0.126 380 952 413 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.126 380 952 413(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
0.126 380 952 413(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: 0.126 380 952 413.

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) 0.126 380 952 413 × 2 = 0 + 0.252 761 904 826;
  • 2) 0.252 761 904 826 × 2 = 0 + 0.505 523 809 652;
  • 3) 0.505 523 809 652 × 2 = 1 + 0.011 047 619 304;
  • 4) 0.011 047 619 304 × 2 = 0 + 0.022 095 238 608;
  • 5) 0.022 095 238 608 × 2 = 0 + 0.044 190 477 216;
  • 6) 0.044 190 477 216 × 2 = 0 + 0.088 380 954 432;
  • 7) 0.088 380 954 432 × 2 = 0 + 0.176 761 908 864;
  • 8) 0.176 761 908 864 × 2 = 0 + 0.353 523 817 728;
  • 9) 0.353 523 817 728 × 2 = 0 + 0.707 047 635 456;
  • 10) 0.707 047 635 456 × 2 = 1 + 0.414 095 270 912;
  • 11) 0.414 095 270 912 × 2 = 0 + 0.828 190 541 824;
  • 12) 0.828 190 541 824 × 2 = 1 + 0.656 381 083 648;
  • 13) 0.656 381 083 648 × 2 = 1 + 0.312 762 167 296;
  • 14) 0.312 762 167 296 × 2 = 0 + 0.625 524 334 592;
  • 15) 0.625 524 334 592 × 2 = 1 + 0.251 048 669 184;
  • 16) 0.251 048 669 184 × 2 = 0 + 0.502 097 338 368;
  • 17) 0.502 097 338 368 × 2 = 1 + 0.004 194 676 736;
  • 18) 0.004 194 676 736 × 2 = 0 + 0.008 389 353 472;
  • 19) 0.008 389 353 472 × 2 = 0 + 0.016 778 706 944;
  • 20) 0.016 778 706 944 × 2 = 0 + 0.033 557 413 888;
  • 21) 0.033 557 413 888 × 2 = 0 + 0.067 114 827 776;
  • 22) 0.067 114 827 776 × 2 = 0 + 0.134 229 655 552;
  • 23) 0.134 229 655 552 × 2 = 0 + 0.268 459 311 104;
  • 24) 0.268 459 311 104 × 2 = 0 + 0.536 918 622 208;
  • 25) 0.536 918 622 208 × 2 = 1 + 0.073 837 244 416;
  • 26) 0.073 837 244 416 × 2 = 0 + 0.147 674 488 832;
  • 27) 0.147 674 488 832 × 2 = 0 + 0.295 348 977 664;
  • 28) 0.295 348 977 664 × 2 = 0 + 0.590 697 955 328;
  • 29) 0.590 697 955 328 × 2 = 1 + 0.181 395 910 656;
  • 30) 0.181 395 910 656 × 2 = 0 + 0.362 791 821 312;
  • 31) 0.362 791 821 312 × 2 = 0 + 0.725 583 642 624;
  • 32) 0.725 583 642 624 × 2 = 1 + 0.451 167 285 248;
  • 33) 0.451 167 285 248 × 2 = 0 + 0.902 334 570 496;
  • 34) 0.902 334 570 496 × 2 = 1 + 0.804 669 140 992;
  • 35) 0.804 669 140 992 × 2 = 1 + 0.609 338 281 984;
  • 36) 0.609 338 281 984 × 2 = 1 + 0.218 676 563 968;
  • 37) 0.218 676 563 968 × 2 = 0 + 0.437 353 127 936;
  • 38) 0.437 353 127 936 × 2 = 0 + 0.874 706 255 872;
  • 39) 0.874 706 255 872 × 2 = 1 + 0.749 412 511 744;
  • 40) 0.749 412 511 744 × 2 = 1 + 0.498 825 023 488;
  • 41) 0.498 825 023 488 × 2 = 0 + 0.997 650 046 976;
  • 42) 0.997 650 046 976 × 2 = 1 + 0.995 300 093 952;
  • 43) 0.995 300 093 952 × 2 = 1 + 0.990 600 187 904;
  • 44) 0.990 600 187 904 × 2 = 1 + 0.981 200 375 808;
  • 45) 0.981 200 375 808 × 2 = 1 + 0.962 400 751 616;
  • 46) 0.962 400 751 616 × 2 = 1 + 0.924 801 503 232;
  • 47) 0.924 801 503 232 × 2 = 1 + 0.849 603 006 464;
  • 48) 0.849 603 006 464 × 2 = 1 + 0.699 206 012 928;
  • 49) 0.699 206 012 928 × 2 = 1 + 0.398 412 025 856;
  • 50) 0.398 412 025 856 × 2 = 0 + 0.796 824 051 712;
  • 51) 0.796 824 051 712 × 2 = 1 + 0.593 648 103 424;
  • 52) 0.593 648 103 424 × 2 = 1 + 0.187 296 206 848;
  • 53) 0.187 296 206 848 × 2 = 0 + 0.374 592 413 696;
  • 54) 0.374 592 413 696 × 2 = 0 + 0.749 184 827 392;
  • 55) 0.749 184 827 392 × 2 = 1 + 0.498 369 654 784;

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:


0.126 380 952 413(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 0111 0011 0111 1111 1011 001(2)

5. Positive number before normalization:

0.126 380 952 413(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 0111 0011 0111 1111 1011 001(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:


0.126 380 952 413(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 0111 0011 0111 1111 1011 001(2) =

0.0010 0000 0101 1010 1000 0000 1000 1001 0111 0011 0111 1111 1011 001(2) × 20 =

1.0000 0010 1101 0100 0000 0100 0100 1011 1001 1011 1111 1101 1001(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 1011 1001 1011 1111 1101 1001


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 1011 1001 1011 1111 1101 1001 =

0000 0010 1101 0100 0000 0100 0100 1011 1001 1011 1111 1101 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 1100

Mantissa (52 bits) =
0000 0010 1101 0100 0000 0100 0100 1011 1001 1011 1111 1101 1001


Decimal number 0.126 380 952 413 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0000 0010 1101 0100 0000 0100 0100 1011 1001 1011 1111 1101 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