1.438 847 939 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.438 847 939 99(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
1.438 847 939 99(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: 0.438 847 939 99.

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) 0.438 847 939 99 × 2 = 0 + 0.877 695 879 98;
  • 2) 0.877 695 879 98 × 2 = 1 + 0.755 391 759 96;
  • 3) 0.755 391 759 96 × 2 = 1 + 0.510 783 519 92;
  • 4) 0.510 783 519 92 × 2 = 1 + 0.021 567 039 84;
  • 5) 0.021 567 039 84 × 2 = 0 + 0.043 134 079 68;
  • 6) 0.043 134 079 68 × 2 = 0 + 0.086 268 159 36;
  • 7) 0.086 268 159 36 × 2 = 0 + 0.172 536 318 72;
  • 8) 0.172 536 318 72 × 2 = 0 + 0.345 072 637 44;
  • 9) 0.345 072 637 44 × 2 = 0 + 0.690 145 274 88;
  • 10) 0.690 145 274 88 × 2 = 1 + 0.380 290 549 76;
  • 11) 0.380 290 549 76 × 2 = 0 + 0.760 581 099 52;
  • 12) 0.760 581 099 52 × 2 = 1 + 0.521 162 199 04;
  • 13) 0.521 162 199 04 × 2 = 1 + 0.042 324 398 08;
  • 14) 0.042 324 398 08 × 2 = 0 + 0.084 648 796 16;
  • 15) 0.084 648 796 16 × 2 = 0 + 0.169 297 592 32;
  • 16) 0.169 297 592 32 × 2 = 0 + 0.338 595 184 64;
  • 17) 0.338 595 184 64 × 2 = 0 + 0.677 190 369 28;
  • 18) 0.677 190 369 28 × 2 = 1 + 0.354 380 738 56;
  • 19) 0.354 380 738 56 × 2 = 0 + 0.708 761 477 12;
  • 20) 0.708 761 477 12 × 2 = 1 + 0.417 522 954 24;
  • 21) 0.417 522 954 24 × 2 = 0 + 0.835 045 908 48;
  • 22) 0.835 045 908 48 × 2 = 1 + 0.670 091 816 96;
  • 23) 0.670 091 816 96 × 2 = 1 + 0.340 183 633 92;
  • 24) 0.340 183 633 92 × 2 = 0 + 0.680 367 267 84;
  • 25) 0.680 367 267 84 × 2 = 1 + 0.360 734 535 68;
  • 26) 0.360 734 535 68 × 2 = 0 + 0.721 469 071 36;
  • 27) 0.721 469 071 36 × 2 = 1 + 0.442 938 142 72;
  • 28) 0.442 938 142 72 × 2 = 0 + 0.885 876 285 44;
  • 29) 0.885 876 285 44 × 2 = 1 + 0.771 752 570 88;
  • 30) 0.771 752 570 88 × 2 = 1 + 0.543 505 141 76;
  • 31) 0.543 505 141 76 × 2 = 1 + 0.087 010 283 52;
  • 32) 0.087 010 283 52 × 2 = 0 + 0.174 020 567 04;
  • 33) 0.174 020 567 04 × 2 = 0 + 0.348 041 134 08;
  • 34) 0.348 041 134 08 × 2 = 0 + 0.696 082 268 16;
  • 35) 0.696 082 268 16 × 2 = 1 + 0.392 164 536 32;
  • 36) 0.392 164 536 32 × 2 = 0 + 0.784 329 072 64;
  • 37) 0.784 329 072 64 × 2 = 1 + 0.568 658 145 28;
  • 38) 0.568 658 145 28 × 2 = 1 + 0.137 316 290 56;
  • 39) 0.137 316 290 56 × 2 = 0 + 0.274 632 581 12;
  • 40) 0.274 632 581 12 × 2 = 0 + 0.549 265 162 24;
  • 41) 0.549 265 162 24 × 2 = 1 + 0.098 530 324 48;
  • 42) 0.098 530 324 48 × 2 = 0 + 0.197 060 648 96;
  • 43) 0.197 060 648 96 × 2 = 0 + 0.394 121 297 92;
  • 44) 0.394 121 297 92 × 2 = 0 + 0.788 242 595 84;
  • 45) 0.788 242 595 84 × 2 = 1 + 0.576 485 191 68;
  • 46) 0.576 485 191 68 × 2 = 1 + 0.152 970 383 36;
  • 47) 0.152 970 383 36 × 2 = 0 + 0.305 940 766 72;
  • 48) 0.305 940 766 72 × 2 = 0 + 0.611 881 533 44;
  • 49) 0.611 881 533 44 × 2 = 1 + 0.223 763 066 88;
  • 50) 0.223 763 066 88 × 2 = 0 + 0.447 526 133 76;
  • 51) 0.447 526 133 76 × 2 = 0 + 0.895 052 267 52;
  • 52) 0.895 052 267 52 × 2 = 1 + 0.790 104 535 04;
  • 53) 0.790 104 535 04 × 2 = 1 + 0.580 209 070 08;

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:


0.438 847 939 99(10) =

0.0111 0000 0101 1000 0101 0110 1010 1110 0010 1100 1000 1100 1001 1(2)

5. Positive number before normalization:

1.438 847 939 99(10) =

1.0111 0000 0101 1000 0101 0110 1010 1110 0010 1100 1000 1100 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:


1.438 847 939 99(10) =

1.0111 0000 0101 1000 0101 0110 1010 1110 0010 1100 1000 1100 1001 1(2) =

1.0111 0000 0101 1000 0101 0110 1010 1110 0010 1100 1000 1100 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):
1.0111 0000 0101 1000 0101 0110 1010 1110 0010 1100 1000 1100 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. 0111 0000 0101 1000 0101 0110 1010 1110 0010 1100 1000 1100 1001 1 =

0111 0000 0101 1000 0101 0110 1010 1110 0010 1100 1000 1100 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) =
0111 0000 0101 1000 0101 0110 1010 1110 0010 1100 1000 1100 1001


Decimal number 1.438 847 939 99 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0111 0000 0101 1000 0101 0110 1010 1110 0010 1100 1000 1100 1001

Share

» Convert decimal 1.438 847 939 81 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: My Manager Only Says “We” When Referring to “My Work”. Gemini Notebook YouTube Video of 11.13 minutes

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