0.000 000 491 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 491(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
0.000 000 491(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: 0.000 000 491.

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) 0.000 000 491 × 2 = 0 + 0.000 000 982;
  • 2) 0.000 000 982 × 2 = 0 + 0.000 001 964;
  • 3) 0.000 001 964 × 2 = 0 + 0.000 003 928;
  • 4) 0.000 003 928 × 2 = 0 + 0.000 007 856;
  • 5) 0.000 007 856 × 2 = 0 + 0.000 015 712;
  • 6) 0.000 015 712 × 2 = 0 + 0.000 031 424;
  • 7) 0.000 031 424 × 2 = 0 + 0.000 062 848;
  • 8) 0.000 062 848 × 2 = 0 + 0.000 125 696;
  • 9) 0.000 125 696 × 2 = 0 + 0.000 251 392;
  • 10) 0.000 251 392 × 2 = 0 + 0.000 502 784;
  • 11) 0.000 502 784 × 2 = 0 + 0.001 005 568;
  • 12) 0.001 005 568 × 2 = 0 + 0.002 011 136;
  • 13) 0.002 011 136 × 2 = 0 + 0.004 022 272;
  • 14) 0.004 022 272 × 2 = 0 + 0.008 044 544;
  • 15) 0.008 044 544 × 2 = 0 + 0.016 089 088;
  • 16) 0.016 089 088 × 2 = 0 + 0.032 178 176;
  • 17) 0.032 178 176 × 2 = 0 + 0.064 356 352;
  • 18) 0.064 356 352 × 2 = 0 + 0.128 712 704;
  • 19) 0.128 712 704 × 2 = 0 + 0.257 425 408;
  • 20) 0.257 425 408 × 2 = 0 + 0.514 850 816;
  • 21) 0.514 850 816 × 2 = 1 + 0.029 701 632;
  • 22) 0.029 701 632 × 2 = 0 + 0.059 403 264;
  • 23) 0.059 403 264 × 2 = 0 + 0.118 806 528;
  • 24) 0.118 806 528 × 2 = 0 + 0.237 613 056;
  • 25) 0.237 613 056 × 2 = 0 + 0.475 226 112;
  • 26) 0.475 226 112 × 2 = 0 + 0.950 452 224;
  • 27) 0.950 452 224 × 2 = 1 + 0.900 904 448;
  • 28) 0.900 904 448 × 2 = 1 + 0.801 808 896;
  • 29) 0.801 808 896 × 2 = 1 + 0.603 617 792;
  • 30) 0.603 617 792 × 2 = 1 + 0.207 235 584;
  • 31) 0.207 235 584 × 2 = 0 + 0.414 471 168;
  • 32) 0.414 471 168 × 2 = 0 + 0.828 942 336;
  • 33) 0.828 942 336 × 2 = 1 + 0.657 884 672;
  • 34) 0.657 884 672 × 2 = 1 + 0.315 769 344;
  • 35) 0.315 769 344 × 2 = 0 + 0.631 538 688;
  • 36) 0.631 538 688 × 2 = 1 + 0.263 077 376;
  • 37) 0.263 077 376 × 2 = 0 + 0.526 154 752;
  • 38) 0.526 154 752 × 2 = 1 + 0.052 309 504;
  • 39) 0.052 309 504 × 2 = 0 + 0.104 619 008;
  • 40) 0.104 619 008 × 2 = 0 + 0.209 238 016;
  • 41) 0.209 238 016 × 2 = 0 + 0.418 476 032;
  • 42) 0.418 476 032 × 2 = 0 + 0.836 952 064;
  • 43) 0.836 952 064 × 2 = 1 + 0.673 904 128;
  • 44) 0.673 904 128 × 2 = 1 + 0.347 808 256;

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:


0.000 000 491(10) =

0.0000 0000 0000 0000 0000 1000 0011 1100 1101 0100 0011(2)

5. Positive number before normalization:

0.000 000 491(10) =

0.0000 0000 0000 0000 0000 1000 0011 1100 1101 0100 0011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 491(10) =

0.0000 0000 0000 0000 0000 1000 0011 1100 1101 0100 0011(2) =

0.0000 0000 0000 0000 0000 1000 0011 1100 1101 0100 0011(2) × 20 =

1.0000 0111 1001 1010 1000 011(2) × 2-21


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): -21

Mantissa (not normalized):
1.0000 0111 1001 1010 1000 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 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) =

106(10) =

0110 1010(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 0011 1100 1101 0100 0011 =

000 0011 1100 1101 0100 0011


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 1010

Mantissa (23 bits) =
000 0011 1100 1101 0100 0011


Decimal number 0.000 000 491 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 000 0011 1100 1101 0100 0011

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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