0.000 000 398 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 398(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
0.000 000 398(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: 0.000 000 398.

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) 0.000 000 398 × 2 = 0 + 0.000 000 796;
  • 2) 0.000 000 796 × 2 = 0 + 0.000 001 592;
  • 3) 0.000 001 592 × 2 = 0 + 0.000 003 184;
  • 4) 0.000 003 184 × 2 = 0 + 0.000 006 368;
  • 5) 0.000 006 368 × 2 = 0 + 0.000 012 736;
  • 6) 0.000 012 736 × 2 = 0 + 0.000 025 472;
  • 7) 0.000 025 472 × 2 = 0 + 0.000 050 944;
  • 8) 0.000 050 944 × 2 = 0 + 0.000 101 888;
  • 9) 0.000 101 888 × 2 = 0 + 0.000 203 776;
  • 10) 0.000 203 776 × 2 = 0 + 0.000 407 552;
  • 11) 0.000 407 552 × 2 = 0 + 0.000 815 104;
  • 12) 0.000 815 104 × 2 = 0 + 0.001 630 208;
  • 13) 0.001 630 208 × 2 = 0 + 0.003 260 416;
  • 14) 0.003 260 416 × 2 = 0 + 0.006 520 832;
  • 15) 0.006 520 832 × 2 = 0 + 0.013 041 664;
  • 16) 0.013 041 664 × 2 = 0 + 0.026 083 328;
  • 17) 0.026 083 328 × 2 = 0 + 0.052 166 656;
  • 18) 0.052 166 656 × 2 = 0 + 0.104 333 312;
  • 19) 0.104 333 312 × 2 = 0 + 0.208 666 624;
  • 20) 0.208 666 624 × 2 = 0 + 0.417 333 248;
  • 21) 0.417 333 248 × 2 = 0 + 0.834 666 496;
  • 22) 0.834 666 496 × 2 = 1 + 0.669 332 992;
  • 23) 0.669 332 992 × 2 = 1 + 0.338 665 984;
  • 24) 0.338 665 984 × 2 = 0 + 0.677 331 968;
  • 25) 0.677 331 968 × 2 = 1 + 0.354 663 936;
  • 26) 0.354 663 936 × 2 = 0 + 0.709 327 872;
  • 27) 0.709 327 872 × 2 = 1 + 0.418 655 744;
  • 28) 0.418 655 744 × 2 = 0 + 0.837 311 488;
  • 29) 0.837 311 488 × 2 = 1 + 0.674 622 976;
  • 30) 0.674 622 976 × 2 = 1 + 0.349 245 952;
  • 31) 0.349 245 952 × 2 = 0 + 0.698 491 904;
  • 32) 0.698 491 904 × 2 = 1 + 0.396 983 808;
  • 33) 0.396 983 808 × 2 = 0 + 0.793 967 616;
  • 34) 0.793 967 616 × 2 = 1 + 0.587 935 232;
  • 35) 0.587 935 232 × 2 = 1 + 0.175 870 464;
  • 36) 0.175 870 464 × 2 = 0 + 0.351 740 928;
  • 37) 0.351 740 928 × 2 = 0 + 0.703 481 856;
  • 38) 0.703 481 856 × 2 = 1 + 0.406 963 712;
  • 39) 0.406 963 712 × 2 = 0 + 0.813 927 424;
  • 40) 0.813 927 424 × 2 = 1 + 0.627 854 848;
  • 41) 0.627 854 848 × 2 = 1 + 0.255 709 696;
  • 42) 0.255 709 696 × 2 = 0 + 0.511 419 392;
  • 43) 0.511 419 392 × 2 = 1 + 0.022 838 784;
  • 44) 0.022 838 784 × 2 = 0 + 0.045 677 568;
  • 45) 0.045 677 568 × 2 = 0 + 0.091 355 136;

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:


0.000 000 398(10) =

0.0000 0000 0000 0000 0000 0110 1010 1101 0110 0101 1010 0(2)

5. Positive number before normalization:

0.000 000 398(10) =

0.0000 0000 0000 0000 0000 0110 1010 1101 0110 0101 1010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 398(10) =

0.0000 0000 0000 0000 0000 0110 1010 1101 0110 0101 1010 0(2) =

0.0000 0000 0000 0000 0000 0110 1010 1101 0110 0101 1010 0(2) × 20 =

1.1010 1011 0101 1001 0110 100(2) × 2-22


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): -22

Mantissa (not normalized):
1.1010 1011 0101 1001 0110 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(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;
  • 105 ÷ 2 = 52 + 1;
  • 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) =

105(10) =

0110 1001(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. 101 0101 1010 1100 1011 0100 =

101 0101 1010 1100 1011 0100


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 1001

Mantissa (23 bits) =
101 0101 1010 1100 1011 0100


Decimal number 0.000 000 398 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 101 0101 1010 1100 1011 0100

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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