-0.080 000 014 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.080 000 014 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
-0.080 000 014 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:

|-0.080 000 014 4| = 0.080 000 014 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: 0.080 000 014 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) 0.080 000 014 4 × 2 = 0 + 0.160 000 028 8;
  • 2) 0.160 000 028 8 × 2 = 0 + 0.320 000 057 6;
  • 3) 0.320 000 057 6 × 2 = 0 + 0.640 000 115 2;
  • 4) 0.640 000 115 2 × 2 = 1 + 0.280 000 230 4;
  • 5) 0.280 000 230 4 × 2 = 0 + 0.560 000 460 8;
  • 6) 0.560 000 460 8 × 2 = 1 + 0.120 000 921 6;
  • 7) 0.120 000 921 6 × 2 = 0 + 0.240 001 843 2;
  • 8) 0.240 001 843 2 × 2 = 0 + 0.480 003 686 4;
  • 9) 0.480 003 686 4 × 2 = 0 + 0.960 007 372 8;
  • 10) 0.960 007 372 8 × 2 = 1 + 0.920 014 745 6;
  • 11) 0.920 014 745 6 × 2 = 1 + 0.840 029 491 2;
  • 12) 0.840 029 491 2 × 2 = 1 + 0.680 058 982 4;
  • 13) 0.680 058 982 4 × 2 = 1 + 0.360 117 964 8;
  • 14) 0.360 117 964 8 × 2 = 0 + 0.720 235 929 6;
  • 15) 0.720 235 929 6 × 2 = 1 + 0.440 471 859 2;
  • 16) 0.440 471 859 2 × 2 = 0 + 0.880 943 718 4;
  • 17) 0.880 943 718 4 × 2 = 1 + 0.761 887 436 8;
  • 18) 0.761 887 436 8 × 2 = 1 + 0.523 774 873 6;
  • 19) 0.523 774 873 6 × 2 = 1 + 0.047 549 747 2;
  • 20) 0.047 549 747 2 × 2 = 0 + 0.095 099 494 4;
  • 21) 0.095 099 494 4 × 2 = 0 + 0.190 198 988 8;
  • 22) 0.190 198 988 8 × 2 = 0 + 0.380 397 977 6;
  • 23) 0.380 397 977 6 × 2 = 0 + 0.760 795 955 2;
  • 24) 0.760 795 955 2 × 2 = 1 + 0.521 591 910 4;
  • 25) 0.521 591 910 4 × 2 = 1 + 0.043 183 820 8;
  • 26) 0.043 183 820 8 × 2 = 0 + 0.086 367 641 6;
  • 27) 0.086 367 641 6 × 2 = 0 + 0.172 735 283 2;
  • 28) 0.172 735 283 2 × 2 = 0 + 0.345 470 566 4;
  • 29) 0.345 470 566 4 × 2 = 0 + 0.690 941 132 8;
  • 30) 0.690 941 132 8 × 2 = 1 + 0.381 882 265 6;
  • 31) 0.381 882 265 6 × 2 = 0 + 0.763 764 531 2;
  • 32) 0.763 764 531 2 × 2 = 1 + 0.527 529 062 4;
  • 33) 0.527 529 062 4 × 2 = 1 + 0.055 058 124 8;
  • 34) 0.055 058 124 8 × 2 = 0 + 0.110 116 249 6;
  • 35) 0.110 116 249 6 × 2 = 0 + 0.220 232 499 2;
  • 36) 0.220 232 499 2 × 2 = 0 + 0.440 464 998 4;
  • 37) 0.440 464 998 4 × 2 = 0 + 0.880 929 996 8;
  • 38) 0.880 929 996 8 × 2 = 1 + 0.761 859 993 6;
  • 39) 0.761 859 993 6 × 2 = 1 + 0.523 719 987 2;
  • 40) 0.523 719 987 2 × 2 = 1 + 0.047 439 974 4;
  • 41) 0.047 439 974 4 × 2 = 0 + 0.094 879 948 8;
  • 42) 0.094 879 948 8 × 2 = 0 + 0.189 759 897 6;
  • 43) 0.189 759 897 6 × 2 = 0 + 0.379 519 795 2;
  • 44) 0.379 519 795 2 × 2 = 0 + 0.759 039 590 4;
  • 45) 0.759 039 590 4 × 2 = 1 + 0.518 079 180 8;
  • 46) 0.518 079 180 8 × 2 = 1 + 0.036 158 361 6;
  • 47) 0.036 158 361 6 × 2 = 0 + 0.072 316 723 2;
  • 48) 0.072 316 723 2 × 2 = 0 + 0.144 633 446 4;
  • 49) 0.144 633 446 4 × 2 = 0 + 0.289 266 892 8;
  • 50) 0.289 266 892 8 × 2 = 0 + 0.578 533 785 6;
  • 51) 0.578 533 785 6 × 2 = 1 + 0.157 067 571 2;
  • 52) 0.157 067 571 2 × 2 = 0 + 0.314 135 142 4;
  • 53) 0.314 135 142 4 × 2 = 0 + 0.628 270 284 8;
  • 54) 0.628 270 284 8 × 2 = 1 + 0.256 540 569 6;
  • 55) 0.256 540 569 6 × 2 = 0 + 0.513 081 139 2;
  • 56) 0.513 081 139 2 × 2 = 1 + 0.026 162 278 4;

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:


0.080 000 014 4(10) =

0.0001 0100 0111 1010 1110 0001 1000 0101 1000 0111 0000 1100 0010 0101(2)

6. Positive number before normalization:

0.080 000 014 4(10) =

0.0001 0100 0111 1010 1110 0001 1000 0101 1000 0111 0000 1100 0010 0101(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:


0.080 000 014 4(10) =

0.0001 0100 0111 1010 1110 0001 1000 0101 1000 0111 0000 1100 0010 0101(2) =

0.0001 0100 0111 1010 1110 0001 1000 0101 1000 0111 0000 1100 0010 0101(2) × 20 =

1.0100 0111 1010 1110 0001 1000 0101 1000 0111 0000 1100 0010 0101(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 1000 0101 1000 0111 0000 1100 0010 0101


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 1000 0101 1000 0111 0000 1100 0010 0101 =

0100 0111 1010 1110 0001 1000 0101 1000 0111 0000 1100 0010 0101


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 1000 0101 1000 0111 0000 1100 0010 0101


Decimal number -0.080 000 014 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0100 0111 1010 1110 0001 1000 0101 1000 0111 0000 1100 0010 0101

» Convert decimal -0.080 000 011 5 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