0.099 999 994 039 535 563 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.099 999 994 039 535 563 1(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
0.099 999 994 039 535 563 1(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: 0.099 999 994 039 535 563 1.

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) 0.099 999 994 039 535 563 1 × 2 = 0 + 0.199 999 988 079 071 126 2;
  • 2) 0.199 999 988 079 071 126 2 × 2 = 0 + 0.399 999 976 158 142 252 4;
  • 3) 0.399 999 976 158 142 252 4 × 2 = 0 + 0.799 999 952 316 284 504 8;
  • 4) 0.799 999 952 316 284 504 8 × 2 = 1 + 0.599 999 904 632 569 009 6;
  • 5) 0.599 999 904 632 569 009 6 × 2 = 1 + 0.199 999 809 265 138 019 2;
  • 6) 0.199 999 809 265 138 019 2 × 2 = 0 + 0.399 999 618 530 276 038 4;
  • 7) 0.399 999 618 530 276 038 4 × 2 = 0 + 0.799 999 237 060 552 076 8;
  • 8) 0.799 999 237 060 552 076 8 × 2 = 1 + 0.599 998 474 121 104 153 6;
  • 9) 0.599 998 474 121 104 153 6 × 2 = 1 + 0.199 996 948 242 208 307 2;
  • 10) 0.199 996 948 242 208 307 2 × 2 = 0 + 0.399 993 896 484 416 614 4;
  • 11) 0.399 993 896 484 416 614 4 × 2 = 0 + 0.799 987 792 968 833 228 8;
  • 12) 0.799 987 792 968 833 228 8 × 2 = 1 + 0.599 975 585 937 666 457 6;
  • 13) 0.599 975 585 937 666 457 6 × 2 = 1 + 0.199 951 171 875 332 915 2;
  • 14) 0.199 951 171 875 332 915 2 × 2 = 0 + 0.399 902 343 750 665 830 4;
  • 15) 0.399 902 343 750 665 830 4 × 2 = 0 + 0.799 804 687 501 331 660 8;
  • 16) 0.799 804 687 501 331 660 8 × 2 = 1 + 0.599 609 375 002 663 321 6;
  • 17) 0.599 609 375 002 663 321 6 × 2 = 1 + 0.199 218 750 005 326 643 2;
  • 18) 0.199 218 750 005 326 643 2 × 2 = 0 + 0.398 437 500 010 653 286 4;
  • 19) 0.398 437 500 010 653 286 4 × 2 = 0 + 0.796 875 000 021 306 572 8;
  • 20) 0.796 875 000 021 306 572 8 × 2 = 1 + 0.593 750 000 042 613 145 6;
  • 21) 0.593 750 000 042 613 145 6 × 2 = 1 + 0.187 500 000 085 226 291 2;
  • 22) 0.187 500 000 085 226 291 2 × 2 = 0 + 0.375 000 000 170 452 582 4;
  • 23) 0.375 000 000 170 452 582 4 × 2 = 0 + 0.750 000 000 340 905 164 8;
  • 24) 0.750 000 000 340 905 164 8 × 2 = 1 + 0.500 000 000 681 810 329 6;
  • 25) 0.500 000 000 681 810 329 6 × 2 = 1 + 0.000 000 001 363 620 659 2;
  • 26) 0.000 000 001 363 620 659 2 × 2 = 0 + 0.000 000 002 727 241 318 4;
  • 27) 0.000 000 002 727 241 318 4 × 2 = 0 + 0.000 000 005 454 482 636 8;

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:


0.099 999 994 039 535 563 1(10) =

0.0001 1001 1001 1001 1001 1001 100(2)

5. Positive number before normalization:

0.099 999 994 039 535 563 1(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:


0.099 999 994 039 535 563 1(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 0.099 999 994 039 535 563 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1011 - 100 1100 1100 1100 1100 1100

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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