0.000 000 212 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 212(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
0.000 000 212(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: 0.000 000 212.

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) 0.000 000 212 × 2 = 0 + 0.000 000 424;
  • 2) 0.000 000 424 × 2 = 0 + 0.000 000 848;
  • 3) 0.000 000 848 × 2 = 0 + 0.000 001 696;
  • 4) 0.000 001 696 × 2 = 0 + 0.000 003 392;
  • 5) 0.000 003 392 × 2 = 0 + 0.000 006 784;
  • 6) 0.000 006 784 × 2 = 0 + 0.000 013 568;
  • 7) 0.000 013 568 × 2 = 0 + 0.000 027 136;
  • 8) 0.000 027 136 × 2 = 0 + 0.000 054 272;
  • 9) 0.000 054 272 × 2 = 0 + 0.000 108 544;
  • 10) 0.000 108 544 × 2 = 0 + 0.000 217 088;
  • 11) 0.000 217 088 × 2 = 0 + 0.000 434 176;
  • 12) 0.000 434 176 × 2 = 0 + 0.000 868 352;
  • 13) 0.000 868 352 × 2 = 0 + 0.001 736 704;
  • 14) 0.001 736 704 × 2 = 0 + 0.003 473 408;
  • 15) 0.003 473 408 × 2 = 0 + 0.006 946 816;
  • 16) 0.006 946 816 × 2 = 0 + 0.013 893 632;
  • 17) 0.013 893 632 × 2 = 0 + 0.027 787 264;
  • 18) 0.027 787 264 × 2 = 0 + 0.055 574 528;
  • 19) 0.055 574 528 × 2 = 0 + 0.111 149 056;
  • 20) 0.111 149 056 × 2 = 0 + 0.222 298 112;
  • 21) 0.222 298 112 × 2 = 0 + 0.444 596 224;
  • 22) 0.444 596 224 × 2 = 0 + 0.889 192 448;
  • 23) 0.889 192 448 × 2 = 1 + 0.778 384 896;
  • 24) 0.778 384 896 × 2 = 1 + 0.556 769 792;
  • 25) 0.556 769 792 × 2 = 1 + 0.113 539 584;
  • 26) 0.113 539 584 × 2 = 0 + 0.227 079 168;
  • 27) 0.227 079 168 × 2 = 0 + 0.454 158 336;
  • 28) 0.454 158 336 × 2 = 0 + 0.908 316 672;
  • 29) 0.908 316 672 × 2 = 1 + 0.816 633 344;
  • 30) 0.816 633 344 × 2 = 1 + 0.633 266 688;
  • 31) 0.633 266 688 × 2 = 1 + 0.266 533 376;
  • 32) 0.266 533 376 × 2 = 0 + 0.533 066 752;
  • 33) 0.533 066 752 × 2 = 1 + 0.066 133 504;
  • 34) 0.066 133 504 × 2 = 0 + 0.132 267 008;
  • 35) 0.132 267 008 × 2 = 0 + 0.264 534 016;
  • 36) 0.264 534 016 × 2 = 0 + 0.529 068 032;
  • 37) 0.529 068 032 × 2 = 1 + 0.058 136 064;
  • 38) 0.058 136 064 × 2 = 0 + 0.116 272 128;
  • 39) 0.116 272 128 × 2 = 0 + 0.232 544 256;
  • 40) 0.232 544 256 × 2 = 0 + 0.465 088 512;
  • 41) 0.465 088 512 × 2 = 0 + 0.930 177 024;
  • 42) 0.930 177 024 × 2 = 1 + 0.860 354 048;
  • 43) 0.860 354 048 × 2 = 1 + 0.720 708 096;
  • 44) 0.720 708 096 × 2 = 1 + 0.441 416 192;
  • 45) 0.441 416 192 × 2 = 0 + 0.882 832 384;
  • 46) 0.882 832 384 × 2 = 1 + 0.765 664 768;

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:


0.000 000 212(10) =

0.0000 0000 0000 0000 0000 0011 1000 1110 1000 1000 0111 01(2)

5. Positive number before normalization:

0.000 000 212(10) =

0.0000 0000 0000 0000 0000 0011 1000 1110 1000 1000 0111 01(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:


0.000 000 212(10) =

0.0000 0000 0000 0000 0000 0011 1000 1110 1000 1000 0111 01(2) =

0.0000 0000 0000 0000 0000 0011 1000 1110 1000 1000 0111 01(2) × 20 =

1.1100 0111 0100 0100 0011 101(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 0111 0100 0100 0011 101


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 0011 1010 0010 0001 1101 =

110 0011 1010 0010 0001 1101


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 0011 1010 0010 0001 1101


Decimal number 0.000 000 212 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 110 0011 1010 0010 0001 1101

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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