0.126 380 952 447 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.126 380 952 447(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
0.126 380 952 447(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: 0.126 380 952 447.

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) 0.126 380 952 447 × 2 = 0 + 0.252 761 904 894;
  • 2) 0.252 761 904 894 × 2 = 0 + 0.505 523 809 788;
  • 3) 0.505 523 809 788 × 2 = 1 + 0.011 047 619 576;
  • 4) 0.011 047 619 576 × 2 = 0 + 0.022 095 239 152;
  • 5) 0.022 095 239 152 × 2 = 0 + 0.044 190 478 304;
  • 6) 0.044 190 478 304 × 2 = 0 + 0.088 380 956 608;
  • 7) 0.088 380 956 608 × 2 = 0 + 0.176 761 913 216;
  • 8) 0.176 761 913 216 × 2 = 0 + 0.353 523 826 432;
  • 9) 0.353 523 826 432 × 2 = 0 + 0.707 047 652 864;
  • 10) 0.707 047 652 864 × 2 = 1 + 0.414 095 305 728;
  • 11) 0.414 095 305 728 × 2 = 0 + 0.828 190 611 456;
  • 12) 0.828 190 611 456 × 2 = 1 + 0.656 381 222 912;
  • 13) 0.656 381 222 912 × 2 = 1 + 0.312 762 445 824;
  • 14) 0.312 762 445 824 × 2 = 0 + 0.625 524 891 648;
  • 15) 0.625 524 891 648 × 2 = 1 + 0.251 049 783 296;
  • 16) 0.251 049 783 296 × 2 = 0 + 0.502 099 566 592;
  • 17) 0.502 099 566 592 × 2 = 1 + 0.004 199 133 184;
  • 18) 0.004 199 133 184 × 2 = 0 + 0.008 398 266 368;
  • 19) 0.008 398 266 368 × 2 = 0 + 0.016 796 532 736;
  • 20) 0.016 796 532 736 × 2 = 0 + 0.033 593 065 472;
  • 21) 0.033 593 065 472 × 2 = 0 + 0.067 186 130 944;
  • 22) 0.067 186 130 944 × 2 = 0 + 0.134 372 261 888;
  • 23) 0.134 372 261 888 × 2 = 0 + 0.268 744 523 776;
  • 24) 0.268 744 523 776 × 2 = 0 + 0.537 489 047 552;
  • 25) 0.537 489 047 552 × 2 = 1 + 0.074 978 095 104;
  • 26) 0.074 978 095 104 × 2 = 0 + 0.149 956 190 208;
  • 27) 0.149 956 190 208 × 2 = 0 + 0.299 912 380 416;
  • 28) 0.299 912 380 416 × 2 = 0 + 0.599 824 760 832;
  • 29) 0.599 824 760 832 × 2 = 1 + 0.199 649 521 664;
  • 30) 0.199 649 521 664 × 2 = 0 + 0.399 299 043 328;
  • 31) 0.399 299 043 328 × 2 = 0 + 0.798 598 086 656;
  • 32) 0.798 598 086 656 × 2 = 1 + 0.597 196 173 312;
  • 33) 0.597 196 173 312 × 2 = 1 + 0.194 392 346 624;
  • 34) 0.194 392 346 624 × 2 = 0 + 0.388 784 693 248;
  • 35) 0.388 784 693 248 × 2 = 0 + 0.777 569 386 496;
  • 36) 0.777 569 386 496 × 2 = 1 + 0.555 138 772 992;
  • 37) 0.555 138 772 992 × 2 = 1 + 0.110 277 545 984;
  • 38) 0.110 277 545 984 × 2 = 0 + 0.220 555 091 968;
  • 39) 0.220 555 091 968 × 2 = 0 + 0.441 110 183 936;
  • 40) 0.441 110 183 936 × 2 = 0 + 0.882 220 367 872;
  • 41) 0.882 220 367 872 × 2 = 1 + 0.764 440 735 744;
  • 42) 0.764 440 735 744 × 2 = 1 + 0.528 881 471 488;
  • 43) 0.528 881 471 488 × 2 = 1 + 0.057 762 942 976;
  • 44) 0.057 762 942 976 × 2 = 0 + 0.115 525 885 952;
  • 45) 0.115 525 885 952 × 2 = 0 + 0.231 051 771 904;
  • 46) 0.231 051 771 904 × 2 = 0 + 0.462 103 543 808;
  • 47) 0.462 103 543 808 × 2 = 0 + 0.924 207 087 616;
  • 48) 0.924 207 087 616 × 2 = 1 + 0.848 414 175 232;
  • 49) 0.848 414 175 232 × 2 = 1 + 0.696 828 350 464;
  • 50) 0.696 828 350 464 × 2 = 1 + 0.393 656 700 928;
  • 51) 0.393 656 700 928 × 2 = 0 + 0.787 313 401 856;
  • 52) 0.787 313 401 856 × 2 = 1 + 0.574 626 803 712;
  • 53) 0.574 626 803 712 × 2 = 1 + 0.149 253 607 424;
  • 54) 0.149 253 607 424 × 2 = 0 + 0.298 507 214 848;
  • 55) 0.298 507 214 848 × 2 = 0 + 0.597 014 429 696;

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:


0.126 380 952 447(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1001 1000 1110 0001 1101 100(2)

5. Positive number before normalization:

0.126 380 952 447(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1001 1000 1110 0001 1101 100(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:


0.126 380 952 447(10) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1001 1000 1110 0001 1101 100(2) =

0.0010 0000 0101 1010 1000 0000 1000 1001 1001 1000 1110 0001 1101 100(2) × 20 =

1.0000 0010 1101 0100 0000 0100 0100 1100 1100 0111 0000 1110 1100(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 1100 1100 0111 0000 1110 1100


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 1100 1100 0111 0000 1110 1100 =

0000 0010 1101 0100 0000 0100 0100 1100 1100 0111 0000 1110 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 1100

Mantissa (52 bits) =
0000 0010 1101 0100 0000 0100 0100 1100 1100 0111 0000 1110 1100


Decimal number 0.126 380 952 447 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0000 0010 1101 0100 0000 0100 0100 1100 1100 0111 0000 1110 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