-0.008 788 423 610 95 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 610 95(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
-0.008 788 423 610 95(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:

|-0.008 788 423 610 95| = 0.008 788 423 610 95


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: 0.008 788 423 610 95.

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) 0.008 788 423 610 95 × 2 = 0 + 0.017 576 847 221 9;
  • 2) 0.017 576 847 221 9 × 2 = 0 + 0.035 153 694 443 8;
  • 3) 0.035 153 694 443 8 × 2 = 0 + 0.070 307 388 887 6;
  • 4) 0.070 307 388 887 6 × 2 = 0 + 0.140 614 777 775 2;
  • 5) 0.140 614 777 775 2 × 2 = 0 + 0.281 229 555 550 4;
  • 6) 0.281 229 555 550 4 × 2 = 0 + 0.562 459 111 100 8;
  • 7) 0.562 459 111 100 8 × 2 = 1 + 0.124 918 222 201 6;
  • 8) 0.124 918 222 201 6 × 2 = 0 + 0.249 836 444 403 2;
  • 9) 0.249 836 444 403 2 × 2 = 0 + 0.499 672 888 806 4;
  • 10) 0.499 672 888 806 4 × 2 = 0 + 0.999 345 777 612 8;
  • 11) 0.999 345 777 612 8 × 2 = 1 + 0.998 691 555 225 6;
  • 12) 0.998 691 555 225 6 × 2 = 1 + 0.997 383 110 451 2;
  • 13) 0.997 383 110 451 2 × 2 = 1 + 0.994 766 220 902 4;
  • 14) 0.994 766 220 902 4 × 2 = 1 + 0.989 532 441 804 8;
  • 15) 0.989 532 441 804 8 × 2 = 1 + 0.979 064 883 609 6;
  • 16) 0.979 064 883 609 6 × 2 = 1 + 0.958 129 767 219 2;
  • 17) 0.958 129 767 219 2 × 2 = 1 + 0.916 259 534 438 4;
  • 18) 0.916 259 534 438 4 × 2 = 1 + 0.832 519 068 876 8;
  • 19) 0.832 519 068 876 8 × 2 = 1 + 0.665 038 137 753 6;
  • 20) 0.665 038 137 753 6 × 2 = 1 + 0.330 076 275 507 2;
  • 21) 0.330 076 275 507 2 × 2 = 0 + 0.660 152 551 014 4;
  • 22) 0.660 152 551 014 4 × 2 = 1 + 0.320 305 102 028 8;
  • 23) 0.320 305 102 028 8 × 2 = 0 + 0.640 610 204 057 6;
  • 24) 0.640 610 204 057 6 × 2 = 1 + 0.281 220 408 115 2;
  • 25) 0.281 220 408 115 2 × 2 = 0 + 0.562 440 816 230 4;
  • 26) 0.562 440 816 230 4 × 2 = 1 + 0.124 881 632 460 8;
  • 27) 0.124 881 632 460 8 × 2 = 0 + 0.249 763 264 921 6;
  • 28) 0.249 763 264 921 6 × 2 = 0 + 0.499 526 529 843 2;
  • 29) 0.499 526 529 843 2 × 2 = 0 + 0.999 053 059 686 4;
  • 30) 0.999 053 059 686 4 × 2 = 1 + 0.998 106 119 372 8;
  • 31) 0.998 106 119 372 8 × 2 = 1 + 0.996 212 238 745 6;
  • 32) 0.996 212 238 745 6 × 2 = 1 + 0.992 424 477 491 2;
  • 33) 0.992 424 477 491 2 × 2 = 1 + 0.984 848 954 982 4;
  • 34) 0.984 848 954 982 4 × 2 = 1 + 0.969 697 909 964 8;
  • 35) 0.969 697 909 964 8 × 2 = 1 + 0.939 395 819 929 6;
  • 36) 0.939 395 819 929 6 × 2 = 1 + 0.878 791 639 859 2;
  • 37) 0.878 791 639 859 2 × 2 = 1 + 0.757 583 279 718 4;
  • 38) 0.757 583 279 718 4 × 2 = 1 + 0.515 166 559 436 8;
  • 39) 0.515 166 559 436 8 × 2 = 1 + 0.030 333 118 873 6;
  • 40) 0.030 333 118 873 6 × 2 = 0 + 0.060 666 237 747 2;
  • 41) 0.060 666 237 747 2 × 2 = 0 + 0.121 332 475 494 4;
  • 42) 0.121 332 475 494 4 × 2 = 0 + 0.242 664 950 988 8;
  • 43) 0.242 664 950 988 8 × 2 = 0 + 0.485 329 901 977 6;
  • 44) 0.485 329 901 977 6 × 2 = 0 + 0.970 659 803 955 2;
  • 45) 0.970 659 803 955 2 × 2 = 1 + 0.941 319 607 910 4;
  • 46) 0.941 319 607 910 4 × 2 = 1 + 0.882 639 215 820 8;
  • 47) 0.882 639 215 820 8 × 2 = 1 + 0.765 278 431 641 6;
  • 48) 0.765 278 431 641 6 × 2 = 1 + 0.530 556 863 283 2;
  • 49) 0.530 556 863 283 2 × 2 = 1 + 0.061 113 726 566 4;
  • 50) 0.061 113 726 566 4 × 2 = 0 + 0.122 227 453 132 8;
  • 51) 0.122 227 453 132 8 × 2 = 0 + 0.244 454 906 265 6;
  • 52) 0.244 454 906 265 6 × 2 = 0 + 0.488 909 812 531 2;
  • 53) 0.488 909 812 531 2 × 2 = 0 + 0.977 819 625 062 4;
  • 54) 0.977 819 625 062 4 × 2 = 1 + 0.955 639 250 124 8;
  • 55) 0.955 639 250 124 8 × 2 = 1 + 0.911 278 500 249 6;
  • 56) 0.911 278 500 249 6 × 2 = 1 + 0.822 557 000 499 2;
  • 57) 0.822 557 000 499 2 × 2 = 1 + 0.645 114 000 998 4;
  • 58) 0.645 114 000 998 4 × 2 = 1 + 0.290 228 001 996 8;
  • 59) 0.290 228 001 996 8 × 2 = 0 + 0.580 456 003 993 6;

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:


0.008 788 423 610 95(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1110 0000 1111 1000 0111 110(2)

6. Positive number before normalization:

0.008 788 423 610 95(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1110 0000 1111 1000 0111 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:


0.008 788 423 610 95(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1110 0000 1111 1000 0111 110(2) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1110 0000 1111 1000 0111 110(2) × 20 =

1.0001 1111 1111 1010 1010 0011 1111 1111 0000 0111 1100 0011 1110(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 0011 1111 1111 0000 0111 1100 0011 1110


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 0011 1111 1111 0000 0111 1100 0011 1110 =

0001 1111 1111 1010 1010 0011 1111 1111 0000 0111 1100 0011 1110


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 0011 1111 1111 0000 0111 1100 0011 1110


Decimal number -0.008 788 423 610 95 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0011 1111 1111 0000 0111 1100 0011 1110

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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