0.000 000 247 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 247(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
0.000 000 247(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: 0.000 000 247.

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) 0.000 000 247 × 2 = 0 + 0.000 000 494;
  • 2) 0.000 000 494 × 2 = 0 + 0.000 000 988;
  • 3) 0.000 000 988 × 2 = 0 + 0.000 001 976;
  • 4) 0.000 001 976 × 2 = 0 + 0.000 003 952;
  • 5) 0.000 003 952 × 2 = 0 + 0.000 007 904;
  • 6) 0.000 007 904 × 2 = 0 + 0.000 015 808;
  • 7) 0.000 015 808 × 2 = 0 + 0.000 031 616;
  • 8) 0.000 031 616 × 2 = 0 + 0.000 063 232;
  • 9) 0.000 063 232 × 2 = 0 + 0.000 126 464;
  • 10) 0.000 126 464 × 2 = 0 + 0.000 252 928;
  • 11) 0.000 252 928 × 2 = 0 + 0.000 505 856;
  • 12) 0.000 505 856 × 2 = 0 + 0.001 011 712;
  • 13) 0.001 011 712 × 2 = 0 + 0.002 023 424;
  • 14) 0.002 023 424 × 2 = 0 + 0.004 046 848;
  • 15) 0.004 046 848 × 2 = 0 + 0.008 093 696;
  • 16) 0.008 093 696 × 2 = 0 + 0.016 187 392;
  • 17) 0.016 187 392 × 2 = 0 + 0.032 374 784;
  • 18) 0.032 374 784 × 2 = 0 + 0.064 749 568;
  • 19) 0.064 749 568 × 2 = 0 + 0.129 499 136;
  • 20) 0.129 499 136 × 2 = 0 + 0.258 998 272;
  • 21) 0.258 998 272 × 2 = 0 + 0.517 996 544;
  • 22) 0.517 996 544 × 2 = 1 + 0.035 993 088;
  • 23) 0.035 993 088 × 2 = 0 + 0.071 986 176;
  • 24) 0.071 986 176 × 2 = 0 + 0.143 972 352;
  • 25) 0.143 972 352 × 2 = 0 + 0.287 944 704;
  • 26) 0.287 944 704 × 2 = 0 + 0.575 889 408;
  • 27) 0.575 889 408 × 2 = 1 + 0.151 778 816;
  • 28) 0.151 778 816 × 2 = 0 + 0.303 557 632;
  • 29) 0.303 557 632 × 2 = 0 + 0.607 115 264;
  • 30) 0.607 115 264 × 2 = 1 + 0.214 230 528;
  • 31) 0.214 230 528 × 2 = 0 + 0.428 461 056;
  • 32) 0.428 461 056 × 2 = 0 + 0.856 922 112;
  • 33) 0.856 922 112 × 2 = 1 + 0.713 844 224;
  • 34) 0.713 844 224 × 2 = 1 + 0.427 688 448;
  • 35) 0.427 688 448 × 2 = 0 + 0.855 376 896;
  • 36) 0.855 376 896 × 2 = 1 + 0.710 753 792;
  • 37) 0.710 753 792 × 2 = 1 + 0.421 507 584;
  • 38) 0.421 507 584 × 2 = 0 + 0.843 015 168;
  • 39) 0.843 015 168 × 2 = 1 + 0.686 030 336;
  • 40) 0.686 030 336 × 2 = 1 + 0.372 060 672;
  • 41) 0.372 060 672 × 2 = 0 + 0.744 121 344;
  • 42) 0.744 121 344 × 2 = 1 + 0.488 242 688;
  • 43) 0.488 242 688 × 2 = 0 + 0.976 485 376;
  • 44) 0.976 485 376 × 2 = 1 + 0.952 970 752;
  • 45) 0.952 970 752 × 2 = 1 + 0.905 941 504;

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:


0.000 000 247(10) =

0.0000 0000 0000 0000 0000 0100 0010 0100 1101 1011 0101 1(2)

5. Positive number before normalization:

0.000 000 247(10) =

0.0000 0000 0000 0000 0000 0100 0010 0100 1101 1011 0101 1(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:


0.000 000 247(10) =

0.0000 0000 0000 0000 0000 0100 0010 0100 1101 1011 0101 1(2) =

0.0000 0000 0000 0000 0000 0100 0010 0100 1101 1011 0101 1(2) × 20 =

1.0000 1001 0011 0110 1101 011(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):
1.0000 1001 0011 0110 1101 011


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. 000 0100 1001 1011 0110 1011 =

000 0100 1001 1011 0110 1011


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) =
000 0100 1001 1011 0110 1011


Decimal number 0.000 000 247 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 000 0100 1001 1011 0110 1011

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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