-0.008 788 423 614 46 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 614 46(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
-0.008 788 423 614 46(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:

|-0.008 788 423 614 46| = 0.008 788 423 614 46


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: 0.008 788 423 614 46.

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) 0.008 788 423 614 46 × 2 = 0 + 0.017 576 847 228 92;
  • 2) 0.017 576 847 228 92 × 2 = 0 + 0.035 153 694 457 84;
  • 3) 0.035 153 694 457 84 × 2 = 0 + 0.070 307 388 915 68;
  • 4) 0.070 307 388 915 68 × 2 = 0 + 0.140 614 777 831 36;
  • 5) 0.140 614 777 831 36 × 2 = 0 + 0.281 229 555 662 72;
  • 6) 0.281 229 555 662 72 × 2 = 0 + 0.562 459 111 325 44;
  • 7) 0.562 459 111 325 44 × 2 = 1 + 0.124 918 222 650 88;
  • 8) 0.124 918 222 650 88 × 2 = 0 + 0.249 836 445 301 76;
  • 9) 0.249 836 445 301 76 × 2 = 0 + 0.499 672 890 603 52;
  • 10) 0.499 672 890 603 52 × 2 = 0 + 0.999 345 781 207 04;
  • 11) 0.999 345 781 207 04 × 2 = 1 + 0.998 691 562 414 08;
  • 12) 0.998 691 562 414 08 × 2 = 1 + 0.997 383 124 828 16;
  • 13) 0.997 383 124 828 16 × 2 = 1 + 0.994 766 249 656 32;
  • 14) 0.994 766 249 656 32 × 2 = 1 + 0.989 532 499 312 64;
  • 15) 0.989 532 499 312 64 × 2 = 1 + 0.979 064 998 625 28;
  • 16) 0.979 064 998 625 28 × 2 = 1 + 0.958 129 997 250 56;
  • 17) 0.958 129 997 250 56 × 2 = 1 + 0.916 259 994 501 12;
  • 18) 0.916 259 994 501 12 × 2 = 1 + 0.832 519 989 002 24;
  • 19) 0.832 519 989 002 24 × 2 = 1 + 0.665 039 978 004 48;
  • 20) 0.665 039 978 004 48 × 2 = 1 + 0.330 079 956 008 96;
  • 21) 0.330 079 956 008 96 × 2 = 0 + 0.660 159 912 017 92;
  • 22) 0.660 159 912 017 92 × 2 = 1 + 0.320 319 824 035 84;
  • 23) 0.320 319 824 035 84 × 2 = 0 + 0.640 639 648 071 68;
  • 24) 0.640 639 648 071 68 × 2 = 1 + 0.281 279 296 143 36;
  • 25) 0.281 279 296 143 36 × 2 = 0 + 0.562 558 592 286 72;
  • 26) 0.562 558 592 286 72 × 2 = 1 + 0.125 117 184 573 44;
  • 27) 0.125 117 184 573 44 × 2 = 0 + 0.250 234 369 146 88;
  • 28) 0.250 234 369 146 88 × 2 = 0 + 0.500 468 738 293 76;
  • 29) 0.500 468 738 293 76 × 2 = 1 + 0.000 937 476 587 52;
  • 30) 0.000 937 476 587 52 × 2 = 0 + 0.001 874 953 175 04;
  • 31) 0.001 874 953 175 04 × 2 = 0 + 0.003 749 906 350 08;
  • 32) 0.003 749 906 350 08 × 2 = 0 + 0.007 499 812 700 16;
  • 33) 0.007 499 812 700 16 × 2 = 0 + 0.014 999 625 400 32;
  • 34) 0.014 999 625 400 32 × 2 = 0 + 0.029 999 250 800 64;
  • 35) 0.029 999 250 800 64 × 2 = 0 + 0.059 998 501 601 28;
  • 36) 0.059 998 501 601 28 × 2 = 0 + 0.119 997 003 202 56;
  • 37) 0.119 997 003 202 56 × 2 = 0 + 0.239 994 006 405 12;
  • 38) 0.239 994 006 405 12 × 2 = 0 + 0.479 988 012 810 24;
  • 39) 0.479 988 012 810 24 × 2 = 0 + 0.959 976 025 620 48;
  • 40) 0.959 976 025 620 48 × 2 = 1 + 0.919 952 051 240 96;
  • 41) 0.919 952 051 240 96 × 2 = 1 + 0.839 904 102 481 92;
  • 42) 0.839 904 102 481 92 × 2 = 1 + 0.679 808 204 963 84;
  • 43) 0.679 808 204 963 84 × 2 = 1 + 0.359 616 409 927 68;
  • 44) 0.359 616 409 927 68 × 2 = 0 + 0.719 232 819 855 36;
  • 45) 0.719 232 819 855 36 × 2 = 1 + 0.438 465 639 710 72;
  • 46) 0.438 465 639 710 72 × 2 = 0 + 0.876 931 279 421 44;
  • 47) 0.876 931 279 421 44 × 2 = 1 + 0.753 862 558 842 88;
  • 48) 0.753 862 558 842 88 × 2 = 1 + 0.507 725 117 685 76;
  • 49) 0.507 725 117 685 76 × 2 = 1 + 0.015 450 235 371 52;
  • 50) 0.015 450 235 371 52 × 2 = 0 + 0.030 900 470 743 04;
  • 51) 0.030 900 470 743 04 × 2 = 0 + 0.061 800 941 486 08;
  • 52) 0.061 800 941 486 08 × 2 = 0 + 0.123 601 882 972 16;
  • 53) 0.123 601 882 972 16 × 2 = 0 + 0.247 203 765 944 32;
  • 54) 0.247 203 765 944 32 × 2 = 0 + 0.494 407 531 888 64;
  • 55) 0.494 407 531 888 64 × 2 = 0 + 0.988 815 063 777 28;
  • 56) 0.988 815 063 777 28 × 2 = 1 + 0.977 630 127 554 56;
  • 57) 0.977 630 127 554 56 × 2 = 1 + 0.955 260 255 109 12;
  • 58) 0.955 260 255 109 12 × 2 = 1 + 0.910 520 510 218 24;
  • 59) 0.910 520 510 218 24 × 2 = 1 + 0.821 041 020 436 48;

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:


0.008 788 423 614 46(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 1110 1011 1000 0001 111(2)

6. Positive number before normalization:

0.008 788 423 614 46(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 1110 1011 1000 0001 111(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:


0.008 788 423 614 46(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 1110 1011 1000 0001 111(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 1110 1011 1000 0001 111(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 1111 0101 1100 0000 1111(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 0000 1111 0101 1100 0000 1111


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 0000 1111 0101 1100 0000 1111 =

0001 1111 1111 1010 1010 0100 0000 0000 1111 0101 1100 0000 1111


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 0000 1111 0101 1100 0000 1111


Decimal number -0.008 788 423 614 46 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 1111 0101 1100 0000 1111

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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