-0.079 999 990 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.079 999 990 6(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
-0.079 999 990 6(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:

|-0.079 999 990 6| = 0.079 999 990 6


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: 0.079 999 990 6.

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) 0.079 999 990 6 × 2 = 0 + 0.159 999 981 2;
  • 2) 0.159 999 981 2 × 2 = 0 + 0.319 999 962 4;
  • 3) 0.319 999 962 4 × 2 = 0 + 0.639 999 924 8;
  • 4) 0.639 999 924 8 × 2 = 1 + 0.279 999 849 6;
  • 5) 0.279 999 849 6 × 2 = 0 + 0.559 999 699 2;
  • 6) 0.559 999 699 2 × 2 = 1 + 0.119 999 398 4;
  • 7) 0.119 999 398 4 × 2 = 0 + 0.239 998 796 8;
  • 8) 0.239 998 796 8 × 2 = 0 + 0.479 997 593 6;
  • 9) 0.479 997 593 6 × 2 = 0 + 0.959 995 187 2;
  • 10) 0.959 995 187 2 × 2 = 1 + 0.919 990 374 4;
  • 11) 0.919 990 374 4 × 2 = 1 + 0.839 980 748 8;
  • 12) 0.839 980 748 8 × 2 = 1 + 0.679 961 497 6;
  • 13) 0.679 961 497 6 × 2 = 1 + 0.359 922 995 2;
  • 14) 0.359 922 995 2 × 2 = 0 + 0.719 845 990 4;
  • 15) 0.719 845 990 4 × 2 = 1 + 0.439 691 980 8;
  • 16) 0.439 691 980 8 × 2 = 0 + 0.879 383 961 6;
  • 17) 0.879 383 961 6 × 2 = 1 + 0.758 767 923 2;
  • 18) 0.758 767 923 2 × 2 = 1 + 0.517 535 846 4;
  • 19) 0.517 535 846 4 × 2 = 1 + 0.035 071 692 8;
  • 20) 0.035 071 692 8 × 2 = 0 + 0.070 143 385 6;
  • 21) 0.070 143 385 6 × 2 = 0 + 0.140 286 771 2;
  • 22) 0.140 286 771 2 × 2 = 0 + 0.280 573 542 4;
  • 23) 0.280 573 542 4 × 2 = 0 + 0.561 147 084 8;
  • 24) 0.561 147 084 8 × 2 = 1 + 0.122 294 169 6;
  • 25) 0.122 294 169 6 × 2 = 0 + 0.244 588 339 2;
  • 26) 0.244 588 339 2 × 2 = 0 + 0.489 176 678 4;
  • 27) 0.489 176 678 4 × 2 = 0 + 0.978 353 356 8;
  • 28) 0.978 353 356 8 × 2 = 1 + 0.956 706 713 6;
  • 29) 0.956 706 713 6 × 2 = 1 + 0.913 413 427 2;
  • 30) 0.913 413 427 2 × 2 = 1 + 0.826 826 854 4;
  • 31) 0.826 826 854 4 × 2 = 1 + 0.653 653 708 8;
  • 32) 0.653 653 708 8 × 2 = 1 + 0.307 307 417 6;
  • 33) 0.307 307 417 6 × 2 = 0 + 0.614 614 835 2;
  • 34) 0.614 614 835 2 × 2 = 1 + 0.229 229 670 4;
  • 35) 0.229 229 670 4 × 2 = 0 + 0.458 459 340 8;
  • 36) 0.458 459 340 8 × 2 = 0 + 0.916 918 681 6;
  • 37) 0.916 918 681 6 × 2 = 1 + 0.833 837 363 2;
  • 38) 0.833 837 363 2 × 2 = 1 + 0.667 674 726 4;
  • 39) 0.667 674 726 4 × 2 = 1 + 0.335 349 452 8;
  • 40) 0.335 349 452 8 × 2 = 0 + 0.670 698 905 6;
  • 41) 0.670 698 905 6 × 2 = 1 + 0.341 397 811 2;
  • 42) 0.341 397 811 2 × 2 = 0 + 0.682 795 622 4;
  • 43) 0.682 795 622 4 × 2 = 1 + 0.365 591 244 8;
  • 44) 0.365 591 244 8 × 2 = 0 + 0.731 182 489 6;
  • 45) 0.731 182 489 6 × 2 = 1 + 0.462 364 979 2;
  • 46) 0.462 364 979 2 × 2 = 0 + 0.924 729 958 4;
  • 47) 0.924 729 958 4 × 2 = 1 + 0.849 459 916 8;
  • 48) 0.849 459 916 8 × 2 = 1 + 0.698 919 833 6;
  • 49) 0.698 919 833 6 × 2 = 1 + 0.397 839 667 2;
  • 50) 0.397 839 667 2 × 2 = 0 + 0.795 679 334 4;
  • 51) 0.795 679 334 4 × 2 = 1 + 0.591 358 668 8;
  • 52) 0.591 358 668 8 × 2 = 1 + 0.182 717 337 6;
  • 53) 0.182 717 337 6 × 2 = 0 + 0.365 434 675 2;
  • 54) 0.365 434 675 2 × 2 = 0 + 0.730 869 350 4;
  • 55) 0.730 869 350 4 × 2 = 1 + 0.461 738 700 8;
  • 56) 0.461 738 700 8 × 2 = 0 + 0.923 477 401 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:


0.079 999 990 6(10) =

0.0001 0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010(2)

6. Positive number before normalization:

0.079 999 990 6(10) =

0.0001 0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.079 999 990 6(10) =

0.0001 0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010(2) =

0.0001 0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010(2) × 20 =

1.0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010(2) × 2-4


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):
1.0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010 =

0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010


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 1011

Mantissa (52 bits) =
0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010


Decimal number -0.079 999 990 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0100 0111 1010 1110 0001 0001 1111 0100 1110 1010 1011 1011 0010

» Convert decimal -0.079 999 998 2 to 64 bit double precision IEEE 754 binary floating point representation standard

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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