0.000 000 177 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 177(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
0.000 000 177(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: 0.000 000 177.

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) 0.000 000 177 × 2 = 0 + 0.000 000 354;
  • 2) 0.000 000 354 × 2 = 0 + 0.000 000 708;
  • 3) 0.000 000 708 × 2 = 0 + 0.000 001 416;
  • 4) 0.000 001 416 × 2 = 0 + 0.000 002 832;
  • 5) 0.000 002 832 × 2 = 0 + 0.000 005 664;
  • 6) 0.000 005 664 × 2 = 0 + 0.000 011 328;
  • 7) 0.000 011 328 × 2 = 0 + 0.000 022 656;
  • 8) 0.000 022 656 × 2 = 0 + 0.000 045 312;
  • 9) 0.000 045 312 × 2 = 0 + 0.000 090 624;
  • 10) 0.000 090 624 × 2 = 0 + 0.000 181 248;
  • 11) 0.000 181 248 × 2 = 0 + 0.000 362 496;
  • 12) 0.000 362 496 × 2 = 0 + 0.000 724 992;
  • 13) 0.000 724 992 × 2 = 0 + 0.001 449 984;
  • 14) 0.001 449 984 × 2 = 0 + 0.002 899 968;
  • 15) 0.002 899 968 × 2 = 0 + 0.005 799 936;
  • 16) 0.005 799 936 × 2 = 0 + 0.011 599 872;
  • 17) 0.011 599 872 × 2 = 0 + 0.023 199 744;
  • 18) 0.023 199 744 × 2 = 0 + 0.046 399 488;
  • 19) 0.046 399 488 × 2 = 0 + 0.092 798 976;
  • 20) 0.092 798 976 × 2 = 0 + 0.185 597 952;
  • 21) 0.185 597 952 × 2 = 0 + 0.371 195 904;
  • 22) 0.371 195 904 × 2 = 0 + 0.742 391 808;
  • 23) 0.742 391 808 × 2 = 1 + 0.484 783 616;
  • 24) 0.484 783 616 × 2 = 0 + 0.969 567 232;
  • 25) 0.969 567 232 × 2 = 1 + 0.939 134 464;
  • 26) 0.939 134 464 × 2 = 1 + 0.878 268 928;
  • 27) 0.878 268 928 × 2 = 1 + 0.756 537 856;
  • 28) 0.756 537 856 × 2 = 1 + 0.513 075 712;
  • 29) 0.513 075 712 × 2 = 1 + 0.026 151 424;
  • 30) 0.026 151 424 × 2 = 0 + 0.052 302 848;
  • 31) 0.052 302 848 × 2 = 0 + 0.104 605 696;
  • 32) 0.104 605 696 × 2 = 0 + 0.209 211 392;
  • 33) 0.209 211 392 × 2 = 0 + 0.418 422 784;
  • 34) 0.418 422 784 × 2 = 0 + 0.836 845 568;
  • 35) 0.836 845 568 × 2 = 1 + 0.673 691 136;
  • 36) 0.673 691 136 × 2 = 1 + 0.347 382 272;
  • 37) 0.347 382 272 × 2 = 0 + 0.694 764 544;
  • 38) 0.694 764 544 × 2 = 1 + 0.389 529 088;
  • 39) 0.389 529 088 × 2 = 0 + 0.779 058 176;
  • 40) 0.779 058 176 × 2 = 1 + 0.558 116 352;
  • 41) 0.558 116 352 × 2 = 1 + 0.116 232 704;
  • 42) 0.116 232 704 × 2 = 0 + 0.232 465 408;
  • 43) 0.232 465 408 × 2 = 0 + 0.464 930 816;
  • 44) 0.464 930 816 × 2 = 0 + 0.929 861 632;
  • 45) 0.929 861 632 × 2 = 1 + 0.859 723 264;
  • 46) 0.859 723 264 × 2 = 1 + 0.719 446 528;

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:


0.000 000 177(10) =

0.0000 0000 0000 0000 0000 0010 1111 1000 0011 0101 1000 11(2)

5. Positive number before normalization:

0.000 000 177(10) =

0.0000 0000 0000 0000 0000 0010 1111 1000 0011 0101 1000 11(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:


0.000 000 177(10) =

0.0000 0000 0000 0000 0000 0010 1111 1000 0011 0101 1000 11(2) =

0.0000 0000 0000 0000 0000 0010 1111 1000 0011 0101 1000 11(2) × 20 =

1.0111 1100 0001 1010 1100 011(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):
1.0111 1100 0001 1010 1100 011


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. 011 1110 0000 1101 0110 0011 =

011 1110 0000 1101 0110 0011


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) =
011 1110 0000 1101 0110 0011


Decimal number 0.000 000 177 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 011 1110 0000 1101 0110 0011

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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