0.000 000 371 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 371(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
0.000 000 371(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: 0.000 000 371.

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) 0.000 000 371 × 2 = 0 + 0.000 000 742;
  • 2) 0.000 000 742 × 2 = 0 + 0.000 001 484;
  • 3) 0.000 001 484 × 2 = 0 + 0.000 002 968;
  • 4) 0.000 002 968 × 2 = 0 + 0.000 005 936;
  • 5) 0.000 005 936 × 2 = 0 + 0.000 011 872;
  • 6) 0.000 011 872 × 2 = 0 + 0.000 023 744;
  • 7) 0.000 023 744 × 2 = 0 + 0.000 047 488;
  • 8) 0.000 047 488 × 2 = 0 + 0.000 094 976;
  • 9) 0.000 094 976 × 2 = 0 + 0.000 189 952;
  • 10) 0.000 189 952 × 2 = 0 + 0.000 379 904;
  • 11) 0.000 379 904 × 2 = 0 + 0.000 759 808;
  • 12) 0.000 759 808 × 2 = 0 + 0.001 519 616;
  • 13) 0.001 519 616 × 2 = 0 + 0.003 039 232;
  • 14) 0.003 039 232 × 2 = 0 + 0.006 078 464;
  • 15) 0.006 078 464 × 2 = 0 + 0.012 156 928;
  • 16) 0.012 156 928 × 2 = 0 + 0.024 313 856;
  • 17) 0.024 313 856 × 2 = 0 + 0.048 627 712;
  • 18) 0.048 627 712 × 2 = 0 + 0.097 255 424;
  • 19) 0.097 255 424 × 2 = 0 + 0.194 510 848;
  • 20) 0.194 510 848 × 2 = 0 + 0.389 021 696;
  • 21) 0.389 021 696 × 2 = 0 + 0.778 043 392;
  • 22) 0.778 043 392 × 2 = 1 + 0.556 086 784;
  • 23) 0.556 086 784 × 2 = 1 + 0.112 173 568;
  • 24) 0.112 173 568 × 2 = 0 + 0.224 347 136;
  • 25) 0.224 347 136 × 2 = 0 + 0.448 694 272;
  • 26) 0.448 694 272 × 2 = 0 + 0.897 388 544;
  • 27) 0.897 388 544 × 2 = 1 + 0.794 777 088;
  • 28) 0.794 777 088 × 2 = 1 + 0.589 554 176;
  • 29) 0.589 554 176 × 2 = 1 + 0.179 108 352;
  • 30) 0.179 108 352 × 2 = 0 + 0.358 216 704;
  • 31) 0.358 216 704 × 2 = 0 + 0.716 433 408;
  • 32) 0.716 433 408 × 2 = 1 + 0.432 866 816;
  • 33) 0.432 866 816 × 2 = 0 + 0.865 733 632;
  • 34) 0.865 733 632 × 2 = 1 + 0.731 467 264;
  • 35) 0.731 467 264 × 2 = 1 + 0.462 934 528;
  • 36) 0.462 934 528 × 2 = 0 + 0.925 869 056;
  • 37) 0.925 869 056 × 2 = 1 + 0.851 738 112;
  • 38) 0.851 738 112 × 2 = 1 + 0.703 476 224;
  • 39) 0.703 476 224 × 2 = 1 + 0.406 952 448;
  • 40) 0.406 952 448 × 2 = 0 + 0.813 904 896;
  • 41) 0.813 904 896 × 2 = 1 + 0.627 809 792;
  • 42) 0.627 809 792 × 2 = 1 + 0.255 619 584;
  • 43) 0.255 619 584 × 2 = 0 + 0.511 239 168;
  • 44) 0.511 239 168 × 2 = 1 + 0.022 478 336;
  • 45) 0.022 478 336 × 2 = 0 + 0.044 956 672;

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:


0.000 000 371(10) =

0.0000 0000 0000 0000 0000 0110 0011 1001 0110 1110 1101 0(2)

5. Positive number before normalization:

0.000 000 371(10) =

0.0000 0000 0000 0000 0000 0110 0011 1001 0110 1110 1101 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:


0.000 000 371(10) =

0.0000 0000 0000 0000 0000 0110 0011 1001 0110 1110 1101 0(2) =

0.0000 0000 0000 0000 0000 0110 0011 1001 0110 1110 1101 0(2) × 20 =

1.1000 1110 0101 1011 1011 010(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):
1.1000 1110 0101 1011 1011 010


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. 100 0111 0010 1101 1101 1010 =

100 0111 0010 1101 1101 1010


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) =
100 0111 0010 1101 1101 1010


Decimal number 0.000 000 371 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 100 0111 0010 1101 1101 1010

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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