-0.066 994 523 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.066 994 523 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
-0.066 994 523 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:

|-0.066 994 523 7| = 0.066 994 523 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: 0.066 994 523 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) 0.066 994 523 7 × 2 = 0 + 0.133 989 047 4;
  • 2) 0.133 989 047 4 × 2 = 0 + 0.267 978 094 8;
  • 3) 0.267 978 094 8 × 2 = 0 + 0.535 956 189 6;
  • 4) 0.535 956 189 6 × 2 = 1 + 0.071 912 379 2;
  • 5) 0.071 912 379 2 × 2 = 0 + 0.143 824 758 4;
  • 6) 0.143 824 758 4 × 2 = 0 + 0.287 649 516 8;
  • 7) 0.287 649 516 8 × 2 = 0 + 0.575 299 033 6;
  • 8) 0.575 299 033 6 × 2 = 1 + 0.150 598 067 2;
  • 9) 0.150 598 067 2 × 2 = 0 + 0.301 196 134 4;
  • 10) 0.301 196 134 4 × 2 = 0 + 0.602 392 268 8;
  • 11) 0.602 392 268 8 × 2 = 1 + 0.204 784 537 6;
  • 12) 0.204 784 537 6 × 2 = 0 + 0.409 569 075 2;
  • 13) 0.409 569 075 2 × 2 = 0 + 0.819 138 150 4;
  • 14) 0.819 138 150 4 × 2 = 1 + 0.638 276 300 8;
  • 15) 0.638 276 300 8 × 2 = 1 + 0.276 552 601 6;
  • 16) 0.276 552 601 6 × 2 = 0 + 0.553 105 203 2;
  • 17) 0.553 105 203 2 × 2 = 1 + 0.106 210 406 4;
  • 18) 0.106 210 406 4 × 2 = 0 + 0.212 420 812 8;
  • 19) 0.212 420 812 8 × 2 = 0 + 0.424 841 625 6;
  • 20) 0.424 841 625 6 × 2 = 0 + 0.849 683 251 2;
  • 21) 0.849 683 251 2 × 2 = 1 + 0.699 366 502 4;
  • 22) 0.699 366 502 4 × 2 = 1 + 0.398 733 004 8;
  • 23) 0.398 733 004 8 × 2 = 0 + 0.797 466 009 6;
  • 24) 0.797 466 009 6 × 2 = 1 + 0.594 932 019 2;
  • 25) 0.594 932 019 2 × 2 = 1 + 0.189 864 038 4;
  • 26) 0.189 864 038 4 × 2 = 0 + 0.379 728 076 8;
  • 27) 0.379 728 076 8 × 2 = 0 + 0.759 456 153 6;
  • 28) 0.759 456 153 6 × 2 = 1 + 0.518 912 307 2;
  • 29) 0.518 912 307 2 × 2 = 1 + 0.037 824 614 4;
  • 30) 0.037 824 614 4 × 2 = 0 + 0.075 649 228 8;
  • 31) 0.075 649 228 8 × 2 = 0 + 0.151 298 457 6;
  • 32) 0.151 298 457 6 × 2 = 0 + 0.302 596 915 2;
  • 33) 0.302 596 915 2 × 2 = 0 + 0.605 193 830 4;
  • 34) 0.605 193 830 4 × 2 = 1 + 0.210 387 660 8;
  • 35) 0.210 387 660 8 × 2 = 0 + 0.420 775 321 6;
  • 36) 0.420 775 321 6 × 2 = 0 + 0.841 550 643 2;
  • 37) 0.841 550 643 2 × 2 = 1 + 0.683 101 286 4;
  • 38) 0.683 101 286 4 × 2 = 1 + 0.366 202 572 8;
  • 39) 0.366 202 572 8 × 2 = 0 + 0.732 405 145 6;
  • 40) 0.732 405 145 6 × 2 = 1 + 0.464 810 291 2;
  • 41) 0.464 810 291 2 × 2 = 0 + 0.929 620 582 4;
  • 42) 0.929 620 582 4 × 2 = 1 + 0.859 241 164 8;
  • 43) 0.859 241 164 8 × 2 = 1 + 0.718 482 329 6;
  • 44) 0.718 482 329 6 × 2 = 1 + 0.436 964 659 2;
  • 45) 0.436 964 659 2 × 2 = 0 + 0.873 929 318 4;
  • 46) 0.873 929 318 4 × 2 = 1 + 0.747 858 636 8;
  • 47) 0.747 858 636 8 × 2 = 1 + 0.495 717 273 6;
  • 48) 0.495 717 273 6 × 2 = 0 + 0.991 434 547 2;
  • 49) 0.991 434 547 2 × 2 = 1 + 0.982 869 094 4;
  • 50) 0.982 869 094 4 × 2 = 1 + 0.965 738 188 8;
  • 51) 0.965 738 188 8 × 2 = 1 + 0.931 476 377 6;
  • 52) 0.931 476 377 6 × 2 = 1 + 0.862 952 755 2;
  • 53) 0.862 952 755 2 × 2 = 1 + 0.725 905 510 4;
  • 54) 0.725 905 510 4 × 2 = 1 + 0.451 811 020 8;
  • 55) 0.451 811 020 8 × 2 = 0 + 0.903 622 041 6;
  • 56) 0.903 622 041 6 × 2 = 1 + 0.807 244 083 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:


0.066 994 523 7(10) =

0.0001 0001 0010 0110 1000 1101 1001 1000 0100 1101 0111 0110 1111 1101(2)

6. Positive number before normalization:

0.066 994 523 7(10) =

0.0001 0001 0010 0110 1000 1101 1001 1000 0100 1101 0111 0110 1111 1101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.066 994 523 7(10) =

0.0001 0001 0010 0110 1000 1101 1001 1000 0100 1101 0111 0110 1111 1101(2) =

0.0001 0001 0010 0110 1000 1101 1001 1000 0100 1101 0111 0110 1111 1101(2) × 20 =

1.0001 0010 0110 1000 1101 1001 1000 0100 1101 0111 0110 1111 1101(2) × 2-4


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):
1.0001 0010 0110 1000 1101 1001 1000 0100 1101 0111 0110 1111 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 0001 0010 0110 1000 1101 1001 1000 0100 1101 0111 0110 1111 1101 =

0001 0010 0110 1000 1101 1001 1000 0100 1101 0111 0110 1111 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 1011

Mantissa (52 bits) =
0001 0010 0110 1000 1101 1001 1000 0100 1101 0111 0110 1111 1101


Decimal number -0.066 994 523 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

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