0.000 000 397 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 397(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
0.000 000 397(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: 0.000 000 397.

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) 0.000 000 397 × 2 = 0 + 0.000 000 794;
  • 2) 0.000 000 794 × 2 = 0 + 0.000 001 588;
  • 3) 0.000 001 588 × 2 = 0 + 0.000 003 176;
  • 4) 0.000 003 176 × 2 = 0 + 0.000 006 352;
  • 5) 0.000 006 352 × 2 = 0 + 0.000 012 704;
  • 6) 0.000 012 704 × 2 = 0 + 0.000 025 408;
  • 7) 0.000 025 408 × 2 = 0 + 0.000 050 816;
  • 8) 0.000 050 816 × 2 = 0 + 0.000 101 632;
  • 9) 0.000 101 632 × 2 = 0 + 0.000 203 264;
  • 10) 0.000 203 264 × 2 = 0 + 0.000 406 528;
  • 11) 0.000 406 528 × 2 = 0 + 0.000 813 056;
  • 12) 0.000 813 056 × 2 = 0 + 0.001 626 112;
  • 13) 0.001 626 112 × 2 = 0 + 0.003 252 224;
  • 14) 0.003 252 224 × 2 = 0 + 0.006 504 448;
  • 15) 0.006 504 448 × 2 = 0 + 0.013 008 896;
  • 16) 0.013 008 896 × 2 = 0 + 0.026 017 792;
  • 17) 0.026 017 792 × 2 = 0 + 0.052 035 584;
  • 18) 0.052 035 584 × 2 = 0 + 0.104 071 168;
  • 19) 0.104 071 168 × 2 = 0 + 0.208 142 336;
  • 20) 0.208 142 336 × 2 = 0 + 0.416 284 672;
  • 21) 0.416 284 672 × 2 = 0 + 0.832 569 344;
  • 22) 0.832 569 344 × 2 = 1 + 0.665 138 688;
  • 23) 0.665 138 688 × 2 = 1 + 0.330 277 376;
  • 24) 0.330 277 376 × 2 = 0 + 0.660 554 752;
  • 25) 0.660 554 752 × 2 = 1 + 0.321 109 504;
  • 26) 0.321 109 504 × 2 = 0 + 0.642 219 008;
  • 27) 0.642 219 008 × 2 = 1 + 0.284 438 016;
  • 28) 0.284 438 016 × 2 = 0 + 0.568 876 032;
  • 29) 0.568 876 032 × 2 = 1 + 0.137 752 064;
  • 30) 0.137 752 064 × 2 = 0 + 0.275 504 128;
  • 31) 0.275 504 128 × 2 = 0 + 0.551 008 256;
  • 32) 0.551 008 256 × 2 = 1 + 0.102 016 512;
  • 33) 0.102 016 512 × 2 = 0 + 0.204 033 024;
  • 34) 0.204 033 024 × 2 = 0 + 0.408 066 048;
  • 35) 0.408 066 048 × 2 = 0 + 0.816 132 096;
  • 36) 0.816 132 096 × 2 = 1 + 0.632 264 192;
  • 37) 0.632 264 192 × 2 = 1 + 0.264 528 384;
  • 38) 0.264 528 384 × 2 = 0 + 0.529 056 768;
  • 39) 0.529 056 768 × 2 = 1 + 0.058 113 536;
  • 40) 0.058 113 536 × 2 = 0 + 0.116 227 072;
  • 41) 0.116 227 072 × 2 = 0 + 0.232 454 144;
  • 42) 0.232 454 144 × 2 = 0 + 0.464 908 288;
  • 43) 0.464 908 288 × 2 = 0 + 0.929 816 576;
  • 44) 0.929 816 576 × 2 = 1 + 0.859 633 152;
  • 45) 0.859 633 152 × 2 = 1 + 0.719 266 304;

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:


0.000 000 397(10) =

0.0000 0000 0000 0000 0000 0110 1010 1001 0001 1010 0001 1(2)

5. Positive number before normalization:

0.000 000 397(10) =

0.0000 0000 0000 0000 0000 0110 1010 1001 0001 1010 0001 1(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:


0.000 000 397(10) =

0.0000 0000 0000 0000 0000 0110 1010 1001 0001 1010 0001 1(2) =

0.0000 0000 0000 0000 0000 0110 1010 1001 0001 1010 0001 1(2) × 20 =

1.1010 1010 0100 0110 1000 011(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 1010 0100 0110 1000 011


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 0010 0011 0100 0011 =

101 0101 0010 0011 0100 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 1001

Mantissa (23 bits) =
101 0101 0010 0011 0100 0011


Decimal number 0.000 000 397 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 101 0101 0010 0011 0100 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