-0.000 002 77 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 002 77(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
-0.000 002 77(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:

|-0.000 002 77| = 0.000 002 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: 0.000 002 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) 0.000 002 77 × 2 = 0 + 0.000 005 54;
  • 2) 0.000 005 54 × 2 = 0 + 0.000 011 08;
  • 3) 0.000 011 08 × 2 = 0 + 0.000 022 16;
  • 4) 0.000 022 16 × 2 = 0 + 0.000 044 32;
  • 5) 0.000 044 32 × 2 = 0 + 0.000 088 64;
  • 6) 0.000 088 64 × 2 = 0 + 0.000 177 28;
  • 7) 0.000 177 28 × 2 = 0 + 0.000 354 56;
  • 8) 0.000 354 56 × 2 = 0 + 0.000 709 12;
  • 9) 0.000 709 12 × 2 = 0 + 0.001 418 24;
  • 10) 0.001 418 24 × 2 = 0 + 0.002 836 48;
  • 11) 0.002 836 48 × 2 = 0 + 0.005 672 96;
  • 12) 0.005 672 96 × 2 = 0 + 0.011 345 92;
  • 13) 0.011 345 92 × 2 = 0 + 0.022 691 84;
  • 14) 0.022 691 84 × 2 = 0 + 0.045 383 68;
  • 15) 0.045 383 68 × 2 = 0 + 0.090 767 36;
  • 16) 0.090 767 36 × 2 = 0 + 0.181 534 72;
  • 17) 0.181 534 72 × 2 = 0 + 0.363 069 44;
  • 18) 0.363 069 44 × 2 = 0 + 0.726 138 88;
  • 19) 0.726 138 88 × 2 = 1 + 0.452 277 76;
  • 20) 0.452 277 76 × 2 = 0 + 0.904 555 52;
  • 21) 0.904 555 52 × 2 = 1 + 0.809 111 04;
  • 22) 0.809 111 04 × 2 = 1 + 0.618 222 08;
  • 23) 0.618 222 08 × 2 = 1 + 0.236 444 16;
  • 24) 0.236 444 16 × 2 = 0 + 0.472 888 32;
  • 25) 0.472 888 32 × 2 = 0 + 0.945 776 64;
  • 26) 0.945 776 64 × 2 = 1 + 0.891 553 28;
  • 27) 0.891 553 28 × 2 = 1 + 0.783 106 56;
  • 28) 0.783 106 56 × 2 = 1 + 0.566 213 12;
  • 29) 0.566 213 12 × 2 = 1 + 0.132 426 24;
  • 30) 0.132 426 24 × 2 = 0 + 0.264 852 48;
  • 31) 0.264 852 48 × 2 = 0 + 0.529 704 96;
  • 32) 0.529 704 96 × 2 = 1 + 0.059 409 92;
  • 33) 0.059 409 92 × 2 = 0 + 0.118 819 84;
  • 34) 0.118 819 84 × 2 = 0 + 0.237 639 68;
  • 35) 0.237 639 68 × 2 = 0 + 0.475 279 36;
  • 36) 0.475 279 36 × 2 = 0 + 0.950 558 72;
  • 37) 0.950 558 72 × 2 = 1 + 0.901 117 44;
  • 38) 0.901 117 44 × 2 = 1 + 0.802 234 88;
  • 39) 0.802 234 88 × 2 = 1 + 0.604 469 76;
  • 40) 0.604 469 76 × 2 = 1 + 0.208 939 52;
  • 41) 0.208 939 52 × 2 = 0 + 0.417 879 04;
  • 42) 0.417 879 04 × 2 = 0 + 0.835 758 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:


0.000 002 77(10) =

0.0000 0000 0000 0000 0010 1110 0111 1001 0000 1111 00(2)

6. Positive number before normalization:

0.000 002 77(10) =

0.0000 0000 0000 0000 0010 1110 0111 1001 0000 1111 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 002 77(10) =

0.0000 0000 0000 0000 0010 1110 0111 1001 0000 1111 00(2) =

0.0000 0000 0000 0000 0010 1110 0111 1001 0000 1111 00(2) × 20 =

1.0111 0011 1100 1000 0111 100(2) × 2-19


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): -19

Mantissa (not normalized):
1.0111 0011 1100 1000 0111 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(8-1) - 1 =

(-19 + 127)(10) =

108(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;
  • 108 ÷ 2 = 54 + 0;
  • 54 ÷ 2 = 27 + 0;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

108(10) =

0110 1100(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 1001 1110 0100 0011 1100 =

011 1001 1110 0100 0011 1100


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 1100

Mantissa (23 bits) =
011 1001 1110 0100 0011 1100


Decimal number -0.000 002 77 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1100 - 011 1001 1110 0100 0011 1100

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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