0.000 000 246 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 246(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
0.000 000 246(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: 0.000 000 246.

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) 0.000 000 246 × 2 = 0 + 0.000 000 492;
  • 2) 0.000 000 492 × 2 = 0 + 0.000 000 984;
  • 3) 0.000 000 984 × 2 = 0 + 0.000 001 968;
  • 4) 0.000 001 968 × 2 = 0 + 0.000 003 936;
  • 5) 0.000 003 936 × 2 = 0 + 0.000 007 872;
  • 6) 0.000 007 872 × 2 = 0 + 0.000 015 744;
  • 7) 0.000 015 744 × 2 = 0 + 0.000 031 488;
  • 8) 0.000 031 488 × 2 = 0 + 0.000 062 976;
  • 9) 0.000 062 976 × 2 = 0 + 0.000 125 952;
  • 10) 0.000 125 952 × 2 = 0 + 0.000 251 904;
  • 11) 0.000 251 904 × 2 = 0 + 0.000 503 808;
  • 12) 0.000 503 808 × 2 = 0 + 0.001 007 616;
  • 13) 0.001 007 616 × 2 = 0 + 0.002 015 232;
  • 14) 0.002 015 232 × 2 = 0 + 0.004 030 464;
  • 15) 0.004 030 464 × 2 = 0 + 0.008 060 928;
  • 16) 0.008 060 928 × 2 = 0 + 0.016 121 856;
  • 17) 0.016 121 856 × 2 = 0 + 0.032 243 712;
  • 18) 0.032 243 712 × 2 = 0 + 0.064 487 424;
  • 19) 0.064 487 424 × 2 = 0 + 0.128 974 848;
  • 20) 0.128 974 848 × 2 = 0 + 0.257 949 696;
  • 21) 0.257 949 696 × 2 = 0 + 0.515 899 392;
  • 22) 0.515 899 392 × 2 = 1 + 0.031 798 784;
  • 23) 0.031 798 784 × 2 = 0 + 0.063 597 568;
  • 24) 0.063 597 568 × 2 = 0 + 0.127 195 136;
  • 25) 0.127 195 136 × 2 = 0 + 0.254 390 272;
  • 26) 0.254 390 272 × 2 = 0 + 0.508 780 544;
  • 27) 0.508 780 544 × 2 = 1 + 0.017 561 088;
  • 28) 0.017 561 088 × 2 = 0 + 0.035 122 176;
  • 29) 0.035 122 176 × 2 = 0 + 0.070 244 352;
  • 30) 0.070 244 352 × 2 = 0 + 0.140 488 704;
  • 31) 0.140 488 704 × 2 = 0 + 0.280 977 408;
  • 32) 0.280 977 408 × 2 = 0 + 0.561 954 816;
  • 33) 0.561 954 816 × 2 = 1 + 0.123 909 632;
  • 34) 0.123 909 632 × 2 = 0 + 0.247 819 264;
  • 35) 0.247 819 264 × 2 = 0 + 0.495 638 528;
  • 36) 0.495 638 528 × 2 = 0 + 0.991 277 056;
  • 37) 0.991 277 056 × 2 = 1 + 0.982 554 112;
  • 38) 0.982 554 112 × 2 = 1 + 0.965 108 224;
  • 39) 0.965 108 224 × 2 = 1 + 0.930 216 448;
  • 40) 0.930 216 448 × 2 = 1 + 0.860 432 896;
  • 41) 0.860 432 896 × 2 = 1 + 0.720 865 792;
  • 42) 0.720 865 792 × 2 = 1 + 0.441 731 584;
  • 43) 0.441 731 584 × 2 = 0 + 0.883 463 168;
  • 44) 0.883 463 168 × 2 = 1 + 0.766 926 336;
  • 45) 0.766 926 336 × 2 = 1 + 0.533 852 672;

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:


0.000 000 246(10) =

0.0000 0000 0000 0000 0000 0100 0010 0000 1000 1111 1101 1(2)

5. Positive number before normalization:

0.000 000 246(10) =

0.0000 0000 0000 0000 0000 0100 0010 0000 1000 1111 1101 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:


0.000 000 246(10) =

0.0000 0000 0000 0000 0000 0100 0010 0000 1000 1111 1101 1(2) =

0.0000 0000 0000 0000 0000 0100 0010 0000 1000 1111 1101 1(2) × 20 =

1.0000 1000 0010 0011 1111 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 1000 0010 0011 1111 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 0001 0001 1111 1011 =

000 0100 0001 0001 1111 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 0001 0001 1111 1011


Decimal number 0.000 000 246 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 000 0100 0001 0001 1111 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