0.126 380 952 378 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.126 380 952 378(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
0.126 380 952 378(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: 0.126 380 952 378.

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) 0.126 380 952 378 × 2 = 0 + 0.252 761 904 756;
  • 2) 0.252 761 904 756 × 2 = 0 + 0.505 523 809 512;
  • 3) 0.505 523 809 512 × 2 = 1 + 0.011 047 619 024;
  • 4) 0.011 047 619 024 × 2 = 0 + 0.022 095 238 048;
  • 5) 0.022 095 238 048 × 2 = 0 + 0.044 190 476 096;
  • 6) 0.044 190 476 096 × 2 = 0 + 0.088 380 952 192;
  • 7) 0.088 380 952 192 × 2 = 0 + 0.176 761 904 384;
  • 8) 0.176 761 904 384 × 2 = 0 + 0.353 523 808 768;
  • 9) 0.353 523 808 768 × 2 = 0 + 0.707 047 617 536;
  • 10) 0.707 047 617 536 × 2 = 1 + 0.414 095 235 072;
  • 11) 0.414 095 235 072 × 2 = 0 + 0.828 190 470 144;
  • 12) 0.828 190 470 144 × 2 = 1 + 0.656 380 940 288;
  • 13) 0.656 380 940 288 × 2 = 1 + 0.312 761 880 576;
  • 14) 0.312 761 880 576 × 2 = 0 + 0.625 523 761 152;
  • 15) 0.625 523 761 152 × 2 = 1 + 0.251 047 522 304;
  • 16) 0.251 047 522 304 × 2 = 0 + 0.502 095 044 608;
  • 17) 0.502 095 044 608 × 2 = 1 + 0.004 190 089 216;
  • 18) 0.004 190 089 216 × 2 = 0 + 0.008 380 178 432;
  • 19) 0.008 380 178 432 × 2 = 0 + 0.016 760 356 864;
  • 20) 0.016 760 356 864 × 2 = 0 + 0.033 520 713 728;
  • 21) 0.033 520 713 728 × 2 = 0 + 0.067 041 427 456;
  • 22) 0.067 041 427 456 × 2 = 0 + 0.134 082 854 912;
  • 23) 0.134 082 854 912 × 2 = 0 + 0.268 165 709 824;
  • 24) 0.268 165 709 824 × 2 = 0 + 0.536 331 419 648;
  • 25) 0.536 331 419 648 × 2 = 1 + 0.072 662 839 296;
  • 26) 0.072 662 839 296 × 2 = 0 + 0.145 325 678 592;
  • 27) 0.145 325 678 592 × 2 = 0 + 0.290 651 357 184;
  • 28) 0.290 651 357 184 × 2 = 0 + 0.581 302 714 368;
  • 29) 0.581 302 714 368 × 2 = 1 + 0.162 605 428 736;
  • 30) 0.162 605 428 736 × 2 = 0 + 0.325 210 857 472;
  • 31) 0.325 210 857 472 × 2 = 0 + 0.650 421 714 944;
  • 32) 0.650 421 714 944 × 2 = 1 + 0.300 843 429 888;
  • 33) 0.300 843 429 888 × 2 = 0 + 0.601 686 859 776;
  • 34) 0.601 686 859 776 × 2 = 1 + 0.203 373 719 552;
  • 35) 0.203 373 719 552 × 2 = 0 + 0.406 747 439 104;
  • 36) 0.406 747 439 104 × 2 = 0 + 0.813 494 878 208;
  • 37) 0.813 494 878 208 × 2 = 1 + 0.626 989 756 416;
  • 38) 0.626 989 756 416 × 2 = 1 + 0.253 979 512 832;
  • 39) 0.253 979 512 832 × 2 = 0 + 0.507 959 025 664;
  • 40) 0.507 959 025 664 × 2 = 1 + 0.015 918 051 328;
  • 41) 0.015 918 051 328 × 2 = 0 + 0.031 836 102 656;
  • 42) 0.031 836 102 656 × 2 = 0 + 0.063 672 205 312;
  • 43) 0.063 672 205 312 × 2 = 0 + 0.127 344 410 624;
  • 44) 0.127 344 410 624 × 2 = 0 + 0.254 688 821 248;
  • 45) 0.254 688 821 248 × 2 = 0 + 0.509 377 642 496;
  • 46) 0.509 377 642 496 × 2 = 1 + 0.018 755 284 992;
  • 47) 0.018 755 284 992 × 2 = 0 + 0.037 510 569 984;
  • 48) 0.037 510 569 984 × 2 = 0 + 0.075 021 139 968;
  • 49) 0.075 021 139 968 × 2 = 0 + 0.150 042 279 936;
  • 50) 0.150 042 279 936 × 2 = 0 + 0.300 084 559 872;
  • 51) 0.300 084 559 872 × 2 = 0 + 0.600 169 119 744;
  • 52) 0.600 169 119 744 × 2 = 1 + 0.200 338 239 488;
  • 53) 0.200 338 239 488 × 2 = 0 + 0.400 676 478 976;
  • 54) 0.400 676 478 976 × 2 = 0 + 0.801 352 957 952;
  • 55) 0.801 352 957 952 × 2 = 1 + 0.602 705 915 904;

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:


0.126 380 952 378(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 0100 1101 0000 0100 0001 001(2)

5. Positive number before normalization:

0.126 380 952 378(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 0100 1101 0000 0100 0001 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:


0.126 380 952 378(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 0100 1101 0000 0100 0001 001(2) =

0.0010 0000 0101 1010 1000 0000 1000 1001 0100 1101 0000 0100 0001 001(2) × 20 =

1.0000 0010 1101 0100 0000 0100 0100 1010 0110 1000 0010 0000 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 1010 0110 1000 0010 0000 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 1010 0110 1000 0010 0000 1001 =

0000 0010 1101 0100 0000 0100 0100 1010 0110 1000 0010 0000 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 1010 0110 1000 0010 0000 1001


Decimal number 0.126 380 952 378 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0000 0010 1101 0100 0000 0100 0100 1010 0110 1000 0010 0000 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