-0.000 130 459 697 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 130 459 697 99(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
-0.000 130 459 697 99(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:

|-0.000 130 459 697 99| = 0.000 130 459 697 99


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: 0.000 130 459 697 99.

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) 0.000 130 459 697 99 × 2 = 0 + 0.000 260 919 395 98;
  • 2) 0.000 260 919 395 98 × 2 = 0 + 0.000 521 838 791 96;
  • 3) 0.000 521 838 791 96 × 2 = 0 + 0.001 043 677 583 92;
  • 4) 0.001 043 677 583 92 × 2 = 0 + 0.002 087 355 167 84;
  • 5) 0.002 087 355 167 84 × 2 = 0 + 0.004 174 710 335 68;
  • 6) 0.004 174 710 335 68 × 2 = 0 + 0.008 349 420 671 36;
  • 7) 0.008 349 420 671 36 × 2 = 0 + 0.016 698 841 342 72;
  • 8) 0.016 698 841 342 72 × 2 = 0 + 0.033 397 682 685 44;
  • 9) 0.033 397 682 685 44 × 2 = 0 + 0.066 795 365 370 88;
  • 10) 0.066 795 365 370 88 × 2 = 0 + 0.133 590 730 741 76;
  • 11) 0.133 590 730 741 76 × 2 = 0 + 0.267 181 461 483 52;
  • 12) 0.267 181 461 483 52 × 2 = 0 + 0.534 362 922 967 04;
  • 13) 0.534 362 922 967 04 × 2 = 1 + 0.068 725 845 934 08;
  • 14) 0.068 725 845 934 08 × 2 = 0 + 0.137 451 691 868 16;
  • 15) 0.137 451 691 868 16 × 2 = 0 + 0.274 903 383 736 32;
  • 16) 0.274 903 383 736 32 × 2 = 0 + 0.549 806 767 472 64;
  • 17) 0.549 806 767 472 64 × 2 = 1 + 0.099 613 534 945 28;
  • 18) 0.099 613 534 945 28 × 2 = 0 + 0.199 227 069 890 56;
  • 19) 0.199 227 069 890 56 × 2 = 0 + 0.398 454 139 781 12;
  • 20) 0.398 454 139 781 12 × 2 = 0 + 0.796 908 279 562 24;
  • 21) 0.796 908 279 562 24 × 2 = 1 + 0.593 816 559 124 48;
  • 22) 0.593 816 559 124 48 × 2 = 1 + 0.187 633 118 248 96;
  • 23) 0.187 633 118 248 96 × 2 = 0 + 0.375 266 236 497 92;
  • 24) 0.375 266 236 497 92 × 2 = 0 + 0.750 532 472 995 84;
  • 25) 0.750 532 472 995 84 × 2 = 1 + 0.501 064 945 991 68;
  • 26) 0.501 064 945 991 68 × 2 = 1 + 0.002 129 891 983 36;
  • 27) 0.002 129 891 983 36 × 2 = 0 + 0.004 259 783 966 72;
  • 28) 0.004 259 783 966 72 × 2 = 0 + 0.008 519 567 933 44;
  • 29) 0.008 519 567 933 44 × 2 = 0 + 0.017 039 135 866 88;
  • 30) 0.017 039 135 866 88 × 2 = 0 + 0.034 078 271 733 76;
  • 31) 0.034 078 271 733 76 × 2 = 0 + 0.068 156 543 467 52;
  • 32) 0.068 156 543 467 52 × 2 = 0 + 0.136 313 086 935 04;
  • 33) 0.136 313 086 935 04 × 2 = 0 + 0.272 626 173 870 08;
  • 34) 0.272 626 173 870 08 × 2 = 0 + 0.545 252 347 740 16;
  • 35) 0.545 252 347 740 16 × 2 = 1 + 0.090 504 695 480 32;
  • 36) 0.090 504 695 480 32 × 2 = 0 + 0.181 009 390 960 64;
  • 37) 0.181 009 390 960 64 × 2 = 0 + 0.362 018 781 921 28;
  • 38) 0.362 018 781 921 28 × 2 = 0 + 0.724 037 563 842 56;
  • 39) 0.724 037 563 842 56 × 2 = 1 + 0.448 075 127 685 12;
  • 40) 0.448 075 127 685 12 × 2 = 0 + 0.896 150 255 370 24;
  • 41) 0.896 150 255 370 24 × 2 = 1 + 0.792 300 510 740 48;
  • 42) 0.792 300 510 740 48 × 2 = 1 + 0.584 601 021 480 96;
  • 43) 0.584 601 021 480 96 × 2 = 1 + 0.169 202 042 961 92;
  • 44) 0.169 202 042 961 92 × 2 = 0 + 0.338 404 085 923 84;
  • 45) 0.338 404 085 923 84 × 2 = 0 + 0.676 808 171 847 68;
  • 46) 0.676 808 171 847 68 × 2 = 1 + 0.353 616 343 695 36;
  • 47) 0.353 616 343 695 36 × 2 = 0 + 0.707 232 687 390 72;
  • 48) 0.707 232 687 390 72 × 2 = 1 + 0.414 465 374 781 44;
  • 49) 0.414 465 374 781 44 × 2 = 0 + 0.828 930 749 562 88;
  • 50) 0.828 930 749 562 88 × 2 = 1 + 0.657 861 499 125 76;
  • 51) 0.657 861 499 125 76 × 2 = 1 + 0.315 722 998 251 52;
  • 52) 0.315 722 998 251 52 × 2 = 0 + 0.631 445 996 503 04;
  • 53) 0.631 445 996 503 04 × 2 = 1 + 0.262 891 993 006 08;
  • 54) 0.262 891 993 006 08 × 2 = 0 + 0.525 783 986 012 16;
  • 55) 0.525 783 986 012 16 × 2 = 1 + 0.051 567 972 024 32;
  • 56) 0.051 567 972 024 32 × 2 = 0 + 0.103 135 944 048 64;
  • 57) 0.103 135 944 048 64 × 2 = 0 + 0.206 271 888 097 28;
  • 58) 0.206 271 888 097 28 × 2 = 0 + 0.412 543 776 194 56;
  • 59) 0.412 543 776 194 56 × 2 = 0 + 0.825 087 552 389 12;
  • 60) 0.825 087 552 389 12 × 2 = 1 + 0.650 175 104 778 24;
  • 61) 0.650 175 104 778 24 × 2 = 1 + 0.300 350 209 556 48;
  • 62) 0.300 350 209 556 48 × 2 = 0 + 0.600 700 419 112 96;
  • 63) 0.600 700 419 112 96 × 2 = 1 + 0.201 400 838 225 92;
  • 64) 0.201 400 838 225 92 × 2 = 0 + 0.402 801 676 451 84;
  • 65) 0.402 801 676 451 84 × 2 = 0 + 0.805 603 352 903 68;

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:


0.000 130 459 697 99(10) =

0.0000 0000 0000 1000 1000 1100 1100 0000 0010 0010 1110 0101 0110 1010 0001 1010 0(2)

6. Positive number before normalization:

0.000 130 459 697 99(10) =

0.0000 0000 0000 1000 1000 1100 1100 0000 0010 0010 1110 0101 0110 1010 0001 1010 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the right, so that only one non zero digit remains to the left of it:


0.000 130 459 697 99(10) =

0.0000 0000 0000 1000 1000 1100 1100 0000 0010 0010 1110 0101 0110 1010 0001 1010 0(2) =

0.0000 0000 0000 1000 1000 1100 1100 0000 0010 0010 1110 0101 0110 1010 0001 1010 0(2) × 20 =

1.0001 0001 1001 1000 0000 0100 0101 1100 1010 1101 0100 0011 0100(2) × 2-13


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): -13

Mantissa (not normalized):
1.0001 0001 1001 1000 0000 0100 0101 1100 1010 1101 0100 0011 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-13 + 2(11-1) - 1 =

(-13 + 1 023)(10) =

1 010(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;
  • 1 010 ÷ 2 = 505 + 0;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1010(10) =

011 1111 0010(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. 0001 0001 1001 1000 0000 0100 0101 1100 1010 1101 0100 0011 0100 =

0001 0001 1001 1000 0000 0100 0101 1100 1010 1101 0100 0011 0100


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 0010

Mantissa (52 bits) =
0001 0001 1001 1000 0000 0100 0101 1100 1010 1101 0100 0011 0100


Decimal number -0.000 130 459 697 99 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0010 - 0001 0001 1001 1000 0000 0100 0101 1100 1010 1101 0100 0011 0100

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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