-0.008 788 423 620 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 620 1(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
-0.008 788 423 620 1(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:

|-0.008 788 423 620 1| = 0.008 788 423 620 1


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: 0.008 788 423 620 1.

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) 0.008 788 423 620 1 × 2 = 0 + 0.017 576 847 240 2;
  • 2) 0.017 576 847 240 2 × 2 = 0 + 0.035 153 694 480 4;
  • 3) 0.035 153 694 480 4 × 2 = 0 + 0.070 307 388 960 8;
  • 4) 0.070 307 388 960 8 × 2 = 0 + 0.140 614 777 921 6;
  • 5) 0.140 614 777 921 6 × 2 = 0 + 0.281 229 555 843 2;
  • 6) 0.281 229 555 843 2 × 2 = 0 + 0.562 459 111 686 4;
  • 7) 0.562 459 111 686 4 × 2 = 1 + 0.124 918 223 372 8;
  • 8) 0.124 918 223 372 8 × 2 = 0 + 0.249 836 446 745 6;
  • 9) 0.249 836 446 745 6 × 2 = 0 + 0.499 672 893 491 2;
  • 10) 0.499 672 893 491 2 × 2 = 0 + 0.999 345 786 982 4;
  • 11) 0.999 345 786 982 4 × 2 = 1 + 0.998 691 573 964 8;
  • 12) 0.998 691 573 964 8 × 2 = 1 + 0.997 383 147 929 6;
  • 13) 0.997 383 147 929 6 × 2 = 1 + 0.994 766 295 859 2;
  • 14) 0.994 766 295 859 2 × 2 = 1 + 0.989 532 591 718 4;
  • 15) 0.989 532 591 718 4 × 2 = 1 + 0.979 065 183 436 8;
  • 16) 0.979 065 183 436 8 × 2 = 1 + 0.958 130 366 873 6;
  • 17) 0.958 130 366 873 6 × 2 = 1 + 0.916 260 733 747 2;
  • 18) 0.916 260 733 747 2 × 2 = 1 + 0.832 521 467 494 4;
  • 19) 0.832 521 467 494 4 × 2 = 1 + 0.665 042 934 988 8;
  • 20) 0.665 042 934 988 8 × 2 = 1 + 0.330 085 869 977 6;
  • 21) 0.330 085 869 977 6 × 2 = 0 + 0.660 171 739 955 2;
  • 22) 0.660 171 739 955 2 × 2 = 1 + 0.320 343 479 910 4;
  • 23) 0.320 343 479 910 4 × 2 = 0 + 0.640 686 959 820 8;
  • 24) 0.640 686 959 820 8 × 2 = 1 + 0.281 373 919 641 6;
  • 25) 0.281 373 919 641 6 × 2 = 0 + 0.562 747 839 283 2;
  • 26) 0.562 747 839 283 2 × 2 = 1 + 0.125 495 678 566 4;
  • 27) 0.125 495 678 566 4 × 2 = 0 + 0.250 991 357 132 8;
  • 28) 0.250 991 357 132 8 × 2 = 0 + 0.501 982 714 265 6;
  • 29) 0.501 982 714 265 6 × 2 = 1 + 0.003 965 428 531 2;
  • 30) 0.003 965 428 531 2 × 2 = 0 + 0.007 930 857 062 4;
  • 31) 0.007 930 857 062 4 × 2 = 0 + 0.015 861 714 124 8;
  • 32) 0.015 861 714 124 8 × 2 = 0 + 0.031 723 428 249 6;
  • 33) 0.031 723 428 249 6 × 2 = 0 + 0.063 446 856 499 2;
  • 34) 0.063 446 856 499 2 × 2 = 0 + 0.126 893 712 998 4;
  • 35) 0.126 893 712 998 4 × 2 = 0 + 0.253 787 425 996 8;
  • 36) 0.253 787 425 996 8 × 2 = 0 + 0.507 574 851 993 6;
  • 37) 0.507 574 851 993 6 × 2 = 1 + 0.015 149 703 987 2;
  • 38) 0.015 149 703 987 2 × 2 = 0 + 0.030 299 407 974 4;
  • 39) 0.030 299 407 974 4 × 2 = 0 + 0.060 598 815 948 8;
  • 40) 0.060 598 815 948 8 × 2 = 0 + 0.121 197 631 897 6;
  • 41) 0.121 197 631 897 6 × 2 = 0 + 0.242 395 263 795 2;
  • 42) 0.242 395 263 795 2 × 2 = 0 + 0.484 790 527 590 4;
  • 43) 0.484 790 527 590 4 × 2 = 0 + 0.969 581 055 180 8;
  • 44) 0.969 581 055 180 8 × 2 = 1 + 0.939 162 110 361 6;
  • 45) 0.939 162 110 361 6 × 2 = 1 + 0.878 324 220 723 2;
  • 46) 0.878 324 220 723 2 × 2 = 1 + 0.756 648 441 446 4;
  • 47) 0.756 648 441 446 4 × 2 = 1 + 0.513 296 882 892 8;
  • 48) 0.513 296 882 892 8 × 2 = 1 + 0.026 593 765 785 6;
  • 49) 0.026 593 765 785 6 × 2 = 0 + 0.053 187 531 571 2;
  • 50) 0.053 187 531 571 2 × 2 = 0 + 0.106 375 063 142 4;
  • 51) 0.106 375 063 142 4 × 2 = 0 + 0.212 750 126 284 8;
  • 52) 0.212 750 126 284 8 × 2 = 0 + 0.425 500 252 569 6;
  • 53) 0.425 500 252 569 6 × 2 = 0 + 0.851 000 505 139 2;
  • 54) 0.851 000 505 139 2 × 2 = 1 + 0.702 001 010 278 4;
  • 55) 0.702 001 010 278 4 × 2 = 1 + 0.404 002 020 556 8;
  • 56) 0.404 002 020 556 8 × 2 = 0 + 0.808 004 041 113 6;
  • 57) 0.808 004 041 113 6 × 2 = 1 + 0.616 008 082 227 2;
  • 58) 0.616 008 082 227 2 × 2 = 1 + 0.232 016 164 454 4;
  • 59) 0.232 016 164 454 4 × 2 = 0 + 0.464 032 328 908 8;

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:


0.008 788 423 620 1(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 1000 0001 1111 0000 0110 110(2)

6. Positive number before normalization:

0.008 788 423 620 1(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 1000 0001 1111 0000 0110 110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.008 788 423 620 1(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 1000 0001 1111 0000 0110 110(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 1000 0001 1111 0000 0110 110(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0100 0000 1111 1000 0011 0110(2) × 2-7


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): -7

Mantissa (not normalized):
1.0001 1111 1111 1010 1010 0100 0000 0100 0000 1111 1000 0011 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1016(10) =

011 1111 1000(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 1111 1111 1010 1010 0100 0000 0100 0000 1111 1000 0011 0110 =

0001 1111 1111 1010 1010 0100 0000 0100 0000 1111 1000 0011 0110


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 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0100 0000 0100 0000 1111 1000 0011 0110


Decimal number -0.008 788 423 620 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0100 0000 1111 1000 0011 0110

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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