-0.008 788 423 613 35 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 613 35(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 35(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 35| = 0.008 788 423 613 35


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 35.

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 35 × 2 = 0 + 0.017 576 847 226 7;
  • 2) 0.017 576 847 226 7 × 2 = 0 + 0.035 153 694 453 4;
  • 3) 0.035 153 694 453 4 × 2 = 0 + 0.070 307 388 906 8;
  • 4) 0.070 307 388 906 8 × 2 = 0 + 0.140 614 777 813 6;
  • 5) 0.140 614 777 813 6 × 2 = 0 + 0.281 229 555 627 2;
  • 6) 0.281 229 555 627 2 × 2 = 0 + 0.562 459 111 254 4;
  • 7) 0.562 459 111 254 4 × 2 = 1 + 0.124 918 222 508 8;
  • 8) 0.124 918 222 508 8 × 2 = 0 + 0.249 836 445 017 6;
  • 9) 0.249 836 445 017 6 × 2 = 0 + 0.499 672 890 035 2;
  • 10) 0.499 672 890 035 2 × 2 = 0 + 0.999 345 780 070 4;
  • 11) 0.999 345 780 070 4 × 2 = 1 + 0.998 691 560 140 8;
  • 12) 0.998 691 560 140 8 × 2 = 1 + 0.997 383 120 281 6;
  • 13) 0.997 383 120 281 6 × 2 = 1 + 0.994 766 240 563 2;
  • 14) 0.994 766 240 563 2 × 2 = 1 + 0.989 532 481 126 4;
  • 15) 0.989 532 481 126 4 × 2 = 1 + 0.979 064 962 252 8;
  • 16) 0.979 064 962 252 8 × 2 = 1 + 0.958 129 924 505 6;
  • 17) 0.958 129 924 505 6 × 2 = 1 + 0.916 259 849 011 2;
  • 18) 0.916 259 849 011 2 × 2 = 1 + 0.832 519 698 022 4;
  • 19) 0.832 519 698 022 4 × 2 = 1 + 0.665 039 396 044 8;
  • 20) 0.665 039 396 044 8 × 2 = 1 + 0.330 078 792 089 6;
  • 21) 0.330 078 792 089 6 × 2 = 0 + 0.660 157 584 179 2;
  • 22) 0.660 157 584 179 2 × 2 = 1 + 0.320 315 168 358 4;
  • 23) 0.320 315 168 358 4 × 2 = 0 + 0.640 630 336 716 8;
  • 24) 0.640 630 336 716 8 × 2 = 1 + 0.281 260 673 433 6;
  • 25) 0.281 260 673 433 6 × 2 = 0 + 0.562 521 346 867 2;
  • 26) 0.562 521 346 867 2 × 2 = 1 + 0.125 042 693 734 4;
  • 27) 0.125 042 693 734 4 × 2 = 0 + 0.250 085 387 468 8;
  • 28) 0.250 085 387 468 8 × 2 = 0 + 0.500 170 774 937 6;
  • 29) 0.500 170 774 937 6 × 2 = 1 + 0.000 341 549 875 2;
  • 30) 0.000 341 549 875 2 × 2 = 0 + 0.000 683 099 750 4;
  • 31) 0.000 683 099 750 4 × 2 = 0 + 0.001 366 199 500 8;
  • 32) 0.001 366 199 500 8 × 2 = 0 + 0.002 732 399 001 6;
  • 33) 0.002 732 399 001 6 × 2 = 0 + 0.005 464 798 003 2;
  • 34) 0.005 464 798 003 2 × 2 = 0 + 0.010 929 596 006 4;
  • 35) 0.010 929 596 006 4 × 2 = 0 + 0.021 859 192 012 8;
  • 36) 0.021 859 192 012 8 × 2 = 0 + 0.043 718 384 025 6;
  • 37) 0.043 718 384 025 6 × 2 = 0 + 0.087 436 768 051 2;
  • 38) 0.087 436 768 051 2 × 2 = 0 + 0.174 873 536 102 4;
  • 39) 0.174 873 536 102 4 × 2 = 0 + 0.349 747 072 204 8;
  • 40) 0.349 747 072 204 8 × 2 = 0 + 0.699 494 144 409 6;
  • 41) 0.699 494 144 409 6 × 2 = 1 + 0.398 988 288 819 2;
  • 42) 0.398 988 288 819 2 × 2 = 0 + 0.797 976 577 638 4;
  • 43) 0.797 976 577 638 4 × 2 = 1 + 0.595 953 155 276 8;
  • 44) 0.595 953 155 276 8 × 2 = 1 + 0.191 906 310 553 6;
  • 45) 0.191 906 310 553 6 × 2 = 0 + 0.383 812 621 107 2;
  • 46) 0.383 812 621 107 2 × 2 = 0 + 0.767 625 242 214 4;
  • 47) 0.767 625 242 214 4 × 2 = 1 + 0.535 250 484 428 8;
  • 48) 0.535 250 484 428 8 × 2 = 1 + 0.070 500 968 857 6;
  • 49) 0.070 500 968 857 6 × 2 = 0 + 0.141 001 937 715 2;
  • 50) 0.141 001 937 715 2 × 2 = 0 + 0.282 003 875 430 4;
  • 51) 0.282 003 875 430 4 × 2 = 0 + 0.564 007 750 860 8;
  • 52) 0.564 007 750 860 8 × 2 = 1 + 0.128 015 501 721 6;
  • 53) 0.128 015 501 721 6 × 2 = 0 + 0.256 031 003 443 2;
  • 54) 0.256 031 003 443 2 × 2 = 0 + 0.512 062 006 886 4;
  • 55) 0.512 062 006 886 4 × 2 = 1 + 0.024 124 013 772 8;
  • 56) 0.024 124 013 772 8 × 2 = 0 + 0.048 248 027 545 6;
  • 57) 0.048 248 027 545 6 × 2 = 0 + 0.096 496 055 091 2;
  • 58) 0.096 496 055 091 2 × 2 = 0 + 0.192 992 110 182 4;
  • 59) 0.192 992 110 182 4 × 2 = 0 + 0.385 984 220 364 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:


0.008 788 423 613 35(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1011 0011 0001 0010 000(2)

6. Positive number before normalization:

0.008 788 423 613 35(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1011 0011 0001 0010 000(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 35(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1011 0011 0001 0010 000(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 1011 0011 0001 0010 000(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0101 1001 1000 1001 0000(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 0101 1001 1000 1001 0000


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 0101 1001 1000 1001 0000 =

0001 1111 1111 1010 1010 0100 0000 0000 0101 1001 1000 1001 0000


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 0101 1001 1000 1001 0000


Decimal number -0.008 788 423 613 35 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 0101 1001 1000 1001 0000

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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