-0.001 035 77 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.001 035 77(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
-0.001 035 77(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:

|-0.001 035 77| = 0.001 035 77


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: 0.001 035 77.

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) 0.001 035 77 × 2 = 0 + 0.002 071 54;
  • 2) 0.002 071 54 × 2 = 0 + 0.004 143 08;
  • 3) 0.004 143 08 × 2 = 0 + 0.008 286 16;
  • 4) 0.008 286 16 × 2 = 0 + 0.016 572 32;
  • 5) 0.016 572 32 × 2 = 0 + 0.033 144 64;
  • 6) 0.033 144 64 × 2 = 0 + 0.066 289 28;
  • 7) 0.066 289 28 × 2 = 0 + 0.132 578 56;
  • 8) 0.132 578 56 × 2 = 0 + 0.265 157 12;
  • 9) 0.265 157 12 × 2 = 0 + 0.530 314 24;
  • 10) 0.530 314 24 × 2 = 1 + 0.060 628 48;
  • 11) 0.060 628 48 × 2 = 0 + 0.121 256 96;
  • 12) 0.121 256 96 × 2 = 0 + 0.242 513 92;
  • 13) 0.242 513 92 × 2 = 0 + 0.485 027 84;
  • 14) 0.485 027 84 × 2 = 0 + 0.970 055 68;
  • 15) 0.970 055 68 × 2 = 1 + 0.940 111 36;
  • 16) 0.940 111 36 × 2 = 1 + 0.880 222 72;
  • 17) 0.880 222 72 × 2 = 1 + 0.760 445 44;
  • 18) 0.760 445 44 × 2 = 1 + 0.520 890 88;
  • 19) 0.520 890 88 × 2 = 1 + 0.041 781 76;
  • 20) 0.041 781 76 × 2 = 0 + 0.083 563 52;
  • 21) 0.083 563 52 × 2 = 0 + 0.167 127 04;
  • 22) 0.167 127 04 × 2 = 0 + 0.334 254 08;
  • 23) 0.334 254 08 × 2 = 0 + 0.668 508 16;
  • 24) 0.668 508 16 × 2 = 1 + 0.337 016 32;
  • 25) 0.337 016 32 × 2 = 0 + 0.674 032 64;
  • 26) 0.674 032 64 × 2 = 1 + 0.348 065 28;
  • 27) 0.348 065 28 × 2 = 0 + 0.696 130 56;
  • 28) 0.696 130 56 × 2 = 1 + 0.392 261 12;
  • 29) 0.392 261 12 × 2 = 0 + 0.784 522 24;
  • 30) 0.784 522 24 × 2 = 1 + 0.569 044 48;
  • 31) 0.569 044 48 × 2 = 1 + 0.138 088 96;
  • 32) 0.138 088 96 × 2 = 0 + 0.276 177 92;
  • 33) 0.276 177 92 × 2 = 0 + 0.552 355 84;
  • 34) 0.552 355 84 × 2 = 1 + 0.104 711 68;
  • 35) 0.104 711 68 × 2 = 0 + 0.209 423 36;
  • 36) 0.209 423 36 × 2 = 0 + 0.418 846 72;
  • 37) 0.418 846 72 × 2 = 0 + 0.837 693 44;
  • 38) 0.837 693 44 × 2 = 1 + 0.675 386 88;
  • 39) 0.675 386 88 × 2 = 1 + 0.350 773 76;
  • 40) 0.350 773 76 × 2 = 0 + 0.701 547 52;
  • 41) 0.701 547 52 × 2 = 1 + 0.403 095 04;
  • 42) 0.403 095 04 × 2 = 0 + 0.806 190 08;
  • 43) 0.806 190 08 × 2 = 1 + 0.612 380 16;
  • 44) 0.612 380 16 × 2 = 1 + 0.224 760 32;
  • 45) 0.224 760 32 × 2 = 0 + 0.449 520 64;
  • 46) 0.449 520 64 × 2 = 0 + 0.899 041 28;
  • 47) 0.899 041 28 × 2 = 1 + 0.798 082 56;
  • 48) 0.798 082 56 × 2 = 1 + 0.596 165 12;
  • 49) 0.596 165 12 × 2 = 1 + 0.192 330 24;
  • 50) 0.192 330 24 × 2 = 0 + 0.384 660 48;
  • 51) 0.384 660 48 × 2 = 0 + 0.769 320 96;
  • 52) 0.769 320 96 × 2 = 1 + 0.538 641 92;
  • 53) 0.538 641 92 × 2 = 1 + 0.077 283 84;
  • 54) 0.077 283 84 × 2 = 0 + 0.154 567 68;
  • 55) 0.154 567 68 × 2 = 0 + 0.309 135 36;
  • 56) 0.309 135 36 × 2 = 0 + 0.618 270 72;
  • 57) 0.618 270 72 × 2 = 1 + 0.236 541 44;
  • 58) 0.236 541 44 × 2 = 0 + 0.473 082 88;
  • 59) 0.473 082 88 × 2 = 0 + 0.946 165 76;
  • 60) 0.946 165 76 × 2 = 1 + 0.892 331 52;
  • 61) 0.892 331 52 × 2 = 1 + 0.784 663 04;
  • 62) 0.784 663 04 × 2 = 1 + 0.569 326 08;

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:


0.001 035 77(10) =

0.0000 0000 0100 0011 1110 0001 0101 0110 0100 0110 1011 0011 1001 1000 1001 11(2)

6. Positive number before normalization:

0.001 035 77(10) =

0.0000 0000 0100 0011 1110 0001 0101 0110 0100 0110 1011 0011 1001 1000 1001 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 035 77(10) =

0.0000 0000 0100 0011 1110 0001 0101 0110 0100 0110 1011 0011 1001 1000 1001 11(2) =

0.0000 0000 0100 0011 1110 0001 0101 0110 0100 0110 1011 0011 1001 1000 1001 11(2) × 20 =

1.0000 1111 1000 0101 0101 1001 0001 1010 1100 1110 0110 0010 0111(2) × 2-10


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): -10

Mantissa (not normalized):
1.0000 1111 1000 0101 0101 1001 0001 1010 1100 1110 0110 0010 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-10 + 2(11-1) - 1 =

(-10 + 1 023)(10) =

1 013(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;
  • 1 013 ÷ 2 = 506 + 1;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1013(10) =

011 1111 0101(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. 0000 1111 1000 0101 0101 1001 0001 1010 1100 1110 0110 0010 0111 =

0000 1111 1000 0101 0101 1001 0001 1010 1100 1110 0110 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 0101

Mantissa (52 bits) =
0000 1111 1000 0101 0101 1001 0001 1010 1100 1110 0110 0010 0111


Decimal number -0.001 035 77 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0101 - 0000 1111 1000 0101 0101 1001 0001 1010 1100 1110 0110 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