0.906 307 787 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.906 307 787 47(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
0.906 307 787 47(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.906 307 787 47.

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) 0.906 307 787 47 × 2 = 1 + 0.812 615 574 94;
  • 2) 0.812 615 574 94 × 2 = 1 + 0.625 231 149 88;
  • 3) 0.625 231 149 88 × 2 = 1 + 0.250 462 299 76;
  • 4) 0.250 462 299 76 × 2 = 0 + 0.500 924 599 52;
  • 5) 0.500 924 599 52 × 2 = 1 + 0.001 849 199 04;
  • 6) 0.001 849 199 04 × 2 = 0 + 0.003 698 398 08;
  • 7) 0.003 698 398 08 × 2 = 0 + 0.007 396 796 16;
  • 8) 0.007 396 796 16 × 2 = 0 + 0.014 793 592 32;
  • 9) 0.014 793 592 32 × 2 = 0 + 0.029 587 184 64;
  • 10) 0.029 587 184 64 × 2 = 0 + 0.059 174 369 28;
  • 11) 0.059 174 369 28 × 2 = 0 + 0.118 348 738 56;
  • 12) 0.118 348 738 56 × 2 = 0 + 0.236 697 477 12;
  • 13) 0.236 697 477 12 × 2 = 0 + 0.473 394 954 24;
  • 14) 0.473 394 954 24 × 2 = 0 + 0.946 789 908 48;
  • 15) 0.946 789 908 48 × 2 = 1 + 0.893 579 816 96;
  • 16) 0.893 579 816 96 × 2 = 1 + 0.787 159 633 92;
  • 17) 0.787 159 633 92 × 2 = 1 + 0.574 319 267 84;
  • 18) 0.574 319 267 84 × 2 = 1 + 0.148 638 535 68;
  • 19) 0.148 638 535 68 × 2 = 0 + 0.297 277 071 36;
  • 20) 0.297 277 071 36 × 2 = 0 + 0.594 554 142 72;
  • 21) 0.594 554 142 72 × 2 = 1 + 0.189 108 285 44;
  • 22) 0.189 108 285 44 × 2 = 0 + 0.378 216 570 88;
  • 23) 0.378 216 570 88 × 2 = 0 + 0.756 433 141 76;
  • 24) 0.756 433 141 76 × 2 = 1 + 0.512 866 283 52;
  • 25) 0.512 866 283 52 × 2 = 1 + 0.025 732 567 04;
  • 26) 0.025 732 567 04 × 2 = 0 + 0.051 465 134 08;
  • 27) 0.051 465 134 08 × 2 = 0 + 0.102 930 268 16;
  • 28) 0.102 930 268 16 × 2 = 0 + 0.205 860 536 32;
  • 29) 0.205 860 536 32 × 2 = 0 + 0.411 721 072 64;
  • 30) 0.411 721 072 64 × 2 = 0 + 0.823 442 145 28;
  • 31) 0.823 442 145 28 × 2 = 1 + 0.646 884 290 56;
  • 32) 0.646 884 290 56 × 2 = 1 + 0.293 768 581 12;
  • 33) 0.293 768 581 12 × 2 = 0 + 0.587 537 162 24;
  • 34) 0.587 537 162 24 × 2 = 1 + 0.175 074 324 48;
  • 35) 0.175 074 324 48 × 2 = 0 + 0.350 148 648 96;
  • 36) 0.350 148 648 96 × 2 = 0 + 0.700 297 297 92;
  • 37) 0.700 297 297 92 × 2 = 1 + 0.400 594 595 84;
  • 38) 0.400 594 595 84 × 2 = 0 + 0.801 189 191 68;
  • 39) 0.801 189 191 68 × 2 = 1 + 0.602 378 383 36;
  • 40) 0.602 378 383 36 × 2 = 1 + 0.204 756 766 72;
  • 41) 0.204 756 766 72 × 2 = 0 + 0.409 513 533 44;
  • 42) 0.409 513 533 44 × 2 = 0 + 0.819 027 066 88;
  • 43) 0.819 027 066 88 × 2 = 1 + 0.638 054 133 76;
  • 44) 0.638 054 133 76 × 2 = 1 + 0.276 108 267 52;
  • 45) 0.276 108 267 52 × 2 = 0 + 0.552 216 535 04;
  • 46) 0.552 216 535 04 × 2 = 1 + 0.104 433 070 08;
  • 47) 0.104 433 070 08 × 2 = 0 + 0.208 866 140 16;
  • 48) 0.208 866 140 16 × 2 = 0 + 0.417 732 280 32;
  • 49) 0.417 732 280 32 × 2 = 0 + 0.835 464 560 64;
  • 50) 0.835 464 560 64 × 2 = 1 + 0.670 929 121 28;
  • 51) 0.670 929 121 28 × 2 = 1 + 0.341 858 242 56;
  • 52) 0.341 858 242 56 × 2 = 0 + 0.683 716 485 12;
  • 53) 0.683 716 485 12 × 2 = 1 + 0.367 432 970 24;

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).


4. 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:


0.906 307 787 47(10) =

0.1110 1000 0000 0011 1100 1001 1000 0011 0100 1011 0011 0100 0110 1(2)

5. Positive number before normalization:

0.906 307 787 47(10) =

0.1110 1000 0000 0011 1100 1001 1000 0011 0100 1011 0011 0100 0110 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.906 307 787 47(10) =

0.1110 1000 0000 0011 1100 1001 1000 0011 0100 1011 0011 0100 0110 1(2) =

0.1110 1000 0000 0011 1100 1001 1000 0011 0100 1011 0011 0100 0110 1(2) × 20 =

1.1101 0000 0000 0111 1001 0011 0000 0110 1001 0110 0110 1000 1101(2) × 2-1


7. 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 0 (a positive number)

Exponent (unadjusted): -1

Mantissa (not normalized):
1.1101 0000 0000 0111 1001 0011 0000 0110 1001 0110 0110 1000 1101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(10)


9. 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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1022(10) =

011 1111 1110(2)


11. 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. 1101 0000 0000 0111 1001 0011 0000 0110 1001 0110 0110 1000 1101 =

1101 0000 0000 0111 1001 0011 0000 0110 1001 0110 0110 1000 1101


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1110

Mantissa (52 bits) =
1101 0000 0000 0111 1001 0011 0000 0110 1001 0110 0110 1000 1101


Decimal number 0.906 307 787 47 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1101 0000 0000 0111 1001 0011 0000 0110 1001 0110 0110 1000 1101

» Convert decimal 0.906 307 787 54 to 64 bit double precision IEEE 754 binary floating point representation standard

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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