0.000 000 000 47 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 47(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
0.000 000 000 47(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: 0.000 000 000 47.

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) 0.000 000 000 47 × 2 = 0 + 0.000 000 000 94;
  • 2) 0.000 000 000 94 × 2 = 0 + 0.000 000 001 88;
  • 3) 0.000 000 001 88 × 2 = 0 + 0.000 000 003 76;
  • 4) 0.000 000 003 76 × 2 = 0 + 0.000 000 007 52;
  • 5) 0.000 000 007 52 × 2 = 0 + 0.000 000 015 04;
  • 6) 0.000 000 015 04 × 2 = 0 + 0.000 000 030 08;
  • 7) 0.000 000 030 08 × 2 = 0 + 0.000 000 060 16;
  • 8) 0.000 000 060 16 × 2 = 0 + 0.000 000 120 32;
  • 9) 0.000 000 120 32 × 2 = 0 + 0.000 000 240 64;
  • 10) 0.000 000 240 64 × 2 = 0 + 0.000 000 481 28;
  • 11) 0.000 000 481 28 × 2 = 0 + 0.000 000 962 56;
  • 12) 0.000 000 962 56 × 2 = 0 + 0.000 001 925 12;
  • 13) 0.000 001 925 12 × 2 = 0 + 0.000 003 850 24;
  • 14) 0.000 003 850 24 × 2 = 0 + 0.000 007 700 48;
  • 15) 0.000 007 700 48 × 2 = 0 + 0.000 015 400 96;
  • 16) 0.000 015 400 96 × 2 = 0 + 0.000 030 801 92;
  • 17) 0.000 030 801 92 × 2 = 0 + 0.000 061 603 84;
  • 18) 0.000 061 603 84 × 2 = 0 + 0.000 123 207 68;
  • 19) 0.000 123 207 68 × 2 = 0 + 0.000 246 415 36;
  • 20) 0.000 246 415 36 × 2 = 0 + 0.000 492 830 72;
  • 21) 0.000 492 830 72 × 2 = 0 + 0.000 985 661 44;
  • 22) 0.000 985 661 44 × 2 = 0 + 0.001 971 322 88;
  • 23) 0.001 971 322 88 × 2 = 0 + 0.003 942 645 76;
  • 24) 0.003 942 645 76 × 2 = 0 + 0.007 885 291 52;
  • 25) 0.007 885 291 52 × 2 = 0 + 0.015 770 583 04;
  • 26) 0.015 770 583 04 × 2 = 0 + 0.031 541 166 08;
  • 27) 0.031 541 166 08 × 2 = 0 + 0.063 082 332 16;
  • 28) 0.063 082 332 16 × 2 = 0 + 0.126 164 664 32;
  • 29) 0.126 164 664 32 × 2 = 0 + 0.252 329 328 64;
  • 30) 0.252 329 328 64 × 2 = 0 + 0.504 658 657 28;
  • 31) 0.504 658 657 28 × 2 = 1 + 0.009 317 314 56;
  • 32) 0.009 317 314 56 × 2 = 0 + 0.018 634 629 12;
  • 33) 0.018 634 629 12 × 2 = 0 + 0.037 269 258 24;
  • 34) 0.037 269 258 24 × 2 = 0 + 0.074 538 516 48;
  • 35) 0.074 538 516 48 × 2 = 0 + 0.149 077 032 96;
  • 36) 0.149 077 032 96 × 2 = 0 + 0.298 154 065 92;
  • 37) 0.298 154 065 92 × 2 = 0 + 0.596 308 131 84;
  • 38) 0.596 308 131 84 × 2 = 1 + 0.192 616 263 68;
  • 39) 0.192 616 263 68 × 2 = 0 + 0.385 232 527 36;
  • 40) 0.385 232 527 36 × 2 = 0 + 0.770 465 054 72;
  • 41) 0.770 465 054 72 × 2 = 1 + 0.540 930 109 44;
  • 42) 0.540 930 109 44 × 2 = 1 + 0.081 860 218 88;
  • 43) 0.081 860 218 88 × 2 = 0 + 0.163 720 437 76;
  • 44) 0.163 720 437 76 × 2 = 0 + 0.327 440 875 52;
  • 45) 0.327 440 875 52 × 2 = 0 + 0.654 881 751 04;
  • 46) 0.654 881 751 04 × 2 = 1 + 0.309 763 502 08;
  • 47) 0.309 763 502 08 × 2 = 0 + 0.619 527 004 16;
  • 48) 0.619 527 004 16 × 2 = 1 + 0.239 054 008 32;
  • 49) 0.239 054 008 32 × 2 = 0 + 0.478 108 016 64;
  • 50) 0.478 108 016 64 × 2 = 0 + 0.956 216 033 28;
  • 51) 0.956 216 033 28 × 2 = 1 + 0.912 432 066 56;
  • 52) 0.912 432 066 56 × 2 = 1 + 0.824 864 133 12;
  • 53) 0.824 864 133 12 × 2 = 1 + 0.649 728 266 24;
  • 54) 0.649 728 266 24 × 2 = 1 + 0.299 456 532 48;

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:


0.000 000 000 47(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0000 0100 1100 0101 0011 11(2)

5. Positive number before normalization:

0.000 000 000 47(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0000 0100 1100 0101 0011 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 47(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0000 0100 1100 0101 0011 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0000 0100 1100 0101 0011 11(2) × 20 =

1.0000 0010 0110 0010 1001 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.0000 0010 0110 0010 1001 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

96(10) =

0110 0000(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. 000 0001 0011 0001 0100 1111 =

000 0001 0011 0001 0100 1111


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 0000

Mantissa (23 bits) =
000 0001 0011 0001 0100 1111


Decimal number 0.000 000 000 47 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0000 - 000 0001 0011 0001 0100 1111

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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