1.005 000 001 66 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.005 000 001 66(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
1.005 000 001 66(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: 1.
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;
  • 1 ÷ 2 = 0 + 1;

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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.005 000 001 66.

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) 0.005 000 001 66 × 2 = 0 + 0.010 000 003 32;
  • 2) 0.010 000 003 32 × 2 = 0 + 0.020 000 006 64;
  • 3) 0.020 000 006 64 × 2 = 0 + 0.040 000 013 28;
  • 4) 0.040 000 013 28 × 2 = 0 + 0.080 000 026 56;
  • 5) 0.080 000 026 56 × 2 = 0 + 0.160 000 053 12;
  • 6) 0.160 000 053 12 × 2 = 0 + 0.320 000 106 24;
  • 7) 0.320 000 106 24 × 2 = 0 + 0.640 000 212 48;
  • 8) 0.640 000 212 48 × 2 = 1 + 0.280 000 424 96;
  • 9) 0.280 000 424 96 × 2 = 0 + 0.560 000 849 92;
  • 10) 0.560 000 849 92 × 2 = 1 + 0.120 001 699 84;
  • 11) 0.120 001 699 84 × 2 = 0 + 0.240 003 399 68;
  • 12) 0.240 003 399 68 × 2 = 0 + 0.480 006 799 36;
  • 13) 0.480 006 799 36 × 2 = 0 + 0.960 013 598 72;
  • 14) 0.960 013 598 72 × 2 = 1 + 0.920 027 197 44;
  • 15) 0.920 027 197 44 × 2 = 1 + 0.840 054 394 88;
  • 16) 0.840 054 394 88 × 2 = 1 + 0.680 108 789 76;
  • 17) 0.680 108 789 76 × 2 = 1 + 0.360 217 579 52;
  • 18) 0.360 217 579 52 × 2 = 0 + 0.720 435 159 04;
  • 19) 0.720 435 159 04 × 2 = 1 + 0.440 870 318 08;
  • 20) 0.440 870 318 08 × 2 = 0 + 0.881 740 636 16;
  • 21) 0.881 740 636 16 × 2 = 1 + 0.763 481 272 32;
  • 22) 0.763 481 272 32 × 2 = 1 + 0.526 962 544 64;
  • 23) 0.526 962 544 64 × 2 = 1 + 0.053 925 089 28;
  • 24) 0.053 925 089 28 × 2 = 0 + 0.107 850 178 56;
  • 25) 0.107 850 178 56 × 2 = 0 + 0.215 700 357 12;
  • 26) 0.215 700 357 12 × 2 = 0 + 0.431 400 714 24;
  • 27) 0.431 400 714 24 × 2 = 0 + 0.862 801 428 48;
  • 28) 0.862 801 428 48 × 2 = 1 + 0.725 602 856 96;
  • 29) 0.725 602 856 96 × 2 = 1 + 0.451 205 713 92;
  • 30) 0.451 205 713 92 × 2 = 0 + 0.902 411 427 84;
  • 31) 0.902 411 427 84 × 2 = 1 + 0.804 822 855 68;
  • 32) 0.804 822 855 68 × 2 = 1 + 0.609 645 711 36;
  • 33) 0.609 645 711 36 × 2 = 1 + 0.219 291 422 72;
  • 34) 0.219 291 422 72 × 2 = 0 + 0.438 582 845 44;
  • 35) 0.438 582 845 44 × 2 = 0 + 0.877 165 690 88;
  • 36) 0.877 165 690 88 × 2 = 1 + 0.754 331 381 76;
  • 37) 0.754 331 381 76 × 2 = 1 + 0.508 662 763 52;
  • 38) 0.508 662 763 52 × 2 = 1 + 0.017 325 527 04;
  • 39) 0.017 325 527 04 × 2 = 0 + 0.034 651 054 08;
  • 40) 0.034 651 054 08 × 2 = 0 + 0.069 302 108 16;
  • 41) 0.069 302 108 16 × 2 = 0 + 0.138 604 216 32;
  • 42) 0.138 604 216 32 × 2 = 0 + 0.277 208 432 64;
  • 43) 0.277 208 432 64 × 2 = 0 + 0.554 416 865 28;
  • 44) 0.554 416 865 28 × 2 = 1 + 0.108 833 730 56;
  • 45) 0.108 833 730 56 × 2 = 0 + 0.217 667 461 12;
  • 46) 0.217 667 461 12 × 2 = 0 + 0.435 334 922 24;
  • 47) 0.435 334 922 24 × 2 = 0 + 0.870 669 844 48;
  • 48) 0.870 669 844 48 × 2 = 1 + 0.741 339 688 96;
  • 49) 0.741 339 688 96 × 2 = 1 + 0.482 679 377 92;
  • 50) 0.482 679 377 92 × 2 = 0 + 0.965 358 755 84;
  • 51) 0.965 358 755 84 × 2 = 1 + 0.930 717 511 68;
  • 52) 0.930 717 511 68 × 2 = 1 + 0.861 435 023 36;
  • 53) 0.861 435 023 36 × 2 = 1 + 0.722 870 046 72;

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:


0.005 000 001 66(10) =

0.0000 0001 0100 0111 1010 1110 0001 1011 1001 1100 0001 0001 1011 1(2)

5. Positive number before normalization:

1.005 000 001 66(10) =

1.0000 0001 0100 0111 1010 1110 0001 1011 1001 1100 0001 0001 1011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.005 000 001 66(10) =

1.0000 0001 0100 0111 1010 1110 0001 1011 1001 1100 0001 0001 1011 1(2) =

1.0000 0001 0100 0111 1010 1110 0001 1011 1001 1100 0001 0001 1011 1(2) × 20


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): 0

Mantissa (not normalized):
1.0000 0001 0100 0111 1010 1110 0001 1011 1001 1100 0001 0001 1011 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 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) =

1023(10) =

011 1111 1111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0000 0001 0100 0111 1010 1110 0001 1011 1001 1100 0001 0001 1011 1 =

0000 0001 0100 0111 1010 1110 0001 1011 1001 1100 0001 0001 1011


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 1111

Mantissa (52 bits) =
0000 0001 0100 0111 1010 1110 0001 1011 1001 1100 0001 0001 1011


Decimal number 1.005 000 001 66 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0000 0001 0100 0111 1010 1110 0001 1011 1001 1100 0001 0001 1011

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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