0.099 999 994 039 535 553 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.099 999 994 039 535 553 3(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
0.099 999 994 039 535 553 3(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: 0.099 999 994 039 535 553 3.

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) 0.099 999 994 039 535 553 3 × 2 = 0 + 0.199 999 988 079 071 106 6;
  • 2) 0.199 999 988 079 071 106 6 × 2 = 0 + 0.399 999 976 158 142 213 2;
  • 3) 0.399 999 976 158 142 213 2 × 2 = 0 + 0.799 999 952 316 284 426 4;
  • 4) 0.799 999 952 316 284 426 4 × 2 = 1 + 0.599 999 904 632 568 852 8;
  • 5) 0.599 999 904 632 568 852 8 × 2 = 1 + 0.199 999 809 265 137 705 6;
  • 6) 0.199 999 809 265 137 705 6 × 2 = 0 + 0.399 999 618 530 275 411 2;
  • 7) 0.399 999 618 530 275 411 2 × 2 = 0 + 0.799 999 237 060 550 822 4;
  • 8) 0.799 999 237 060 550 822 4 × 2 = 1 + 0.599 998 474 121 101 644 8;
  • 9) 0.599 998 474 121 101 644 8 × 2 = 1 + 0.199 996 948 242 203 289 6;
  • 10) 0.199 996 948 242 203 289 6 × 2 = 0 + 0.399 993 896 484 406 579 2;
  • 11) 0.399 993 896 484 406 579 2 × 2 = 0 + 0.799 987 792 968 813 158 4;
  • 12) 0.799 987 792 968 813 158 4 × 2 = 1 + 0.599 975 585 937 626 316 8;
  • 13) 0.599 975 585 937 626 316 8 × 2 = 1 + 0.199 951 171 875 252 633 6;
  • 14) 0.199 951 171 875 252 633 6 × 2 = 0 + 0.399 902 343 750 505 267 2;
  • 15) 0.399 902 343 750 505 267 2 × 2 = 0 + 0.799 804 687 501 010 534 4;
  • 16) 0.799 804 687 501 010 534 4 × 2 = 1 + 0.599 609 375 002 021 068 8;
  • 17) 0.599 609 375 002 021 068 8 × 2 = 1 + 0.199 218 750 004 042 137 6;
  • 18) 0.199 218 750 004 042 137 6 × 2 = 0 + 0.398 437 500 008 084 275 2;
  • 19) 0.398 437 500 008 084 275 2 × 2 = 0 + 0.796 875 000 016 168 550 4;
  • 20) 0.796 875 000 016 168 550 4 × 2 = 1 + 0.593 750 000 032 337 100 8;
  • 21) 0.593 750 000 032 337 100 8 × 2 = 1 + 0.187 500 000 064 674 201 6;
  • 22) 0.187 500 000 064 674 201 6 × 2 = 0 + 0.375 000 000 129 348 403 2;
  • 23) 0.375 000 000 129 348 403 2 × 2 = 0 + 0.750 000 000 258 696 806 4;
  • 24) 0.750 000 000 258 696 806 4 × 2 = 1 + 0.500 000 000 517 393 612 8;
  • 25) 0.500 000 000 517 393 612 8 × 2 = 1 + 0.000 000 001 034 787 225 6;
  • 26) 0.000 000 001 034 787 225 6 × 2 = 0 + 0.000 000 002 069 574 451 2;
  • 27) 0.000 000 002 069 574 451 2 × 2 = 0 + 0.000 000 004 139 148 902 4;

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:


0.099 999 994 039 535 553 3(10) =

0.0001 1001 1001 1001 1001 1001 100(2)

5. Positive number before normalization:

0.099 999 994 039 535 553 3(10) =

0.0001 1001 1001 1001 1001 1001 100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.099 999 994 039 535 553 3(10) =

0.0001 1001 1001 1001 1001 1001 100(2) =

0.0001 1001 1001 1001 1001 1001 100(2) × 20 =

1.1001 1001 1001 1001 1001 100(2) × 2-4


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): -4

Mantissa (not normalized):
1.1001 1001 1001 1001 1001 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(8-1) - 1 =

(-4 + 127)(10) =

123(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;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

123(10) =

0111 1011(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. 100 1100 1100 1100 1100 1100 =

100 1100 1100 1100 1100 1100


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) =
0111 1011

Mantissa (23 bits) =
100 1100 1100 1100 1100 1100


Decimal number 0.099 999 994 039 535 553 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1011 - 100 1100 1100 1100 1100 1100

Share

» Convert decimal 0.099 999 994 039 535 558 3 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 minutes

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