0.000 000 213 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 213(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
0.000 000 213(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: 0.000 000 213.

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) 0.000 000 213 × 2 = 0 + 0.000 000 426;
  • 2) 0.000 000 426 × 2 = 0 + 0.000 000 852;
  • 3) 0.000 000 852 × 2 = 0 + 0.000 001 704;
  • 4) 0.000 001 704 × 2 = 0 + 0.000 003 408;
  • 5) 0.000 003 408 × 2 = 0 + 0.000 006 816;
  • 6) 0.000 006 816 × 2 = 0 + 0.000 013 632;
  • 7) 0.000 013 632 × 2 = 0 + 0.000 027 264;
  • 8) 0.000 027 264 × 2 = 0 + 0.000 054 528;
  • 9) 0.000 054 528 × 2 = 0 + 0.000 109 056;
  • 10) 0.000 109 056 × 2 = 0 + 0.000 218 112;
  • 11) 0.000 218 112 × 2 = 0 + 0.000 436 224;
  • 12) 0.000 436 224 × 2 = 0 + 0.000 872 448;
  • 13) 0.000 872 448 × 2 = 0 + 0.001 744 896;
  • 14) 0.001 744 896 × 2 = 0 + 0.003 489 792;
  • 15) 0.003 489 792 × 2 = 0 + 0.006 979 584;
  • 16) 0.006 979 584 × 2 = 0 + 0.013 959 168;
  • 17) 0.013 959 168 × 2 = 0 + 0.027 918 336;
  • 18) 0.027 918 336 × 2 = 0 + 0.055 836 672;
  • 19) 0.055 836 672 × 2 = 0 + 0.111 673 344;
  • 20) 0.111 673 344 × 2 = 0 + 0.223 346 688;
  • 21) 0.223 346 688 × 2 = 0 + 0.446 693 376;
  • 22) 0.446 693 376 × 2 = 0 + 0.893 386 752;
  • 23) 0.893 386 752 × 2 = 1 + 0.786 773 504;
  • 24) 0.786 773 504 × 2 = 1 + 0.573 547 008;
  • 25) 0.573 547 008 × 2 = 1 + 0.147 094 016;
  • 26) 0.147 094 016 × 2 = 0 + 0.294 188 032;
  • 27) 0.294 188 032 × 2 = 0 + 0.588 376 064;
  • 28) 0.588 376 064 × 2 = 1 + 0.176 752 128;
  • 29) 0.176 752 128 × 2 = 0 + 0.353 504 256;
  • 30) 0.353 504 256 × 2 = 0 + 0.707 008 512;
  • 31) 0.707 008 512 × 2 = 1 + 0.414 017 024;
  • 32) 0.414 017 024 × 2 = 0 + 0.828 034 048;
  • 33) 0.828 034 048 × 2 = 1 + 0.656 068 096;
  • 34) 0.656 068 096 × 2 = 1 + 0.312 136 192;
  • 35) 0.312 136 192 × 2 = 0 + 0.624 272 384;
  • 36) 0.624 272 384 × 2 = 1 + 0.248 544 768;
  • 37) 0.248 544 768 × 2 = 0 + 0.497 089 536;
  • 38) 0.497 089 536 × 2 = 0 + 0.994 179 072;
  • 39) 0.994 179 072 × 2 = 1 + 0.988 358 144;
  • 40) 0.988 358 144 × 2 = 1 + 0.976 716 288;
  • 41) 0.976 716 288 × 2 = 1 + 0.953 432 576;
  • 42) 0.953 432 576 × 2 = 1 + 0.906 865 152;
  • 43) 0.906 865 152 × 2 = 1 + 0.813 730 304;
  • 44) 0.813 730 304 × 2 = 1 + 0.627 460 608;
  • 45) 0.627 460 608 × 2 = 1 + 0.254 921 216;
  • 46) 0.254 921 216 × 2 = 0 + 0.509 842 432;

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:


0.000 000 213(10) =

0.0000 0000 0000 0000 0000 0011 1001 0010 1101 0011 1111 10(2)

5. Positive number before normalization:

0.000 000 213(10) =

0.0000 0000 0000 0000 0000 0011 1001 0010 1101 0011 1111 10(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:


0.000 000 213(10) =

0.0000 0000 0000 0000 0000 0011 1001 0010 1101 0011 1111 10(2) =

0.0000 0000 0000 0000 0000 0011 1001 0010 1101 0011 1111 10(2) × 20 =

1.1100 1001 0110 1001 1111 110(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):
1.1100 1001 0110 1001 1111 110


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. 110 0100 1011 0100 1111 1110 =

110 0100 1011 0100 1111 1110


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) =
110 0100 1011 0100 1111 1110


Decimal number 0.000 000 213 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 110 0100 1011 0100 1111 1110

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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