0.000 000 355 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 355(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
0.000 000 355(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: 0.000 000 355.

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) 0.000 000 355 × 2 = 0 + 0.000 000 71;
  • 2) 0.000 000 71 × 2 = 0 + 0.000 001 42;
  • 3) 0.000 001 42 × 2 = 0 + 0.000 002 84;
  • 4) 0.000 002 84 × 2 = 0 + 0.000 005 68;
  • 5) 0.000 005 68 × 2 = 0 + 0.000 011 36;
  • 6) 0.000 011 36 × 2 = 0 + 0.000 022 72;
  • 7) 0.000 022 72 × 2 = 0 + 0.000 045 44;
  • 8) 0.000 045 44 × 2 = 0 + 0.000 090 88;
  • 9) 0.000 090 88 × 2 = 0 + 0.000 181 76;
  • 10) 0.000 181 76 × 2 = 0 + 0.000 363 52;
  • 11) 0.000 363 52 × 2 = 0 + 0.000 727 04;
  • 12) 0.000 727 04 × 2 = 0 + 0.001 454 08;
  • 13) 0.001 454 08 × 2 = 0 + 0.002 908 16;
  • 14) 0.002 908 16 × 2 = 0 + 0.005 816 32;
  • 15) 0.005 816 32 × 2 = 0 + 0.011 632 64;
  • 16) 0.011 632 64 × 2 = 0 + 0.023 265 28;
  • 17) 0.023 265 28 × 2 = 0 + 0.046 530 56;
  • 18) 0.046 530 56 × 2 = 0 + 0.093 061 12;
  • 19) 0.093 061 12 × 2 = 0 + 0.186 122 24;
  • 20) 0.186 122 24 × 2 = 0 + 0.372 244 48;
  • 21) 0.372 244 48 × 2 = 0 + 0.744 488 96;
  • 22) 0.744 488 96 × 2 = 1 + 0.488 977 92;
  • 23) 0.488 977 92 × 2 = 0 + 0.977 955 84;
  • 24) 0.977 955 84 × 2 = 1 + 0.955 911 68;
  • 25) 0.955 911 68 × 2 = 1 + 0.911 823 36;
  • 26) 0.911 823 36 × 2 = 1 + 0.823 646 72;
  • 27) 0.823 646 72 × 2 = 1 + 0.647 293 44;
  • 28) 0.647 293 44 × 2 = 1 + 0.294 586 88;
  • 29) 0.294 586 88 × 2 = 0 + 0.589 173 76;
  • 30) 0.589 173 76 × 2 = 1 + 0.178 347 52;
  • 31) 0.178 347 52 × 2 = 0 + 0.356 695 04;
  • 32) 0.356 695 04 × 2 = 0 + 0.713 390 08;
  • 33) 0.713 390 08 × 2 = 1 + 0.426 780 16;
  • 34) 0.426 780 16 × 2 = 0 + 0.853 560 32;
  • 35) 0.853 560 32 × 2 = 1 + 0.707 120 64;
  • 36) 0.707 120 64 × 2 = 1 + 0.414 241 28;
  • 37) 0.414 241 28 × 2 = 0 + 0.828 482 56;
  • 38) 0.828 482 56 × 2 = 1 + 0.656 965 12;
  • 39) 0.656 965 12 × 2 = 1 + 0.313 930 24;
  • 40) 0.313 930 24 × 2 = 0 + 0.627 860 48;
  • 41) 0.627 860 48 × 2 = 1 + 0.255 720 96;
  • 42) 0.255 720 96 × 2 = 0 + 0.511 441 92;
  • 43) 0.511 441 92 × 2 = 1 + 0.022 883 84;
  • 44) 0.022 883 84 × 2 = 0 + 0.045 767 68;
  • 45) 0.045 767 68 × 2 = 0 + 0.091 535 36;

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:


0.000 000 355(10) =

0.0000 0000 0000 0000 0000 0101 1111 0100 1011 0110 1010 0(2)

5. Positive number before normalization:

0.000 000 355(10) =

0.0000 0000 0000 0000 0000 0101 1111 0100 1011 0110 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:


0.000 000 355(10) =

0.0000 0000 0000 0000 0000 0101 1111 0100 1011 0110 1010 0(2) =

0.0000 0000 0000 0000 0000 0101 1111 0100 1011 0110 1010 0(2) × 20 =

1.0111 1101 0010 1101 1010 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):
1.0111 1101 0010 1101 1010 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. 011 1110 1001 0110 1101 0100 =

011 1110 1001 0110 1101 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) =
011 1110 1001 0110 1101 0100


Decimal number 0.000 000 355 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 011 1110 1001 0110 1101 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