0.000 000 233 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 233(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
0.000 000 233(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: 0.000 000 233.

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) 0.000 000 233 × 2 = 0 + 0.000 000 466;
  • 2) 0.000 000 466 × 2 = 0 + 0.000 000 932;
  • 3) 0.000 000 932 × 2 = 0 + 0.000 001 864;
  • 4) 0.000 001 864 × 2 = 0 + 0.000 003 728;
  • 5) 0.000 003 728 × 2 = 0 + 0.000 007 456;
  • 6) 0.000 007 456 × 2 = 0 + 0.000 014 912;
  • 7) 0.000 014 912 × 2 = 0 + 0.000 029 824;
  • 8) 0.000 029 824 × 2 = 0 + 0.000 059 648;
  • 9) 0.000 059 648 × 2 = 0 + 0.000 119 296;
  • 10) 0.000 119 296 × 2 = 0 + 0.000 238 592;
  • 11) 0.000 238 592 × 2 = 0 + 0.000 477 184;
  • 12) 0.000 477 184 × 2 = 0 + 0.000 954 368;
  • 13) 0.000 954 368 × 2 = 0 + 0.001 908 736;
  • 14) 0.001 908 736 × 2 = 0 + 0.003 817 472;
  • 15) 0.003 817 472 × 2 = 0 + 0.007 634 944;
  • 16) 0.007 634 944 × 2 = 0 + 0.015 269 888;
  • 17) 0.015 269 888 × 2 = 0 + 0.030 539 776;
  • 18) 0.030 539 776 × 2 = 0 + 0.061 079 552;
  • 19) 0.061 079 552 × 2 = 0 + 0.122 159 104;
  • 20) 0.122 159 104 × 2 = 0 + 0.244 318 208;
  • 21) 0.244 318 208 × 2 = 0 + 0.488 636 416;
  • 22) 0.488 636 416 × 2 = 0 + 0.977 272 832;
  • 23) 0.977 272 832 × 2 = 1 + 0.954 545 664;
  • 24) 0.954 545 664 × 2 = 1 + 0.909 091 328;
  • 25) 0.909 091 328 × 2 = 1 + 0.818 182 656;
  • 26) 0.818 182 656 × 2 = 1 + 0.636 365 312;
  • 27) 0.636 365 312 × 2 = 1 + 0.272 730 624;
  • 28) 0.272 730 624 × 2 = 0 + 0.545 461 248;
  • 29) 0.545 461 248 × 2 = 1 + 0.090 922 496;
  • 30) 0.090 922 496 × 2 = 0 + 0.181 844 992;
  • 31) 0.181 844 992 × 2 = 0 + 0.363 689 984;
  • 32) 0.363 689 984 × 2 = 0 + 0.727 379 968;
  • 33) 0.727 379 968 × 2 = 1 + 0.454 759 936;
  • 34) 0.454 759 936 × 2 = 0 + 0.909 519 872;
  • 35) 0.909 519 872 × 2 = 1 + 0.819 039 744;
  • 36) 0.819 039 744 × 2 = 1 + 0.638 079 488;
  • 37) 0.638 079 488 × 2 = 1 + 0.276 158 976;
  • 38) 0.276 158 976 × 2 = 0 + 0.552 317 952;
  • 39) 0.552 317 952 × 2 = 1 + 0.104 635 904;
  • 40) 0.104 635 904 × 2 = 0 + 0.209 271 808;
  • 41) 0.209 271 808 × 2 = 0 + 0.418 543 616;
  • 42) 0.418 543 616 × 2 = 0 + 0.837 087 232;
  • 43) 0.837 087 232 × 2 = 1 + 0.674 174 464;
  • 44) 0.674 174 464 × 2 = 1 + 0.348 348 928;
  • 45) 0.348 348 928 × 2 = 0 + 0.696 697 856;
  • 46) 0.696 697 856 × 2 = 1 + 0.393 395 712;

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:


0.000 000 233(10) =

0.0000 0000 0000 0000 0000 0011 1110 1000 1011 1010 0011 01(2)

5. Positive number before normalization:

0.000 000 233(10) =

0.0000 0000 0000 0000 0000 0011 1110 1000 1011 1010 0011 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 233(10) =

0.0000 0000 0000 0000 0000 0011 1110 1000 1011 1010 0011 01(2) =

0.0000 0000 0000 0000 0000 0011 1110 1000 1011 1010 0011 01(2) × 20 =

1.1111 0100 0101 1101 0001 101(2) × 2-23


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): -23

Mantissa (not normalized):
1.1111 0100 0101 1101 0001 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(8-1) - 1 =

(-23 + 127)(10) =

104(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;
  • 104 ÷ 2 = 52 + 0;
  • 52 ÷ 2 = 26 + 0;
  • 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) =

104(10) =

0110 1000(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. 111 1010 0010 1110 1000 1101 =

111 1010 0010 1110 1000 1101


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 1000

Mantissa (23 bits) =
111 1010 0010 1110 1000 1101


Decimal number 0.000 000 233 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 111 1010 0010 1110 1000 1101

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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