057 231 052 8;
  • 50) 0.327 057 231 052 8 × 2 = 0 + 0.654 114 462 105 6;
  • 51) 0.654 114 462 105 6 × 2 = 1 + 0.308 228 924 211 2;
  • 52) 0.308 228 924 211 2 × 2 = 0 + 0.616 457 848 422 4;
  • 53) 0.616 457 848 422 4 × 2 = 1 + 0.232 915 696 844 8;
  • 54) 0.232 915 696 844 8 × 2 = 0 + 0.465 831 393 689 6;
  • 55) 0.465 831 393 689 6 × 2 = 0 + 0.931 662 787 379 2;
  • 56) 0.931 662 787 379 2 × 2 = 1 + 0.863 325 574 758 4;
  • 57) 0.863 325 574 758 4 × 2 = 1 + 0.726 651 149 516 8;
  • 58) 0.726 651 149 516 8 × 2 = 1 + 0.453 302 299 033 6;
  • 59) 0.453 302 299 033 6 × 2 = 0 + 0.906 604 598 067 2;
  • 60) 0.906 604 598 067 2 × 2 = 1 + 0.813 209 196 134 4;
  • 61) 0.813 209 196 134 4 × 2 = 1 + 0.626 418 392 268 8;
  • 62) 0.626 418 392 268 8 × 2 = 1 + 0.252 836 784 537 6;
  • 63) 0.252 836 784 537 6 × 2 = 0 + 0.505 673 569 075 2;
  • 64) 0.505 673 569 075 2 × 2 = 1 + 0.011 347 138 150 4;

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:


0.000 282 005 913 15(10) =

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

6. Positive number before normalization:

0.000 282 005 913 15(10) =

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


0.000 282 005 913 15(10) =

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

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

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


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 1100 1001 1100 0010 1001 1101 1101 =

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


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 1100 1001 1100 0010 1001 1101 1101


Decimal number -0.000 282 005 913 15 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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