-0.008 788 423 613 95 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

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

|-0.008 788 423 613 95| = 0.008 788 423 613 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: 0.008 788 423 613 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) 0.008 788 423 613 95 × 2 = 0 + 0.017 576 847 227 9;
  • 2) 0.017 576 847 227 9 × 2 = 0 + 0.035 153 694 455 8;
  • 3) 0.035 153 694 455 8 × 2 = 0 + 0.070 307 388 911 6;
  • 4) 0.070 307 388 911 6 × 2 = 0 + 0.140 614 777 823 2;
  • 5) 0.140 614 777 823 2 × 2 = 0 + 0.281 229 555 646 4;
  • 6) 0.281 229 555 646 4 × 2 = 0 + 0.562 459 111 292 8;
  • 7) 0.562 459 111 292 8 × 2 = 1 + 0.124 918 222 585 6;
  • 8) 0.124 918 222 585 6 × 2 = 0 + 0.249 836 445 171 2;
  • 9) 0.249 836 445 171 2 × 2 = 0 + 0.499 672 890 342 4;
  • 10) 0.499 672 890 342 4 × 2 = 0 + 0.999 345 780 684 8;
  • 11) 0.999 345 780 684 8 × 2 = 1 + 0.998 691 561 369 6;
  • 12) 0.998 691 561 369 6 × 2 = 1 + 0.997 383 122 739 2;
  • 13) 0.997 383 122 739 2 × 2 = 1 + 0.994 766 245 478 4;
  • 14) 0.994 766 245 478 4 × 2 = 1 + 0.989 532 490 956 8;
  • 15) 0.989 532 490 956 8 × 2 = 1 + 0.979 064 981 913 6;
  • 16) 0.979 064 981 913 6 × 2 = 1 + 0.958 129 963 827 2;
  • 17) 0.958 129 963 827 2 × 2 = 1 + 0.916 259 927 654 4;
  • 18) 0.916 259 927 654 4 × 2 = 1 + 0.832 519 855 308 8;
  • 19) 0.832 519 855 308 8 × 2 = 1 + 0.665 039 710 617 6;
  • 20) 0.665 039 710 617 6 × 2 = 1 + 0.330 079 421 235 2;
  • 21) 0.330 079 421 235 2 × 2 = 0 + 0.660 158 842 470 4;
  • 22) 0.660 158 842 470 4 × 2 = 1 + 0.320 317 684 940 8;
  • 23) 0.320 317 684 940 8 × 2 = 0 + 0.640 635 369 881 6;
  • 24) 0.640 635 369 881 6 × 2 = 1 + 0.281 270 739 763 2;
  • 25) 0.281 270 739 763 2 × 2 = 0 + 0.562 541 479 526 4;
  • 26) 0.562 541 479 526 4 × 2 = 1 + 0.125 082 959 052 8;
  • 27) 0.125 082 959 052 8 × 2 = 0 + 0.250 165 918 105 6;
  • 28) 0.250 165 918 105 6 × 2 = 0 + 0.500 331 836 211 2;
  • 29) 0.500 331 836 211 2 × 2 = 1 + 0.000 663 672 422 4;
  • 30) 0.000 663 672 422 4 × 2 = 0 + 0.001 327 344 844 8;
  • 31) 0.001 327 344 844 8 × 2 = 0 + 0.002 654 689 689 6;
  • 32) 0.002 654 689 689 6 × 2 = 0 + 0.005 309 379 379 2;
  • 33) 0.005 309 379 379 2 × 2 = 0 + 0.010 618 758 758 4;
  • 34) 0.010 618 758 758 4 × 2 = 0 + 0.021 237 517 516 8;
  • 35) 0.021 237 517 516 8 × 2 = 0 + 0.042 475 035 033 6;
  • 36) 0.042 475 035 033 6 × 2 = 0 + 0.084 950 070 067 2;
  • 37) 0.084 950 070 067 2 × 2 = 0 + 0.169 900 140 134 4;
  • 38) 0.169 900 140 134 4 × 2 = 0 + 0.339 800 280 268 8;
  • 39) 0.339 800 280 268 8 × 2 = 0 + 0.679 600 560 537 6;
  • 40) 0.679 600 560 537 6 × 2 = 1 + 0.359 201 121 075 2;
  • 41) 0.359 201 121 075 2 × 2 = 0 + 0.718 402 242 150 4;
  • 42) 0.718 402 242 150 4 × 2 = 1 + 0.436 804 484 300 8;
  • 43) 0.436 804 484 300 8 × 2 = 0 + 0.873 608 968 601 6;
  • 44) 0.873 608 968 601 6 × 2 = 1 + 0.747 217 937 203 2;
  • 45) 0.747 217 937 203 2 × 2 = 1 + 0.494 435 874 406 4;
  • 46) 0.494 435 874 406 4 × 2 = 0 + 0.988 871 748 812 8;
  • 47) 0.988 871 748 812 8 × 2 = 1 + 0.977 743 497 625 6;
  • 48) 0.977 743 497 625 6 × 2 = 1 + 0.955 486 995 251 2;
  • 49) 0.955 486 995 251 2 × 2 = 1 + 0.910 973 990 502 4;
  • 50) 0.910 973 990 502 4 × 2 = 1 + 0.821 947 981 004 8;
  • 51) 0.821 947 981 004 8 × 2 = 1 + 0.643 895 962 009 6;
  • 52) 0.643 895 962 009 6 × 2 = 1 + 0.287 791 924 019 2;
  • 53) 0.287 791 924 019 2 × 2 = 0 + 0.575 583 848 038 4;
  • 54) 0.575 583 848 038 4 × 2 = 1 + 0.151 167 696 076 8;
  • 55) 0.151 167 696 076 8 × 2 = 0 + 0.302 335 392 153 6;
  • 56) 0.302 335 392 153 6 × 2 = 0 + 0.604 670 784 307 2;
  • 57) 0.604 670 784 307 2 × 2 = 1 + 0.209 341 568 614 4;
  • 58) 0.209 341 568 614 4 × 2 = 0 + 0.418 683 137 228 8;
  • 59) 0.418 683 137 228 8 × 2 = 0 + 0.837 366 274 457 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:


0.008 788 423 613 95(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0101 1011 1111 0100 100(2)

6. Positive number before normalization:

0.008 788 423 613 95(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0101 1011 1111 0100 100(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 95(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0101 1011 1111 0100 100(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0101 1011 1111 0100 100(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 1010 1101 1111 1010 0100(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 1010 1101 1111 1010 0100


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 1010 1101 1111 1010 0100 =

0001 1111 1111 1010 1010 0100 0000 0000 1010 1101 1111 1010 0100


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 1010 1101 1111 1010 0100


Decimal number -0.008 788 423 613 95 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 1010 1101 1111 1010 0100

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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