0.123 456 789 198 31 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.123 456 789 198 31(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
0.123 456 789 198 31(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: 0.123 456 789 198 31.

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) 0.123 456 789 198 31 × 2 = 0 + 0.246 913 578 396 62;
  • 2) 0.246 913 578 396 62 × 2 = 0 + 0.493 827 156 793 24;
  • 3) 0.493 827 156 793 24 × 2 = 0 + 0.987 654 313 586 48;
  • 4) 0.987 654 313 586 48 × 2 = 1 + 0.975 308 627 172 96;
  • 5) 0.975 308 627 172 96 × 2 = 1 + 0.950 617 254 345 92;
  • 6) 0.950 617 254 345 92 × 2 = 1 + 0.901 234 508 691 84;
  • 7) 0.901 234 508 691 84 × 2 = 1 + 0.802 469 017 383 68;
  • 8) 0.802 469 017 383 68 × 2 = 1 + 0.604 938 034 767 36;
  • 9) 0.604 938 034 767 36 × 2 = 1 + 0.209 876 069 534 72;
  • 10) 0.209 876 069 534 72 × 2 = 0 + 0.419 752 139 069 44;
  • 11) 0.419 752 139 069 44 × 2 = 0 + 0.839 504 278 138 88;
  • 12) 0.839 504 278 138 88 × 2 = 1 + 0.679 008 556 277 76;
  • 13) 0.679 008 556 277 76 × 2 = 1 + 0.358 017 112 555 52;
  • 14) 0.358 017 112 555 52 × 2 = 0 + 0.716 034 225 111 04;
  • 15) 0.716 034 225 111 04 × 2 = 1 + 0.432 068 450 222 08;
  • 16) 0.432 068 450 222 08 × 2 = 0 + 0.864 136 900 444 16;
  • 17) 0.864 136 900 444 16 × 2 = 1 + 0.728 273 800 888 32;
  • 18) 0.728 273 800 888 32 × 2 = 1 + 0.456 547 601 776 64;
  • 19) 0.456 547 601 776 64 × 2 = 0 + 0.913 095 203 553 28;
  • 20) 0.913 095 203 553 28 × 2 = 1 + 0.826 190 407 106 56;
  • 21) 0.826 190 407 106 56 × 2 = 1 + 0.652 380 814 213 12;
  • 22) 0.652 380 814 213 12 × 2 = 1 + 0.304 761 628 426 24;
  • 23) 0.304 761 628 426 24 × 2 = 0 + 0.609 523 256 852 48;
  • 24) 0.609 523 256 852 48 × 2 = 1 + 0.219 046 513 704 96;
  • 25) 0.219 046 513 704 96 × 2 = 0 + 0.438 093 027 409 92;
  • 26) 0.438 093 027 409 92 × 2 = 0 + 0.876 186 054 819 84;
  • 27) 0.876 186 054 819 84 × 2 = 1 + 0.752 372 109 639 68;
  • 28) 0.752 372 109 639 68 × 2 = 1 + 0.504 744 219 279 36;
  • 29) 0.504 744 219 279 36 × 2 = 1 + 0.009 488 438 558 72;
  • 30) 0.009 488 438 558 72 × 2 = 0 + 0.018 976 877 117 44;
  • 31) 0.018 976 877 117 44 × 2 = 0 + 0.037 953 754 234 88;
  • 32) 0.037 953 754 234 88 × 2 = 0 + 0.075 907 508 469 76;
  • 33) 0.075 907 508 469 76 × 2 = 0 + 0.151 815 016 939 52;
  • 34) 0.151 815 016 939 52 × 2 = 0 + 0.303 630 033 879 04;
  • 35) 0.303 630 033 879 04 × 2 = 0 + 0.607 260 067 758 08;
  • 36) 0.607 260 067 758 08 × 2 = 1 + 0.214 520 135 516 16;
  • 37) 0.214 520 135 516 16 × 2 = 0 + 0.429 040 271 032 32;
  • 38) 0.429 040 271 032 32 × 2 = 0 + 0.858 080 542 064 64;
  • 39) 0.858 080 542 064 64 × 2 = 1 + 0.716 161 084 129 28;
  • 40) 0.716 161 084 129 28 × 2 = 1 + 0.432 322 168 258 56;
  • 41) 0.432 322 168 258 56 × 2 = 0 + 0.864 644 336 517 12;
  • 42) 0.864 644 336 517 12 × 2 = 1 + 0.729 288 673 034 24;
  • 43) 0.729 288 673 034 24 × 2 = 1 + 0.458 577 346 068 48;
  • 44) 0.458 577 346 068 48 × 2 = 0 + 0.917 154 692 136 96;
  • 45) 0.917 154 692 136 96 × 2 = 1 + 0.834 309 384 273 92;
  • 46) 0.834 309 384 273 92 × 2 = 1 + 0.668 618 768 547 84;
  • 47) 0.668 618 768 547 84 × 2 = 1 + 0.337 237 537 095 68;
  • 48) 0.337 237 537 095 68 × 2 = 0 + 0.674 475 074 191 36;
  • 49) 0.674 475 074 191 36 × 2 = 1 + 0.348 950 148 382 72;
  • 50) 0.348 950 148 382 72 × 2 = 0 + 0.697 900 296 765 44;
  • 51) 0.697 900 296 765 44 × 2 = 1 + 0.395 800 593 530 88;
  • 52) 0.395 800 593 530 88 × 2 = 0 + 0.791 601 187 061 76;
  • 53) 0.791 601 187 061 76 × 2 = 1 + 0.583 202 374 123 52;
  • 54) 0.583 202 374 123 52 × 2 = 1 + 0.166 404 748 247 04;
  • 55) 0.166 404 748 247 04 × 2 = 0 + 0.332 809 496 494 08;
  • 56) 0.332 809 496 494 08 × 2 = 0 + 0.665 618 992 988 16;

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:


0.123 456 789 198 31(10) =

0.0001 1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100(2)

5. Positive number before normalization:

0.123 456 789 198 31(10) =

0.0001 1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.123 456 789 198 31(10) =

0.0001 1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100(2) =

0.0001 1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100(2) × 20 =

1.1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100(2) × 2-4


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): -4

Mantissa (not normalized):
1.1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1019(10) =

011 1111 1011(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. 1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100 =

1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100


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 1011

Mantissa (52 bits) =
1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100


Decimal number 0.123 456 789 198 31 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1111 1001 1010 1101 1101 0011 1000 0001 0011 0110 1110 1010 1100

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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