-0.000 164 767 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 164 767 5(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
-0.000 164 767 5(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:

|-0.000 164 767 5| = 0.000 164 767 5


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: 0.000 164 767 5.

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) 0.000 164 767 5 × 2 = 0 + 0.000 329 535;
  • 2) 0.000 329 535 × 2 = 0 + 0.000 659 07;
  • 3) 0.000 659 07 × 2 = 0 + 0.001 318 14;
  • 4) 0.001 318 14 × 2 = 0 + 0.002 636 28;
  • 5) 0.002 636 28 × 2 = 0 + 0.005 272 56;
  • 6) 0.005 272 56 × 2 = 0 + 0.010 545 12;
  • 7) 0.010 545 12 × 2 = 0 + 0.021 090 24;
  • 8) 0.021 090 24 × 2 = 0 + 0.042 180 48;
  • 9) 0.042 180 48 × 2 = 0 + 0.084 360 96;
  • 10) 0.084 360 96 × 2 = 0 + 0.168 721 92;
  • 11) 0.168 721 92 × 2 = 0 + 0.337 443 84;
  • 12) 0.337 443 84 × 2 = 0 + 0.674 887 68;
  • 13) 0.674 887 68 × 2 = 1 + 0.349 775 36;
  • 14) 0.349 775 36 × 2 = 0 + 0.699 550 72;
  • 15) 0.699 550 72 × 2 = 1 + 0.399 101 44;
  • 16) 0.399 101 44 × 2 = 0 + 0.798 202 88;
  • 17) 0.798 202 88 × 2 = 1 + 0.596 405 76;
  • 18) 0.596 405 76 × 2 = 1 + 0.192 811 52;
  • 19) 0.192 811 52 × 2 = 0 + 0.385 623 04;
  • 20) 0.385 623 04 × 2 = 0 + 0.771 246 08;
  • 21) 0.771 246 08 × 2 = 1 + 0.542 492 16;
  • 22) 0.542 492 16 × 2 = 1 + 0.084 984 32;
  • 23) 0.084 984 32 × 2 = 0 + 0.169 968 64;
  • 24) 0.169 968 64 × 2 = 0 + 0.339 937 28;
  • 25) 0.339 937 28 × 2 = 0 + 0.679 874 56;
  • 26) 0.679 874 56 × 2 = 1 + 0.359 749 12;
  • 27) 0.359 749 12 × 2 = 0 + 0.719 498 24;
  • 28) 0.719 498 24 × 2 = 1 + 0.438 996 48;
  • 29) 0.438 996 48 × 2 = 0 + 0.877 992 96;
  • 30) 0.877 992 96 × 2 = 1 + 0.755 985 92;
  • 31) 0.755 985 92 × 2 = 1 + 0.511 971 84;
  • 32) 0.511 971 84 × 2 = 1 + 0.023 943 68;
  • 33) 0.023 943 68 × 2 = 0 + 0.047 887 36;
  • 34) 0.047 887 36 × 2 = 0 + 0.095 774 72;
  • 35) 0.095 774 72 × 2 = 0 + 0.191 549 44;
  • 36) 0.191 549 44 × 2 = 0 + 0.383 098 88;
  • 37) 0.383 098 88 × 2 = 0 + 0.766 197 76;
  • 38) 0.766 197 76 × 2 = 1 + 0.532 395 52;
  • 39) 0.532 395 52 × 2 = 1 + 0.064 791 04;
  • 40) 0.064 791 04 × 2 = 0 + 0.129 582 08;
  • 41) 0.129 582 08 × 2 = 0 + 0.259 164 16;
  • 42) 0.259 164 16 × 2 = 0 + 0.518 328 32;
  • 43) 0.518 328 32 × 2 = 1 + 0.036 656 64;
  • 44) 0.036 656 64 × 2 = 0 + 0.073 313 28;
  • 45) 0.073 313 28 × 2 = 0 + 0.146 626 56;
  • 46) 0.146 626 56 × 2 = 0 + 0.293 253 12;
  • 47) 0.293 253 12 × 2 = 0 + 0.586 506 24;
  • 48) 0.586 506 24 × 2 = 1 + 0.173 012 48;
  • 49) 0.173 012 48 × 2 = 0 + 0.346 024 96;
  • 50) 0.346 024 96 × 2 = 0 + 0.692 049 92;
  • 51) 0.692 049 92 × 2 = 1 + 0.384 099 84;
  • 52) 0.384 099 84 × 2 = 0 + 0.768 199 68;
  • 53) 0.768 199 68 × 2 = 1 + 0.536 399 36;
  • 54) 0.536 399 36 × 2 = 1 + 0.072 798 72;
  • 55) 0.072 798 72 × 2 = 0 + 0.145 597 44;
  • 56) 0.145 597 44 × 2 = 0 + 0.291 194 88;
  • 57) 0.291 194 88 × 2 = 0 + 0.582 389 76;
  • 58) 0.582 389 76 × 2 = 1 + 0.164 779 52;
  • 59) 0.164 779 52 × 2 = 0 + 0.329 559 04;
  • 60) 0.329 559 04 × 2 = 0 + 0.659 118 08;
  • 61) 0.659 118 08 × 2 = 1 + 0.318 236 16;
  • 62) 0.318 236 16 × 2 = 0 + 0.636 472 32;
  • 63) 0.636 472 32 × 2 = 1 + 0.272 944 64;
  • 64) 0.272 944 64 × 2 = 0 + 0.545 889 28;
  • 65) 0.545 889 28 × 2 = 1 + 0.091 778 56;

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:


0.000 164 767 5(10) =

0.0000 0000 0000 1010 1100 1100 0101 0111 0000 0110 0010 0001 0010 1100 0100 1010 1(2)

6. Positive number before normalization:

0.000 164 767 5(10) =

0.0000 0000 0000 1010 1100 1100 0101 0111 0000 0110 0010 0001 0010 1100 0100 1010 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the right, so that only one non zero digit remains to the left of it:


0.000 164 767 5(10) =

0.0000 0000 0000 1010 1100 1100 0101 0111 0000 0110 0010 0001 0010 1100 0100 1010 1(2) =

0.0000 0000 0000 1010 1100 1100 0101 0111 0000 0110 0010 0001 0010 1100 0100 1010 1(2) × 20 =

1.0101 1001 1000 1010 1110 0000 1100 0100 0010 0101 1000 1001 0101(2) × 2-13


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): -13

Mantissa (not normalized):
1.0101 1001 1000 1010 1110 0000 1100 0100 0010 0101 1000 1001 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-13 + 2(11-1) - 1 =

(-13 + 1 023)(10) =

1 010(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;
  • 1 010 ÷ 2 = 505 + 0;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 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) =

1010(10) =

011 1111 0010(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. 0101 1001 1000 1010 1110 0000 1100 0100 0010 0101 1000 1001 0101 =

0101 1001 1000 1010 1110 0000 1100 0100 0010 0101 1000 1001 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 0010

Mantissa (52 bits) =
0101 1001 1000 1010 1110 0000 1100 0100 0010 0101 1000 1001 0101


Decimal number -0.000 164 767 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0010 - 0101 1001 1000 1010 1110 0000 1100 0100 0010 0101 1000 1001 0101

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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