1.797 693 134 71 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.797 693 134 71(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
1.797 693 134 71(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: 0.797 693 134 71.

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) 0.797 693 134 71 × 2 = 1 + 0.595 386 269 42;
  • 2) 0.595 386 269 42 × 2 = 1 + 0.190 772 538 84;
  • 3) 0.190 772 538 84 × 2 = 0 + 0.381 545 077 68;
  • 4) 0.381 545 077 68 × 2 = 0 + 0.763 090 155 36;
  • 5) 0.763 090 155 36 × 2 = 1 + 0.526 180 310 72;
  • 6) 0.526 180 310 72 × 2 = 1 + 0.052 360 621 44;
  • 7) 0.052 360 621 44 × 2 = 0 + 0.104 721 242 88;
  • 8) 0.104 721 242 88 × 2 = 0 + 0.209 442 485 76;
  • 9) 0.209 442 485 76 × 2 = 0 + 0.418 884 971 52;
  • 10) 0.418 884 971 52 × 2 = 0 + 0.837 769 943 04;
  • 11) 0.837 769 943 04 × 2 = 1 + 0.675 539 886 08;
  • 12) 0.675 539 886 08 × 2 = 1 + 0.351 079 772 16;
  • 13) 0.351 079 772 16 × 2 = 0 + 0.702 159 544 32;
  • 14) 0.702 159 544 32 × 2 = 1 + 0.404 319 088 64;
  • 15) 0.404 319 088 64 × 2 = 0 + 0.808 638 177 28;
  • 16) 0.808 638 177 28 × 2 = 1 + 0.617 276 354 56;
  • 17) 0.617 276 354 56 × 2 = 1 + 0.234 552 709 12;
  • 18) 0.234 552 709 12 × 2 = 0 + 0.469 105 418 24;
  • 19) 0.469 105 418 24 × 2 = 0 + 0.938 210 836 48;
  • 20) 0.938 210 836 48 × 2 = 1 + 0.876 421 672 96;
  • 21) 0.876 421 672 96 × 2 = 1 + 0.752 843 345 92;
  • 22) 0.752 843 345 92 × 2 = 1 + 0.505 686 691 84;
  • 23) 0.505 686 691 84 × 2 = 1 + 0.011 373 383 68;
  • 24) 0.011 373 383 68 × 2 = 0 + 0.022 746 767 36;
  • 25) 0.022 746 767 36 × 2 = 0 + 0.045 493 534 72;
  • 26) 0.045 493 534 72 × 2 = 0 + 0.090 987 069 44;
  • 27) 0.090 987 069 44 × 2 = 0 + 0.181 974 138 88;
  • 28) 0.181 974 138 88 × 2 = 0 + 0.363 948 277 76;
  • 29) 0.363 948 277 76 × 2 = 0 + 0.727 896 555 52;
  • 30) 0.727 896 555 52 × 2 = 1 + 0.455 793 111 04;
  • 31) 0.455 793 111 04 × 2 = 0 + 0.911 586 222 08;
  • 32) 0.911 586 222 08 × 2 = 1 + 0.823 172 444 16;
  • 33) 0.823 172 444 16 × 2 = 1 + 0.646 344 888 32;
  • 34) 0.646 344 888 32 × 2 = 1 + 0.292 689 776 64;
  • 35) 0.292 689 776 64 × 2 = 0 + 0.585 379 553 28;
  • 36) 0.585 379 553 28 × 2 = 1 + 0.170 759 106 56;
  • 37) 0.170 759 106 56 × 2 = 0 + 0.341 518 213 12;
  • 38) 0.341 518 213 12 × 2 = 0 + 0.683 036 426 24;
  • 39) 0.683 036 426 24 × 2 = 1 + 0.366 072 852 48;
  • 40) 0.366 072 852 48 × 2 = 0 + 0.732 145 704 96;
  • 41) 0.732 145 704 96 × 2 = 1 + 0.464 291 409 92;
  • 42) 0.464 291 409 92 × 2 = 0 + 0.928 582 819 84;
  • 43) 0.928 582 819 84 × 2 = 1 + 0.857 165 639 68;
  • 44) 0.857 165 639 68 × 2 = 1 + 0.714 331 279 36;
  • 45) 0.714 331 279 36 × 2 = 1 + 0.428 662 558 72;
  • 46) 0.428 662 558 72 × 2 = 0 + 0.857 325 117 44;
  • 47) 0.857 325 117 44 × 2 = 1 + 0.714 650 234 88;
  • 48) 0.714 650 234 88 × 2 = 1 + 0.429 300 469 76;
  • 49) 0.429 300 469 76 × 2 = 0 + 0.858 600 939 52;
  • 50) 0.858 600 939 52 × 2 = 1 + 0.717 201 879 04;
  • 51) 0.717 201 879 04 × 2 = 1 + 0.434 403 758 08;
  • 52) 0.434 403 758 08 × 2 = 0 + 0.868 807 516 16;
  • 53) 0.868 807 516 16 × 2 = 1 + 0.737 615 032 32;

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:


0.797 693 134 71(10) =

0.1100 1100 0011 0101 1001 1110 0000 0101 1101 0010 1011 1011 0110 1(2)

5. Positive number before normalization:

1.797 693 134 71(10) =

1.1100 1100 0011 0101 1001 1110 0000 0101 1101 0010 1011 1011 0110 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:


1.797 693 134 71(10) =

1.1100 1100 0011 0101 1001 1110 0000 0101 1101 0010 1011 1011 0110 1(2) =

1.1100 1100 0011 0101 1001 1110 0000 0101 1101 0010 1011 1011 0110 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):
1.1100 1100 0011 0101 1001 1110 0000 0101 1101 0010 1011 1011 0110 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. 1100 1100 0011 0101 1001 1110 0000 0101 1101 0010 1011 1011 0110 1 =

1100 1100 0011 0101 1001 1110 0000 0101 1101 0010 1011 1011 0110


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) =
1100 1100 0011 0101 1001 1110 0000 0101 1101 0010 1011 1011 0110


Decimal number 1.797 693 134 71 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1100 1100 0011 0101 1001 1110 0000 0101 1101 0010 1011 1011 0110

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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