1.644 934 068 44 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.644 934 068 44(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
1.644 934 068 44(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: 0.644 934 068 44.

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) 0.644 934 068 44 × 2 = 1 + 0.289 868 136 88;
  • 2) 0.289 868 136 88 × 2 = 0 + 0.579 736 273 76;
  • 3) 0.579 736 273 76 × 2 = 1 + 0.159 472 547 52;
  • 4) 0.159 472 547 52 × 2 = 0 + 0.318 945 095 04;
  • 5) 0.318 945 095 04 × 2 = 0 + 0.637 890 190 08;
  • 6) 0.637 890 190 08 × 2 = 1 + 0.275 780 380 16;
  • 7) 0.275 780 380 16 × 2 = 0 + 0.551 560 760 32;
  • 8) 0.551 560 760 32 × 2 = 1 + 0.103 121 520 64;
  • 9) 0.103 121 520 64 × 2 = 0 + 0.206 243 041 28;
  • 10) 0.206 243 041 28 × 2 = 0 + 0.412 486 082 56;
  • 11) 0.412 486 082 56 × 2 = 0 + 0.824 972 165 12;
  • 12) 0.824 972 165 12 × 2 = 1 + 0.649 944 330 24;
  • 13) 0.649 944 330 24 × 2 = 1 + 0.299 888 660 48;
  • 14) 0.299 888 660 48 × 2 = 0 + 0.599 777 320 96;
  • 15) 0.599 777 320 96 × 2 = 1 + 0.199 554 641 92;
  • 16) 0.199 554 641 92 × 2 = 0 + 0.399 109 283 84;
  • 17) 0.399 109 283 84 × 2 = 0 + 0.798 218 567 68;
  • 18) 0.798 218 567 68 × 2 = 1 + 0.596 437 135 36;
  • 19) 0.596 437 135 36 × 2 = 1 + 0.192 874 270 72;
  • 20) 0.192 874 270 72 × 2 = 0 + 0.385 748 541 44;
  • 21) 0.385 748 541 44 × 2 = 0 + 0.771 497 082 88;
  • 22) 0.771 497 082 88 × 2 = 1 + 0.542 994 165 76;
  • 23) 0.542 994 165 76 × 2 = 1 + 0.085 988 331 52;
  • 24) 0.085 988 331 52 × 2 = 0 + 0.171 976 663 04;
  • 25) 0.171 976 663 04 × 2 = 0 + 0.343 953 326 08;
  • 26) 0.343 953 326 08 × 2 = 0 + 0.687 906 652 16;
  • 27) 0.687 906 652 16 × 2 = 1 + 0.375 813 304 32;
  • 28) 0.375 813 304 32 × 2 = 0 + 0.751 626 608 64;
  • 29) 0.751 626 608 64 × 2 = 1 + 0.503 253 217 28;
  • 30) 0.503 253 217 28 × 2 = 1 + 0.006 506 434 56;
  • 31) 0.006 506 434 56 × 2 = 0 + 0.013 012 869 12;
  • 32) 0.013 012 869 12 × 2 = 0 + 0.026 025 738 24;
  • 33) 0.026 025 738 24 × 2 = 0 + 0.052 051 476 48;
  • 34) 0.052 051 476 48 × 2 = 0 + 0.104 102 952 96;
  • 35) 0.104 102 952 96 × 2 = 0 + 0.208 205 905 92;
  • 36) 0.208 205 905 92 × 2 = 0 + 0.416 411 811 84;
  • 37) 0.416 411 811 84 × 2 = 0 + 0.832 823 623 68;
  • 38) 0.832 823 623 68 × 2 = 1 + 0.665 647 247 36;
  • 39) 0.665 647 247 36 × 2 = 1 + 0.331 294 494 72;
  • 40) 0.331 294 494 72 × 2 = 0 + 0.662 588 989 44;
  • 41) 0.662 588 989 44 × 2 = 1 + 0.325 177 978 88;
  • 42) 0.325 177 978 88 × 2 = 0 + 0.650 355 957 76;
  • 43) 0.650 355 957 76 × 2 = 1 + 0.300 711 915 52;
  • 44) 0.300 711 915 52 × 2 = 0 + 0.601 423 831 04;
  • 45) 0.601 423 831 04 × 2 = 1 + 0.202 847 662 08;
  • 46) 0.202 847 662 08 × 2 = 0 + 0.405 695 324 16;
  • 47) 0.405 695 324 16 × 2 = 0 + 0.811 390 648 32;
  • 48) 0.811 390 648 32 × 2 = 1 + 0.622 781 296 64;
  • 49) 0.622 781 296 64 × 2 = 1 + 0.245 562 593 28;
  • 50) 0.245 562 593 28 × 2 = 0 + 0.491 125 186 56;
  • 51) 0.491 125 186 56 × 2 = 0 + 0.982 250 373 12;
  • 52) 0.982 250 373 12 × 2 = 1 + 0.964 500 746 24;
  • 53) 0.964 500 746 24 × 2 = 1 + 0.929 001 492 48;

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:


0.644 934 068 44(10) =

0.1010 0101 0001 1010 0110 0110 0010 1100 0000 0110 1010 1001 1001 1(2)

5. Positive number before normalization:

1.644 934 068 44(10) =

1.1010 0101 0001 1010 0110 0110 0010 1100 0000 0110 1010 1001 1001 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:


1.644 934 068 44(10) =

1.1010 0101 0001 1010 0110 0110 0010 1100 0000 0110 1010 1001 1001 1(2) =

1.1010 0101 0001 1010 0110 0110 0010 1100 0000 0110 1010 1001 1001 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):
1.1010 0101 0001 1010 0110 0110 0010 1100 0000 0110 1010 1001 1001 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. 1010 0101 0001 1010 0110 0110 0010 1100 0000 0110 1010 1001 1001 1 =

1010 0101 0001 1010 0110 0110 0010 1100 0000 0110 1010 1001 1001


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) =
1010 0101 0001 1010 0110 0110 0010 1100 0000 0110 1010 1001 1001


Decimal number 1.644 934 068 44 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1010 0101 0001 1010 0110 0110 0010 1100 0000 0110 1010 1001 1001

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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