-0.000 000 000 66 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 66(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 66(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 66| = 0.000 000 000 66


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: 0.000 000 000 66.

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) 0.000 000 000 66 × 2 = 0 + 0.000 000 001 32;
  • 2) 0.000 000 001 32 × 2 = 0 + 0.000 000 002 64;
  • 3) 0.000 000 002 64 × 2 = 0 + 0.000 000 005 28;
  • 4) 0.000 000 005 28 × 2 = 0 + 0.000 000 010 56;
  • 5) 0.000 000 010 56 × 2 = 0 + 0.000 000 021 12;
  • 6) 0.000 000 021 12 × 2 = 0 + 0.000 000 042 24;
  • 7) 0.000 000 042 24 × 2 = 0 + 0.000 000 084 48;
  • 8) 0.000 000 084 48 × 2 = 0 + 0.000 000 168 96;
  • 9) 0.000 000 168 96 × 2 = 0 + 0.000 000 337 92;
  • 10) 0.000 000 337 92 × 2 = 0 + 0.000 000 675 84;
  • 11) 0.000 000 675 84 × 2 = 0 + 0.000 001 351 68;
  • 12) 0.000 001 351 68 × 2 = 0 + 0.000 002 703 36;
  • 13) 0.000 002 703 36 × 2 = 0 + 0.000 005 406 72;
  • 14) 0.000 005 406 72 × 2 = 0 + 0.000 010 813 44;
  • 15) 0.000 010 813 44 × 2 = 0 + 0.000 021 626 88;
  • 16) 0.000 021 626 88 × 2 = 0 + 0.000 043 253 76;
  • 17) 0.000 043 253 76 × 2 = 0 + 0.000 086 507 52;
  • 18) 0.000 086 507 52 × 2 = 0 + 0.000 173 015 04;
  • 19) 0.000 173 015 04 × 2 = 0 + 0.000 346 030 08;
  • 20) 0.000 346 030 08 × 2 = 0 + 0.000 692 060 16;
  • 21) 0.000 692 060 16 × 2 = 0 + 0.001 384 120 32;
  • 22) 0.001 384 120 32 × 2 = 0 + 0.002 768 240 64;
  • 23) 0.002 768 240 64 × 2 = 0 + 0.005 536 481 28;
  • 24) 0.005 536 481 28 × 2 = 0 + 0.011 072 962 56;
  • 25) 0.011 072 962 56 × 2 = 0 + 0.022 145 925 12;
  • 26) 0.022 145 925 12 × 2 = 0 + 0.044 291 850 24;
  • 27) 0.044 291 850 24 × 2 = 0 + 0.088 583 700 48;
  • 28) 0.088 583 700 48 × 2 = 0 + 0.177 167 400 96;
  • 29) 0.177 167 400 96 × 2 = 0 + 0.354 334 801 92;
  • 30) 0.354 334 801 92 × 2 = 0 + 0.708 669 603 84;
  • 31) 0.708 669 603 84 × 2 = 1 + 0.417 339 207 68;
  • 32) 0.417 339 207 68 × 2 = 0 + 0.834 678 415 36;
  • 33) 0.834 678 415 36 × 2 = 1 + 0.669 356 830 72;
  • 34) 0.669 356 830 72 × 2 = 1 + 0.338 713 661 44;
  • 35) 0.338 713 661 44 × 2 = 0 + 0.677 427 322 88;
  • 36) 0.677 427 322 88 × 2 = 1 + 0.354 854 645 76;
  • 37) 0.354 854 645 76 × 2 = 0 + 0.709 709 291 52;
  • 38) 0.709 709 291 52 × 2 = 1 + 0.419 418 583 04;
  • 39) 0.419 418 583 04 × 2 = 0 + 0.838 837 166 08;
  • 40) 0.838 837 166 08 × 2 = 1 + 0.677 674 332 16;
  • 41) 0.677 674 332 16 × 2 = 1 + 0.355 348 664 32;
  • 42) 0.355 348 664 32 × 2 = 0 + 0.710 697 328 64;
  • 43) 0.710 697 328 64 × 2 = 1 + 0.421 394 657 28;
  • 44) 0.421 394 657 28 × 2 = 0 + 0.842 789 314 56;
  • 45) 0.842 789 314 56 × 2 = 1 + 0.685 578 629 12;
  • 46) 0.685 578 629 12 × 2 = 1 + 0.371 157 258 24;
  • 47) 0.371 157 258 24 × 2 = 0 + 0.742 314 516 48;
  • 48) 0.742 314 516 48 × 2 = 1 + 0.484 629 032 96;
  • 49) 0.484 629 032 96 × 2 = 0 + 0.969 258 065 92;
  • 50) 0.969 258 065 92 × 2 = 1 + 0.938 516 131 84;
  • 51) 0.938 516 131 84 × 2 = 1 + 0.877 032 263 68;
  • 52) 0.877 032 263 68 × 2 = 1 + 0.754 064 527 36;
  • 53) 0.754 064 527 36 × 2 = 1 + 0.508 129 054 72;
  • 54) 0.508 129 054 72 × 2 = 1 + 0.016 258 109 44;

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:


0.000 000 000 66(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1101 0101 1010 1101 0111 11(2)

6. Positive number before normalization:

0.000 000 000 66(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1101 0101 1010 1101 0111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 66(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1101 0101 1010 1101 0111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1101 0101 1010 1101 0111 11(2) × 20 =

1.0110 1010 1101 0110 1011 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.0110 1010 1101 0110 1011 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

96(10) =

0110 0000(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 011 0101 0110 1011 0101 1111 =

011 0101 0110 1011 0101 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
011 0101 0110 1011 0101 1111


Decimal number -0.000 000 000 66 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 011 0101 0110 1011 0101 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111