1.234 567 890 123 456 79 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.234 567 890 123 456 79(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
1.234 567 890 123 456 79(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: 0.234 567 890 123 456 79.

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) 0.234 567 890 123 456 79 × 2 = 0 + 0.469 135 780 246 913 58;
  • 2) 0.469 135 780 246 913 58 × 2 = 0 + 0.938 271 560 493 827 16;
  • 3) 0.938 271 560 493 827 16 × 2 = 1 + 0.876 543 120 987 654 32;
  • 4) 0.876 543 120 987 654 32 × 2 = 1 + 0.753 086 241 975 308 64;
  • 5) 0.753 086 241 975 308 64 × 2 = 1 + 0.506 172 483 950 617 28;
  • 6) 0.506 172 483 950 617 28 × 2 = 1 + 0.012 344 967 901 234 56;
  • 7) 0.012 344 967 901 234 56 × 2 = 0 + 0.024 689 935 802 469 12;
  • 8) 0.024 689 935 802 469 12 × 2 = 0 + 0.049 379 871 604 938 24;
  • 9) 0.049 379 871 604 938 24 × 2 = 0 + 0.098 759 743 209 876 48;
  • 10) 0.098 759 743 209 876 48 × 2 = 0 + 0.197 519 486 419 752 96;
  • 11) 0.197 519 486 419 752 96 × 2 = 0 + 0.395 038 972 839 505 92;
  • 12) 0.395 038 972 839 505 92 × 2 = 0 + 0.790 077 945 679 011 84;
  • 13) 0.790 077 945 679 011 84 × 2 = 1 + 0.580 155 891 358 023 68;
  • 14) 0.580 155 891 358 023 68 × 2 = 1 + 0.160 311 782 716 047 36;
  • 15) 0.160 311 782 716 047 36 × 2 = 0 + 0.320 623 565 432 094 72;
  • 16) 0.320 623 565 432 094 72 × 2 = 0 + 0.641 247 130 864 189 44;
  • 17) 0.641 247 130 864 189 44 × 2 = 1 + 0.282 494 261 728 378 88;
  • 18) 0.282 494 261 728 378 88 × 2 = 0 + 0.564 988 523 456 757 76;
  • 19) 0.564 988 523 456 757 76 × 2 = 1 + 0.129 977 046 913 515 52;
  • 20) 0.129 977 046 913 515 52 × 2 = 0 + 0.259 954 093 827 031 04;
  • 21) 0.259 954 093 827 031 04 × 2 = 0 + 0.519 908 187 654 062 08;
  • 22) 0.519 908 187 654 062 08 × 2 = 1 + 0.039 816 375 308 124 16;
  • 23) 0.039 816 375 308 124 16 × 2 = 0 + 0.079 632 750 616 248 32;
  • 24) 0.079 632 750 616 248 32 × 2 = 0 + 0.159 265 501 232 496 64;
  • 25) 0.159 265 501 232 496 64 × 2 = 0 + 0.318 531 002 464 993 28;
  • 26) 0.318 531 002 464 993 28 × 2 = 0 + 0.637 062 004 929 986 56;
  • 27) 0.637 062 004 929 986 56 × 2 = 1 + 0.274 124 009 859 973 12;
  • 28) 0.274 124 009 859 973 12 × 2 = 0 + 0.548 248 019 719 946 24;
  • 29) 0.548 248 019 719 946 24 × 2 = 1 + 0.096 496 039 439 892 48;
  • 30) 0.096 496 039 439 892 48 × 2 = 0 + 0.192 992 078 879 784 96;
  • 31) 0.192 992 078 879 784 96 × 2 = 0 + 0.385 984 157 759 569 92;
  • 32) 0.385 984 157 759 569 92 × 2 = 0 + 0.771 968 315 519 139 84;
  • 33) 0.771 968 315 519 139 84 × 2 = 1 + 0.543 936 631 038 279 68;
  • 34) 0.543 936 631 038 279 68 × 2 = 1 + 0.087 873 262 076 559 36;
  • 35) 0.087 873 262 076 559 36 × 2 = 0 + 0.175 746 524 153 118 72;
  • 36) 0.175 746 524 153 118 72 × 2 = 0 + 0.351 493 048 306 237 44;
  • 37) 0.351 493 048 306 237 44 × 2 = 0 + 0.702 986 096 612 474 88;
  • 38) 0.702 986 096 612 474 88 × 2 = 1 + 0.405 972 193 224 949 76;
  • 39) 0.405 972 193 224 949 76 × 2 = 0 + 0.811 944 386 449 899 52;
  • 40) 0.811 944 386 449 899 52 × 2 = 1 + 0.623 888 772 899 799 04;
  • 41) 0.623 888 772 899 799 04 × 2 = 1 + 0.247 777 545 799 598 08;
  • 42) 0.247 777 545 799 598 08 × 2 = 0 + 0.495 555 091 599 196 16;
  • 43) 0.495 555 091 599 196 16 × 2 = 0 + 0.991 110 183 198 392 32;
  • 44) 0.991 110 183 198 392 32 × 2 = 1 + 0.982 220 366 396 784 64;
  • 45) 0.982 220 366 396 784 64 × 2 = 1 + 0.964 440 732 793 569 28;
  • 46) 0.964 440 732 793 569 28 × 2 = 1 + 0.928 881 465 587 138 56;
  • 47) 0.928 881 465 587 138 56 × 2 = 1 + 0.857 762 931 174 277 12;
  • 48) 0.857 762 931 174 277 12 × 2 = 1 + 0.715 525 862 348 554 24;
  • 49) 0.715 525 862 348 554 24 × 2 = 1 + 0.431 051 724 697 108 48;
  • 50) 0.431 051 724 697 108 48 × 2 = 0 + 0.862 103 449 394 216 96;
  • 51) 0.862 103 449 394 216 96 × 2 = 1 + 0.724 206 898 788 433 92;
  • 52) 0.724 206 898 788 433 92 × 2 = 1 + 0.448 413 797 576 867 84;
  • 53) 0.448 413 797 576 867 84 × 2 = 0 + 0.896 827 595 153 735 68;

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:


0.234 567 890 123 456 79(10) =

0.0011 1100 0000 1100 1010 0100 0010 1000 1100 0101 1001 1111 1011 0(2)

5. Positive number before normalization:

1.234 567 890 123 456 79(10) =

1.0011 1100 0000 1100 1010 0100 0010 1000 1100 0101 1001 1111 1011 0(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:


1.234 567 890 123 456 79(10) =

1.0011 1100 0000 1100 1010 0100 0010 1000 1100 0101 1001 1111 1011 0(2) =

1.0011 1100 0000 1100 1010 0100 0010 1000 1100 0101 1001 1111 1011 0(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):
1.0011 1100 0000 1100 1010 0100 0010 1000 1100 0101 1001 1111 1011 0


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. 0011 1100 0000 1100 1010 0100 0010 1000 1100 0101 1001 1111 1011 0 =

0011 1100 0000 1100 1010 0100 0010 1000 1100 0101 1001 1111 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) =
0011 1100 0000 1100 1010 0100 0010 1000 1100 0101 1001 1111 1011


Decimal number 1.234 567 890 123 456 79 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0011 1100 0000 1100 1010 0100 0010 1000 1100 0101 1001 1111 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