32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 000 101 123 9 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 000 101 123 9(10) converted and written in 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: 0.000 000 101 123 9.

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) 0.000 000 101 123 9 × 2 = 0 + 0.000 000 202 247 8;
  • 2) 0.000 000 202 247 8 × 2 = 0 + 0.000 000 404 495 6;
  • 3) 0.000 000 404 495 6 × 2 = 0 + 0.000 000 808 991 2;
  • 4) 0.000 000 808 991 2 × 2 = 0 + 0.000 001 617 982 4;
  • 5) 0.000 001 617 982 4 × 2 = 0 + 0.000 003 235 964 8;
  • 6) 0.000 003 235 964 8 × 2 = 0 + 0.000 006 471 929 6;
  • 7) 0.000 006 471 929 6 × 2 = 0 + 0.000 012 943 859 2;
  • 8) 0.000 012 943 859 2 × 2 = 0 + 0.000 025 887 718 4;
  • 9) 0.000 025 887 718 4 × 2 = 0 + 0.000 051 775 436 8;
  • 10) 0.000 051 775 436 8 × 2 = 0 + 0.000 103 550 873 6;
  • 11) 0.000 103 550 873 6 × 2 = 0 + 0.000 207 101 747 2;
  • 12) 0.000 207 101 747 2 × 2 = 0 + 0.000 414 203 494 4;
  • 13) 0.000 414 203 494 4 × 2 = 0 + 0.000 828 406 988 8;
  • 14) 0.000 828 406 988 8 × 2 = 0 + 0.001 656 813 977 6;
  • 15) 0.001 656 813 977 6 × 2 = 0 + 0.003 313 627 955 2;
  • 16) 0.003 313 627 955 2 × 2 = 0 + 0.006 627 255 910 4;
  • 17) 0.006 627 255 910 4 × 2 = 0 + 0.013 254 511 820 8;
  • 18) 0.013 254 511 820 8 × 2 = 0 + 0.026 509 023 641 6;
  • 19) 0.026 509 023 641 6 × 2 = 0 + 0.053 018 047 283 2;
  • 20) 0.053 018 047 283 2 × 2 = 0 + 0.106 036 094 566 4;
  • 21) 0.106 036 094 566 4 × 2 = 0 + 0.212 072 189 132 8;
  • 22) 0.212 072 189 132 8 × 2 = 0 + 0.424 144 378 265 6;
  • 23) 0.424 144 378 265 6 × 2 = 0 + 0.848 288 756 531 2;
  • 24) 0.848 288 756 531 2 × 2 = 1 + 0.696 577 513 062 4;
  • 25) 0.696 577 513 062 4 × 2 = 1 + 0.393 155 026 124 8;
  • 26) 0.393 155 026 124 8 × 2 = 0 + 0.786 310 052 249 6;
  • 27) 0.786 310 052 249 6 × 2 = 1 + 0.572 620 104 499 2;
  • 28) 0.572 620 104 499 2 × 2 = 1 + 0.145 240 208 998 4;
  • 29) 0.145 240 208 998 4 × 2 = 0 + 0.290 480 417 996 8;
  • 30) 0.290 480 417 996 8 × 2 = 0 + 0.580 960 835 993 6;
  • 31) 0.580 960 835 993 6 × 2 = 1 + 0.161 921 671 987 2;
  • 32) 0.161 921 671 987 2 × 2 = 0 + 0.323 843 343 974 4;
  • 33) 0.323 843 343 974 4 × 2 = 0 + 0.647 686 687 948 8;
  • 34) 0.647 686 687 948 8 × 2 = 1 + 0.295 373 375 897 6;
  • 35) 0.295 373 375 897 6 × 2 = 0 + 0.590 746 751 795 2;
  • 36) 0.590 746 751 795 2 × 2 = 1 + 0.181 493 503 590 4;
  • 37) 0.181 493 503 590 4 × 2 = 0 + 0.362 987 007 180 8;
  • 38) 0.362 987 007 180 8 × 2 = 0 + 0.725 974 014 361 6;
  • 39) 0.725 974 014 361 6 × 2 = 1 + 0.451 948 028 723 2;
  • 40) 0.451 948 028 723 2 × 2 = 0 + 0.903 896 057 446 4;
  • 41) 0.903 896 057 446 4 × 2 = 1 + 0.807 792 114 892 8;
  • 42) 0.807 792 114 892 8 × 2 = 1 + 0.615 584 229 785 6;
  • 43) 0.615 584 229 785 6 × 2 = 1 + 0.231 168 459 571 2;
  • 44) 0.231 168 459 571 2 × 2 = 0 + 0.462 336 919 142 4;
  • 45) 0.462 336 919 142 4 × 2 = 0 + 0.924 673 838 284 8;
  • 46) 0.924 673 838 284 8 × 2 = 1 + 0.849 347 676 569 6;
  • 47) 0.849 347 676 569 6 × 2 = 1 + 0.698 695 353 139 2;

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...)


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:


0.000 000 101 123 9(10) =

0.0000 0000 0000 0000 0000 0001 1011 0010 0101 0010 1110 011(2)


5. Positive number before normalization:

0.000 000 101 123 9(10) =

0.0000 0000 0000 0000 0000 0001 1011 0010 0101 0010 1110 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 101 123 9(10) =

0.0000 0000 0000 0000 0000 0001 1011 0010 0101 0010 1110 011(2) =

0.0000 0000 0000 0000 0000 0001 1011 0010 0101 0010 1110 011(2) × 20 =

1.1011 0010 0101 0010 1110 011(2) × 2-24


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): -24

Mantissa (not normalized):
1.1011 0010 0101 0010 1110 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

103(10) =

0110 0111(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. 101 1001 0010 1001 0111 0011 =

101 1001 0010 1001 0111 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 0111

Mantissa (23 bits) =
101 1001 0010 1001 0111 0011


The base ten decimal number 0.000 000 101 123 9 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0110 0111 - 101 1001 0010 1001 0111 0011

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

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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