-0.000 281 994 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 994 6(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
-0.000 281 994 6(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:

|-0.000 281 994 6| = 0.000 281 994 6


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: 0.000 281 994 6.

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) 0.000 281 994 6 × 2 = 0 + 0.000 563 989 2;
  • 2) 0.000 563 989 2 × 2 = 0 + 0.001 127 978 4;
  • 3) 0.001 127 978 4 × 2 = 0 + 0.002 255 956 8;
  • 4) 0.002 255 956 8 × 2 = 0 + 0.004 511 913 6;
  • 5) 0.004 511 913 6 × 2 = 0 + 0.009 023 827 2;
  • 6) 0.009 023 827 2 × 2 = 0 + 0.018 047 654 4;
  • 7) 0.018 047 654 4 × 2 = 0 + 0.036 095 308 8;
  • 8) 0.036 095 308 8 × 2 = 0 + 0.072 190 617 6;
  • 9) 0.072 190 617 6 × 2 = 0 + 0.144 381 235 2;
  • 10) 0.144 381 235 2 × 2 = 0 + 0.288 762 470 4;
  • 11) 0.288 762 470 4 × 2 = 0 + 0.577 524 940 8;
  • 12) 0.577 524 940 8 × 2 = 1 + 0.155 049 881 6;
  • 13) 0.155 049 881 6 × 2 = 0 + 0.310 099 763 2;
  • 14) 0.310 099 763 2 × 2 = 0 + 0.620 199 526 4;
  • 15) 0.620 199 526 4 × 2 = 1 + 0.240 399 052 8;
  • 16) 0.240 399 052 8 × 2 = 0 + 0.480 798 105 6;
  • 17) 0.480 798 105 6 × 2 = 0 + 0.961 596 211 2;
  • 18) 0.961 596 211 2 × 2 = 1 + 0.923 192 422 4;
  • 19) 0.923 192 422 4 × 2 = 1 + 0.846 384 844 8;
  • 20) 0.846 384 844 8 × 2 = 1 + 0.692 769 689 6;
  • 21) 0.692 769 689 6 × 2 = 1 + 0.385 539 379 2;
  • 22) 0.385 539 379 2 × 2 = 0 + 0.771 078 758 4;
  • 23) 0.771 078 758 4 × 2 = 1 + 0.542 157 516 8;
  • 24) 0.542 157 516 8 × 2 = 1 + 0.084 315 033 6;
  • 25) 0.084 315 033 6 × 2 = 0 + 0.168 630 067 2;
  • 26) 0.168 630 067 2 × 2 = 0 + 0.337 260 134 4;
  • 27) 0.337 260 134 4 × 2 = 0 + 0.674 520 268 8;
  • 28) 0.674 520 268 8 × 2 = 1 + 0.349 040 537 6;
  • 29) 0.349 040 537 6 × 2 = 0 + 0.698 081 075 2;
  • 30) 0.698 081 075 2 × 2 = 1 + 0.396 162 150 4;
  • 31) 0.396 162 150 4 × 2 = 0 + 0.792 324 300 8;
  • 32) 0.792 324 300 8 × 2 = 1 + 0.584 648 601 6;
  • 33) 0.584 648 601 6 × 2 = 1 + 0.169 297 203 2;
  • 34) 0.169 297 203 2 × 2 = 0 + 0.338 594 406 4;
  • 35) 0.338 594 406 4 × 2 = 0 + 0.677 188 812 8;
  • 36) 0.677 188 812 8 × 2 = 1 + 0.354 377 625 6;
  • 37) 0.354 377 625 6 × 2 = 0 + 0.708 755 251 2;
  • 38) 0.708 755 251 2 × 2 = 1 + 0.417 510 502 4;
  • 39) 0.417 510 502 4 × 2 = 0 + 0.835 021 004 8;
  • 40) 0.835 021 004 8 × 2 = 1 + 0.670 042 009 6;
  • 41) 0.670 042 009 6 × 2 = 1 + 0.340 084 019 2;
  • 42) 0.340 084 019 2 × 2 = 0 + 0.680 168 038 4;
  • 43) 0.680 168 038 4 × 2 = 1 + 0.360 336 076 8;
  • 44) 0.360 336 076 8 × 2 = 0 + 0.720 672 153 6;
  • 45) 0.720 672 153 6 × 2 = 1 + 0.441 344 307 2;
  • 46) 0.441 344 307 2 × 2 = 0 + 0.882 688 614 4;
  • 47) 0.882 688 614 4 × 2 = 1 + 0.765 377 228 8;
  • 48) 0.765 377 228 8 × 2 = 1 + 0.530 754 457 6;
  • 49) 0.530 754 457 6 × 2 = 1 + 0.061 508 915 2;
  • 50) 0.061 508 915 2 × 2 = 0 + 0.123 017 830 4;
  • 51) 0.123 017 830 4 × 2 = 0 + 0.246 035 660 8;
  • 52) 0.246 035 660 8 × 2 = 0 + 0.492 071 321 6;
  • 53) 0.492 071 321 6 × 2 = 0 + 0.984 142 643 2;
  • 54) 0.984 142 643 2 × 2 = 1 + 0.968 285 286 4;
  • 55) 0.968 285 286 4 × 2 = 1 + 0.936 570 572 8;
  • 56) 0.936 570 572 8 × 2 = 1 + 0.873 141 145 6;
  • 57) 0.873 141 145 6 × 2 = 1 + 0.746 282 291 2;
  • 58) 0.746 282 291 2 × 2 = 1 + 0.492 564 582 4;
  • 59) 0.492 564 582 4 × 2 = 0 + 0.985 129 164 8;
  • 60) 0.985 129 164 8 × 2 = 1 + 0.970 258 329 6;
  • 61) 0.970 258 329 6 × 2 = 1 + 0.940 516 659 2;
  • 62) 0.940 516 659 2 × 2 = 1 + 0.881 033 318 4;
  • 63) 0.881 033 318 4 × 2 = 1 + 0.762 066 636 8;
  • 64) 0.762 066 636 8 × 2 = 1 + 0.524 133 273 6;

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:


0.000 281 994 6(10) =

0.0000 0000 0001 0010 0111 1011 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111(2)

6. Positive number before normalization:

0.000 281 994 6(10) =

0.0000 0000 0001 0010 0111 1011 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111(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:


0.000 281 994 6(10) =

0.0000 0000 0001 0010 0111 1011 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111(2) =

0.0000 0000 0001 0010 0111 1011 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111(2) × 20 =

1.0010 0111 1011 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111(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 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111


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 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111 =

0010 0111 1011 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111


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 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111


Decimal number -0.000 281 994 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0001 0101 1001 0101 1010 1011 1000 0111 1101 1111

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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