-0.008 788 423 613 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 613 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
-0.008 788 423 613 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:

|-0.008 788 423 613 4| = 0.008 788 423 613 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: 0.008 788 423 613 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) 0.008 788 423 613 4 × 2 = 0 + 0.017 576 847 226 8;
  • 2) 0.017 576 847 226 8 × 2 = 0 + 0.035 153 694 453 6;
  • 3) 0.035 153 694 453 6 × 2 = 0 + 0.070 307 388 907 2;
  • 4) 0.070 307 388 907 2 × 2 = 0 + 0.140 614 777 814 4;
  • 5) 0.140 614 777 814 4 × 2 = 0 + 0.281 229 555 628 8;
  • 6) 0.281 229 555 628 8 × 2 = 0 + 0.562 459 111 257 6;
  • 7) 0.562 459 111 257 6 × 2 = 1 + 0.124 918 222 515 2;
  • 8) 0.124 918 222 515 2 × 2 = 0 + 0.249 836 445 030 4;
  • 9) 0.249 836 445 030 4 × 2 = 0 + 0.499 672 890 060 8;
  • 10) 0.499 672 890 060 8 × 2 = 0 + 0.999 345 780 121 6;
  • 11) 0.999 345 780 121 6 × 2 = 1 + 0.998 691 560 243 2;
  • 12) 0.998 691 560 243 2 × 2 = 1 + 0.997 383 120 486 4;
  • 13) 0.997 383 120 486 4 × 2 = 1 + 0.994 766 240 972 8;
  • 14) 0.994 766 240 972 8 × 2 = 1 + 0.989 532 481 945 6;
  • 15) 0.989 532 481 945 6 × 2 = 1 + 0.979 064 963 891 2;
  • 16) 0.979 064 963 891 2 × 2 = 1 + 0.958 129 927 782 4;
  • 17) 0.958 129 927 782 4 × 2 = 1 + 0.916 259 855 564 8;
  • 18) 0.916 259 855 564 8 × 2 = 1 + 0.832 519 711 129 6;
  • 19) 0.832 519 711 129 6 × 2 = 1 + 0.665 039 422 259 2;
  • 20) 0.665 039 422 259 2 × 2 = 1 + 0.330 078 844 518 4;
  • 21) 0.330 078 844 518 4 × 2 = 0 + 0.660 157 689 036 8;
  • 22) 0.660 157 689 036 8 × 2 = 1 + 0.320 315 378 073 6;
  • 23) 0.320 315 378 073 6 × 2 = 0 + 0.640 630 756 147 2;
  • 24) 0.640 630 756 147 2 × 2 = 1 + 0.281 261 512 294 4;
  • 25) 0.281 261 512 294 4 × 2 = 0 + 0.562 523 024 588 8;
  • 26) 0.562 523 024 588 8 × 2 = 1 + 0.125 046 049 177 6;
  • 27) 0.125 046 049 177 6 × 2 = 0 + 0.250 092 098 355 2;
  • 28) 0.250 092 098 355 2 × 2 = 0 + 0.500 184 196 710 4;
  • 29) 0.500 184 196 710 4 × 2 = 1 + 0.000 368 393 420 8;
  • 30) 0.000 368 393 420 8 × 2 = 0 + 0.000 736 786 841 6;
  • 31) 0.000 736 786 841 6 × 2 = 0 + 0.001 473 573 683 2;
  • 32) 0.001 473 573 683 2 × 2 = 0 + 0.002 947 147 366 4;
  • 33) 0.002 947 147 366 4 × 2 = 0 + 0.005 894 294 732 8;
  • 34) 0.005 894 294 732 8 × 2 = 0 + 0.011 788 589 465 6;
  • 35) 0.011 788 589 465 6 × 2 = 0 + 0.023 577 178 931 2;
  • 36) 0.023 577 178 931 2 × 2 = 0 + 0.047 154 357 862 4;
  • 37) 0.047 154 357 862 4 × 2 = 0 + 0.094 308 715 724 8;
  • 38) 0.094 308 715 724 8 × 2 = 0 + 0.188 617 431 449 6;
  • 39) 0.188 617 431 449 6 × 2 = 0 + 0.377 234 862 899 2;
  • 40) 0.377 234 862 899 2 × 2 = 0 + 0.754 469 725 798 4;
  • 41) 0.754 469 725 798 4 × 2 = 1 + 0.508 939 451 596 8;
  • 42) 0.508 939 451 596 8 × 2 = 1 + 0.017 878 903 193 6;
  • 43) 0.017 878 903 193 6 × 2 = 0 + 0.035 757 806 387 2;
  • 44) 0.035 757 806 387 2 × 2 = 0 + 0.071 515 612 774 4;
  • 45) 0.071 515 612 774 4 × 2 = 0 + 0.143 031 225 548 8;
  • 46) 0.143 031 225 548 8 × 2 = 0 + 0.286 062 451 097 6;
  • 47) 0.286 062 451 097 6 × 2 = 0 + 0.572 124 902 195 2;
  • 48) 0.572 124 902 195 2 × 2 = 1 + 0.144 249 804 390 4;
  • 49) 0.144 249 804 390 4 × 2 = 0 + 0.288 499 608 780 8;
  • 50) 0.288 499 608 780 8 × 2 = 0 + 0.576 999 217 561 6;
  • 51) 0.576 999 217 561 6 × 2 = 1 + 0.153 998 435 123 2;
  • 52) 0.153 998 435 123 2 × 2 = 0 + 0.307 996 870 246 4;
  • 53) 0.307 996 870 246 4 × 2 = 0 + 0.615 993 740 492 8;
  • 54) 0.615 993 740 492 8 × 2 = 1 + 0.231 987 480 985 6;
  • 55) 0.231 987 480 985 6 × 2 = 0 + 0.463 974 961 971 2;
  • 56) 0.463 974 961 971 2 × 2 = 0 + 0.927 949 923 942 4;
  • 57) 0.927 949 923 942 4 × 2 = 1 + 0.855 899 847 884 8;
  • 58) 0.855 899 847 884 8 × 2 = 1 + 0.711 799 695 769 6;
  • 59) 0.711 799 695 769 6 × 2 = 1 + 0.423 599 391 539 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:


0.008 788 423 613 4(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1100 0001 0010 0100 111(2)

6. Positive number before normalization:

0.008 788 423 613 4(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1100 0001 0010 0100 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:


0.008 788 423 613 4(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1100 0001 0010 0100 111(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1100 0001 0010 0100 111(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0110 0000 1001 0010 0111(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 0110 0000 1001 0010 0111


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 0110 0000 1001 0010 0111 =

0001 1111 1111 1010 1010 0100 0000 0000 0110 0000 1001 0010 0111


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 0110 0000 1001 0010 0111


Decimal number -0.008 788 423 613 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 0110 0000 1001 0010 0111

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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