-0.129 999 999 999 971 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.129 999 999 999 971 4(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
-0.129 999 999 999 971 4(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. Start with the positive version of the number:

|-0.129 999 999 999 971 4| = 0.129 999 999 999 971 4


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.129 999 999 999 971 4.

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) 0.129 999 999 999 971 4 × 2 = 0 + 0.259 999 999 999 942 8;
  • 2) 0.259 999 999 999 942 8 × 2 = 0 + 0.519 999 999 999 885 6;
  • 3) 0.519 999 999 999 885 6 × 2 = 1 + 0.039 999 999 999 771 2;
  • 4) 0.039 999 999 999 771 2 × 2 = 0 + 0.079 999 999 999 542 4;
  • 5) 0.079 999 999 999 542 4 × 2 = 0 + 0.159 999 999 999 084 8;
  • 6) 0.159 999 999 999 084 8 × 2 = 0 + 0.319 999 999 998 169 6;
  • 7) 0.319 999 999 998 169 6 × 2 = 0 + 0.639 999 999 996 339 2;
  • 8) 0.639 999 999 996 339 2 × 2 = 1 + 0.279 999 999 992 678 4;
  • 9) 0.279 999 999 992 678 4 × 2 = 0 + 0.559 999 999 985 356 8;
  • 10) 0.559 999 999 985 356 8 × 2 = 1 + 0.119 999 999 970 713 6;
  • 11) 0.119 999 999 970 713 6 × 2 = 0 + 0.239 999 999 941 427 2;
  • 12) 0.239 999 999 941 427 2 × 2 = 0 + 0.479 999 999 882 854 4;
  • 13) 0.479 999 999 882 854 4 × 2 = 0 + 0.959 999 999 765 708 8;
  • 14) 0.959 999 999 765 708 8 × 2 = 1 + 0.919 999 999 531 417 6;
  • 15) 0.919 999 999 531 417 6 × 2 = 1 + 0.839 999 999 062 835 2;
  • 16) 0.839 999 999 062 835 2 × 2 = 1 + 0.679 999 998 125 670 4;
  • 17) 0.679 999 998 125 670 4 × 2 = 1 + 0.359 999 996 251 340 8;
  • 18) 0.359 999 996 251 340 8 × 2 = 0 + 0.719 999 992 502 681 6;
  • 19) 0.719 999 992 502 681 6 × 2 = 1 + 0.439 999 985 005 363 2;
  • 20) 0.439 999 985 005 363 2 × 2 = 0 + 0.879 999 970 010 726 4;
  • 21) 0.879 999 970 010 726 4 × 2 = 1 + 0.759 999 940 021 452 8;
  • 22) 0.759 999 940 021 452 8 × 2 = 1 + 0.519 999 880 042 905 6;
  • 23) 0.519 999 880 042 905 6 × 2 = 1 + 0.039 999 760 085 811 2;
  • 24) 0.039 999 760 085 811 2 × 2 = 0 + 0.079 999 520 171 622 4;
  • 25) 0.079 999 520 171 622 4 × 2 = 0 + 0.159 999 040 343 244 8;
  • 26) 0.159 999 040 343 244 8 × 2 = 0 + 0.319 998 080 686 489 6;
  • 27) 0.319 998 080 686 489 6 × 2 = 0 + 0.639 996 161 372 979 2;
  • 28) 0.639 996 161 372 979 2 × 2 = 1 + 0.279 992 322 745 958 4;
  • 29) 0.279 992 322 745 958 4 × 2 = 0 + 0.559 984 645 491 916 8;
  • 30) 0.559 984 645 491 916 8 × 2 = 1 + 0.119 969 290 983 833 6;
  • 31) 0.119 969 290 983 833 6 × 2 = 0 + 0.239 938 581 967 667 2;
  • 32) 0.239 938 581 967 667 2 × 2 = 0 + 0.479 877 163 935 334 4;
  • 33) 0.479 877 163 935 334 4 × 2 = 0 + 0.959 754 327 870 668 8;
  • 34) 0.959 754 327 870 668 8 × 2 = 1 + 0.919 508 655 741 337 6;
  • 35) 0.919 508 655 741 337 6 × 2 = 1 + 0.839 017 311 482 675 2;
  • 36) 0.839 017 311 482 675 2 × 2 = 1 + 0.678 034 622 965 350 4;
  • 37) 0.678 034 622 965 350 4 × 2 = 1 + 0.356 069 245 930 700 8;
  • 38) 0.356 069 245 930 700 8 × 2 = 0 + 0.712 138 491 861 401 6;
  • 39) 0.712 138 491 861 401 6 × 2 = 1 + 0.424 276 983 722 803 2;
  • 40) 0.424 276 983 722 803 2 × 2 = 0 + 0.848 553 967 445 606 4;
  • 41) 0.848 553 967 445 606 4 × 2 = 1 + 0.697 107 934 891 212 8;
  • 42) 0.697 107 934 891 212 8 × 2 = 1 + 0.394 215 869 782 425 6;
  • 43) 0.394 215 869 782 425 6 × 2 = 0 + 0.788 431 739 564 851 2;
  • 44) 0.788 431 739 564 851 2 × 2 = 1 + 0.576 863 479 129 702 4;
  • 45) 0.576 863 479 129 702 4 × 2 = 1 + 0.153 726 958 259 404 8;
  • 46) 0.153 726 958 259 404 8 × 2 = 0 + 0.307 453 916 518 809 6;
  • 47) 0.307 453 916 518 809 6 × 2 = 0 + 0.614 907 833 037 619 2;
  • 48) 0.614 907 833 037 619 2 × 2 = 1 + 0.229 815 666 075 238 4;
  • 49) 0.229 815 666 075 238 4 × 2 = 0 + 0.459 631 332 150 476 8;
  • 50) 0.459 631 332 150 476 8 × 2 = 0 + 0.919 262 664 300 953 6;
  • 51) 0.919 262 664 300 953 6 × 2 = 1 + 0.838 525 328 601 907 2;
  • 52) 0.838 525 328 601 907 2 × 2 = 1 + 0.677 050 657 203 814 4;
  • 53) 0.677 050 657 203 814 4 × 2 = 1 + 0.354 101 314 407 628 8;
  • 54) 0.354 101 314 407 628 8 × 2 = 0 + 0.708 202 628 815 257 6;
  • 55) 0.708 202 628 815 257 6 × 2 = 1 + 0.416 405 257 630 515 2;

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).


5. 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:


0.129 999 999 999 971 4(10) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 1001 0011 101(2)

6. Positive number before normalization:

0.129 999 999 999 971 4(10) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 1001 0011 101(2)

7. 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:


0.129 999 999 999 971 4(10) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 1001 0011 101(2) =

0.0010 0001 0100 0111 1010 1110 0001 0100 0111 1010 1101 1001 0011 101(2) × 20 =

1.0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 1001 1101(2) × 2-3


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): -3

Mantissa (not normalized):
1.0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 1001 1101


9. 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)


10. 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;

11. 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)


12. 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 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 1001 1101 =

0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 1001 1101


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1100

Mantissa (52 bits) =
0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 1001 1101


Decimal number -0.129 999 999 999 971 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1100 1001 1101

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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