0.000 000 261 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 261(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
0.000 000 261(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: 0.000 000 261.

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) 0.000 000 261 × 2 = 0 + 0.000 000 522;
  • 2) 0.000 000 522 × 2 = 0 + 0.000 001 044;
  • 3) 0.000 001 044 × 2 = 0 + 0.000 002 088;
  • 4) 0.000 002 088 × 2 = 0 + 0.000 004 176;
  • 5) 0.000 004 176 × 2 = 0 + 0.000 008 352;
  • 6) 0.000 008 352 × 2 = 0 + 0.000 016 704;
  • 7) 0.000 016 704 × 2 = 0 + 0.000 033 408;
  • 8) 0.000 033 408 × 2 = 0 + 0.000 066 816;
  • 9) 0.000 066 816 × 2 = 0 + 0.000 133 632;
  • 10) 0.000 133 632 × 2 = 0 + 0.000 267 264;
  • 11) 0.000 267 264 × 2 = 0 + 0.000 534 528;
  • 12) 0.000 534 528 × 2 = 0 + 0.001 069 056;
  • 13) 0.001 069 056 × 2 = 0 + 0.002 138 112;
  • 14) 0.002 138 112 × 2 = 0 + 0.004 276 224;
  • 15) 0.004 276 224 × 2 = 0 + 0.008 552 448;
  • 16) 0.008 552 448 × 2 = 0 + 0.017 104 896;
  • 17) 0.017 104 896 × 2 = 0 + 0.034 209 792;
  • 18) 0.034 209 792 × 2 = 0 + 0.068 419 584;
  • 19) 0.068 419 584 × 2 = 0 + 0.136 839 168;
  • 20) 0.136 839 168 × 2 = 0 + 0.273 678 336;
  • 21) 0.273 678 336 × 2 = 0 + 0.547 356 672;
  • 22) 0.547 356 672 × 2 = 1 + 0.094 713 344;
  • 23) 0.094 713 344 × 2 = 0 + 0.189 426 688;
  • 24) 0.189 426 688 × 2 = 0 + 0.378 853 376;
  • 25) 0.378 853 376 × 2 = 0 + 0.757 706 752;
  • 26) 0.757 706 752 × 2 = 1 + 0.515 413 504;
  • 27) 0.515 413 504 × 2 = 1 + 0.030 827 008;
  • 28) 0.030 827 008 × 2 = 0 + 0.061 654 016;
  • 29) 0.061 654 016 × 2 = 0 + 0.123 308 032;
  • 30) 0.123 308 032 × 2 = 0 + 0.246 616 064;
  • 31) 0.246 616 064 × 2 = 0 + 0.493 232 128;
  • 32) 0.493 232 128 × 2 = 0 + 0.986 464 256;
  • 33) 0.986 464 256 × 2 = 1 + 0.972 928 512;
  • 34) 0.972 928 512 × 2 = 1 + 0.945 857 024;
  • 35) 0.945 857 024 × 2 = 1 + 0.891 714 048;
  • 36) 0.891 714 048 × 2 = 1 + 0.783 428 096;
  • 37) 0.783 428 096 × 2 = 1 + 0.566 856 192;
  • 38) 0.566 856 192 × 2 = 1 + 0.133 712 384;
  • 39) 0.133 712 384 × 2 = 0 + 0.267 424 768;
  • 40) 0.267 424 768 × 2 = 0 + 0.534 849 536;
  • 41) 0.534 849 536 × 2 = 1 + 0.069 699 072;
  • 42) 0.069 699 072 × 2 = 0 + 0.139 398 144;
  • 43) 0.139 398 144 × 2 = 0 + 0.278 796 288;
  • 44) 0.278 796 288 × 2 = 0 + 0.557 592 576;
  • 45) 0.557 592 576 × 2 = 1 + 0.115 185 152;

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:


0.000 000 261(10) =

0.0000 0000 0000 0000 0000 0100 0110 0000 1111 1100 1000 1(2)

5. Positive number before normalization:

0.000 000 261(10) =

0.0000 0000 0000 0000 0000 0100 0110 0000 1111 1100 1000 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:


0.000 000 261(10) =

0.0000 0000 0000 0000 0000 0100 0110 0000 1111 1100 1000 1(2) =

0.0000 0000 0000 0000 0000 0100 0110 0000 1111 1100 1000 1(2) × 20 =

1.0001 1000 0011 1111 0010 001(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):
1.0001 1000 0011 1111 0010 001


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 1100 0001 1111 1001 0001 =

000 1100 0001 1111 1001 0001


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 1100 0001 1111 1001 0001


Decimal number 0.000 000 261 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 000 1100 0001 1111 1001 0001

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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