0.000 000 557 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 557(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
0.000 000 557(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 557.

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) 0.000 000 557 × 2 = 0 + 0.000 001 114;
  • 2) 0.000 001 114 × 2 = 0 + 0.000 002 228;
  • 3) 0.000 002 228 × 2 = 0 + 0.000 004 456;
  • 4) 0.000 004 456 × 2 = 0 + 0.000 008 912;
  • 5) 0.000 008 912 × 2 = 0 + 0.000 017 824;
  • 6) 0.000 017 824 × 2 = 0 + 0.000 035 648;
  • 7) 0.000 035 648 × 2 = 0 + 0.000 071 296;
  • 8) 0.000 071 296 × 2 = 0 + 0.000 142 592;
  • 9) 0.000 142 592 × 2 = 0 + 0.000 285 184;
  • 10) 0.000 285 184 × 2 = 0 + 0.000 570 368;
  • 11) 0.000 570 368 × 2 = 0 + 0.001 140 736;
  • 12) 0.001 140 736 × 2 = 0 + 0.002 281 472;
  • 13) 0.002 281 472 × 2 = 0 + 0.004 562 944;
  • 14) 0.004 562 944 × 2 = 0 + 0.009 125 888;
  • 15) 0.009 125 888 × 2 = 0 + 0.018 251 776;
  • 16) 0.018 251 776 × 2 = 0 + 0.036 503 552;
  • 17) 0.036 503 552 × 2 = 0 + 0.073 007 104;
  • 18) 0.073 007 104 × 2 = 0 + 0.146 014 208;
  • 19) 0.146 014 208 × 2 = 0 + 0.292 028 416;
  • 20) 0.292 028 416 × 2 = 0 + 0.584 056 832;
  • 21) 0.584 056 832 × 2 = 1 + 0.168 113 664;
  • 22) 0.168 113 664 × 2 = 0 + 0.336 227 328;
  • 23) 0.336 227 328 × 2 = 0 + 0.672 454 656;
  • 24) 0.672 454 656 × 2 = 1 + 0.344 909 312;
  • 25) 0.344 909 312 × 2 = 0 + 0.689 818 624;
  • 26) 0.689 818 624 × 2 = 1 + 0.379 637 248;
  • 27) 0.379 637 248 × 2 = 0 + 0.759 274 496;
  • 28) 0.759 274 496 × 2 = 1 + 0.518 548 992;
  • 29) 0.518 548 992 × 2 = 1 + 0.037 097 984;
  • 30) 0.037 097 984 × 2 = 0 + 0.074 195 968;
  • 31) 0.074 195 968 × 2 = 0 + 0.148 391 936;
  • 32) 0.148 391 936 × 2 = 0 + 0.296 783 872;
  • 33) 0.296 783 872 × 2 = 0 + 0.593 567 744;
  • 34) 0.593 567 744 × 2 = 1 + 0.187 135 488;
  • 35) 0.187 135 488 × 2 = 0 + 0.374 270 976;
  • 36) 0.374 270 976 × 2 = 0 + 0.748 541 952;
  • 37) 0.748 541 952 × 2 = 1 + 0.497 083 904;
  • 38) 0.497 083 904 × 2 = 0 + 0.994 167 808;
  • 39) 0.994 167 808 × 2 = 1 + 0.988 335 616;
  • 40) 0.988 335 616 × 2 = 1 + 0.976 671 232;
  • 41) 0.976 671 232 × 2 = 1 + 0.953 342 464;
  • 42) 0.953 342 464 × 2 = 1 + 0.906 684 928;
  • 43) 0.906 684 928 × 2 = 1 + 0.813 369 856;
  • 44) 0.813 369 856 × 2 = 1 + 0.626 739 712;

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).


4. 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:


0.000 000 557(10) =

0.0000 0000 0000 0000 0000 1001 0101 1000 0100 1011 1111(2)

5. Positive number before normalization:

0.000 000 557(10) =

0.0000 0000 0000 0000 0000 1001 0101 1000 0100 1011 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 557(10) =

0.0000 0000 0000 0000 0000 1001 0101 1000 0100 1011 1111(2) =

0.0000 0000 0000 0000 0000 1001 0101 1000 0100 1011 1111(2) × 20 =

1.0010 1011 0000 1001 0111 111(2) × 2-21


7. 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 0 (a positive number)

Exponent (unadjusted): -21

Mantissa (not normalized):
1.0010 1011 0000 1001 0111 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(10)


9. 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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

106(10) =

0110 1010(2)


11. 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. 001 0101 1000 0100 1011 1111 =

001 0101 1000 0100 1011 1111


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

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 1010

Mantissa (23 bits) =
001 0101 1000 0100 1011 1111


Decimal number 0.000 000 557 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 001 0101 1000 0100 1011 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