0.126 380 952 461 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.126 380 952 461(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
0.126 380 952 461(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: 0.126 380 952 461.

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) 0.126 380 952 461 × 2 = 0 + 0.252 761 904 922;
  • 2) 0.252 761 904 922 × 2 = 0 + 0.505 523 809 844;
  • 3) 0.505 523 809 844 × 2 = 1 + 0.011 047 619 688;
  • 4) 0.011 047 619 688 × 2 = 0 + 0.022 095 239 376;
  • 5) 0.022 095 239 376 × 2 = 0 + 0.044 190 478 752;
  • 6) 0.044 190 478 752 × 2 = 0 + 0.088 380 957 504;
  • 7) 0.088 380 957 504 × 2 = 0 + 0.176 761 915 008;
  • 8) 0.176 761 915 008 × 2 = 0 + 0.353 523 830 016;
  • 9) 0.353 523 830 016 × 2 = 0 + 0.707 047 660 032;
  • 10) 0.707 047 660 032 × 2 = 1 + 0.414 095 320 064;
  • 11) 0.414 095 320 064 × 2 = 0 + 0.828 190 640 128;
  • 12) 0.828 190 640 128 × 2 = 1 + 0.656 381 280 256;
  • 13) 0.656 381 280 256 × 2 = 1 + 0.312 762 560 512;
  • 14) 0.312 762 560 512 × 2 = 0 + 0.625 525 121 024;
  • 15) 0.625 525 121 024 × 2 = 1 + 0.251 050 242 048;
  • 16) 0.251 050 242 048 × 2 = 0 + 0.502 100 484 096;
  • 17) 0.502 100 484 096 × 2 = 1 + 0.004 200 968 192;
  • 18) 0.004 200 968 192 × 2 = 0 + 0.008 401 936 384;
  • 19) 0.008 401 936 384 × 2 = 0 + 0.016 803 872 768;
  • 20) 0.016 803 872 768 × 2 = 0 + 0.033 607 745 536;
  • 21) 0.033 607 745 536 × 2 = 0 + 0.067 215 491 072;
  • 22) 0.067 215 491 072 × 2 = 0 + 0.134 430 982 144;
  • 23) 0.134 430 982 144 × 2 = 0 + 0.268 861 964 288;
  • 24) 0.268 861 964 288 × 2 = 0 + 0.537 723 928 576;
  • 25) 0.537 723 928 576 × 2 = 1 + 0.075 447 857 152;
  • 26) 0.075 447 857 152 × 2 = 0 + 0.150 895 714 304;
  • 27) 0.150 895 714 304 × 2 = 0 + 0.301 791 428 608;
  • 28) 0.301 791 428 608 × 2 = 0 + 0.603 582 857 216;
  • 29) 0.603 582 857 216 × 2 = 1 + 0.207 165 714 432;
  • 30) 0.207 165 714 432 × 2 = 0 + 0.414 331 428 864;
  • 31) 0.414 331 428 864 × 2 = 0 + 0.828 662 857 728;
  • 32) 0.828 662 857 728 × 2 = 1 + 0.657 325 715 456;
  • 33) 0.657 325 715 456 × 2 = 1 + 0.314 651 430 912;
  • 34) 0.314 651 430 912 × 2 = 0 + 0.629 302 861 824;
  • 35) 0.629 302 861 824 × 2 = 1 + 0.258 605 723 648;
  • 36) 0.258 605 723 648 × 2 = 0 + 0.517 211 447 296;
  • 37) 0.517 211 447 296 × 2 = 1 + 0.034 422 894 592;
  • 38) 0.034 422 894 592 × 2 = 0 + 0.068 845 789 184;
  • 39) 0.068 845 789 184 × 2 = 0 + 0.137 691 578 368;
  • 40) 0.137 691 578 368 × 2 = 0 + 0.275 383 156 736;
  • 41) 0.275 383 156 736 × 2 = 0 + 0.550 766 313 472;
  • 42) 0.550 766 313 472 × 2 = 1 + 0.101 532 626 944;
  • 43) 0.101 532 626 944 × 2 = 0 + 0.203 065 253 888;
  • 44) 0.203 065 253 888 × 2 = 0 + 0.406 130 507 776;
  • 45) 0.406 130 507 776 × 2 = 0 + 0.812 261 015 552;
  • 46) 0.812 261 015 552 × 2 = 1 + 0.624 522 031 104;
  • 47) 0.624 522 031 104 × 2 = 1 + 0.249 044 062 208;
  • 48) 0.249 044 062 208 × 2 = 0 + 0.498 088 124 416;
  • 49) 0.498 088 124 416 × 2 = 0 + 0.996 176 248 832;
  • 50) 0.996 176 248 832 × 2 = 1 + 0.992 352 497 664;
  • 51) 0.992 352 497 664 × 2 = 1 + 0.984 704 995 328;
  • 52) 0.984 704 995 328 × 2 = 1 + 0.969 409 990 656;
  • 53) 0.969 409 990 656 × 2 = 1 + 0.938 819 981 312;
  • 54) 0.938 819 981 312 × 2 = 1 + 0.877 639 962 624;
  • 55) 0.877 639 962 624 × 2 = 1 + 0.755 279 925 248;

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:


0.126 380 952 461(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1010 1000 0100 0110 0111 111(2)

5. Positive number before normalization:

0.126 380 952 461(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1010 1000 0100 0110 0111 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.126 380 952 461(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1010 1000 0100 0110 0111 111(2) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1010 1000 0100 0110 0111 111(2) × 20 =

1.0000 0010 1101 0100 0000 0100 0100 1101 0100 0010 0011 0011 1111(2) × 2-3


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): -3

Mantissa (not normalized):
1.0000 0010 1101 0100 0000 0100 0100 1101 0100 0010 0011 0011 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 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) =

1020(10) =

011 1111 1100(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. 0000 0010 1101 0100 0000 0100 0100 1101 0100 0010 0011 0011 1111 =

0000 0010 1101 0100 0000 0100 0100 1101 0100 0010 0011 0011 1111


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 1100

Mantissa (52 bits) =
0000 0010 1101 0100 0000 0100 0100 1101 0100 0010 0011 0011 1111


Decimal number 0.126 380 952 461 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0000 0010 1101 0100 0000 0100 0100 1101 0100 0010 0011 0011 1111

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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