32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.001 037 597 3 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.001 037 597 3(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: 0.001 037 597 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) 0.001 037 597 3 × 2 = 0 + 0.002 075 194 6;
  • 2) 0.002 075 194 6 × 2 = 0 + 0.004 150 389 2;
  • 3) 0.004 150 389 2 × 2 = 0 + 0.008 300 778 4;
  • 4) 0.008 300 778 4 × 2 = 0 + 0.016 601 556 8;
  • 5) 0.016 601 556 8 × 2 = 0 + 0.033 203 113 6;
  • 6) 0.033 203 113 6 × 2 = 0 + 0.066 406 227 2;
  • 7) 0.066 406 227 2 × 2 = 0 + 0.132 812 454 4;
  • 8) 0.132 812 454 4 × 2 = 0 + 0.265 624 908 8;
  • 9) 0.265 624 908 8 × 2 = 0 + 0.531 249 817 6;
  • 10) 0.531 249 817 6 × 2 = 1 + 0.062 499 635 2;
  • 11) 0.062 499 635 2 × 2 = 0 + 0.124 999 270 4;
  • 12) 0.124 999 270 4 × 2 = 0 + 0.249 998 540 8;
  • 13) 0.249 998 540 8 × 2 = 0 + 0.499 997 081 6;
  • 14) 0.499 997 081 6 × 2 = 0 + 0.999 994 163 2;
  • 15) 0.999 994 163 2 × 2 = 1 + 0.999 988 326 4;
  • 16) 0.999 988 326 4 × 2 = 1 + 0.999 976 652 8;
  • 17) 0.999 976 652 8 × 2 = 1 + 0.999 953 305 6;
  • 18) 0.999 953 305 6 × 2 = 1 + 0.999 906 611 2;
  • 19) 0.999 906 611 2 × 2 = 1 + 0.999 813 222 4;
  • 20) 0.999 813 222 4 × 2 = 1 + 0.999 626 444 8;
  • 21) 0.999 626 444 8 × 2 = 1 + 0.999 252 889 6;
  • 22) 0.999 252 889 6 × 2 = 1 + 0.998 505 779 2;
  • 23) 0.998 505 779 2 × 2 = 1 + 0.997 011 558 4;
  • 24) 0.997 011 558 4 × 2 = 1 + 0.994 023 116 8;
  • 25) 0.994 023 116 8 × 2 = 1 + 0.988 046 233 6;
  • 26) 0.988 046 233 6 × 2 = 1 + 0.976 092 467 2;
  • 27) 0.976 092 467 2 × 2 = 1 + 0.952 184 934 4;
  • 28) 0.952 184 934 4 × 2 = 1 + 0.904 369 868 8;
  • 29) 0.904 369 868 8 × 2 = 1 + 0.808 739 737 6;
  • 30) 0.808 739 737 6 × 2 = 1 + 0.617 479 475 2;
  • 31) 0.617 479 475 2 × 2 = 1 + 0.234 958 950 4;
  • 32) 0.234 958 950 4 × 2 = 0 + 0.469 917 900 8;
  • 33) 0.469 917 900 8 × 2 = 0 + 0.939 835 801 6;

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:


0.001 037 597 3(10) =

0.0000 0000 0100 0011 1111 1111 1111 1110 0(2)


5. Positive number before normalization:

0.001 037 597 3(10) =

0.0000 0000 0100 0011 1111 1111 1111 1110 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 037 597 3(10) =

0.0000 0000 0100 0011 1111 1111 1111 1110 0(2) =

0.0000 0000 0100 0011 1111 1111 1111 1110 0(2) × 20 =

1.0000 1111 1111 1111 1111 100(2) × 2-10


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): -10

Mantissa (not normalized):
1.0000 1111 1111 1111 1111 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-10 + 2(8-1) - 1 =

(-10 + 127)(10) =

117(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;
  • 117 ÷ 2 = 58 + 1;
  • 58 ÷ 2 = 29 + 0;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 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) =

117(10) =

0111 0101(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 0111 1111 1111 1111 1100 =

000 0111 1111 1111 1111 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 0101

Mantissa (23 bits) =
000 0111 1111 1111 1111 1100


The base ten decimal number 0.001 037 597 3 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 0101 - 000 0111 1111 1111 1111 1100

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

» Number 0.001 037 597 2 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number 0.001 037 597 4 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

The latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

Number 357.16 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:08 UTC (GMT)
Number 5 505 027 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:08 UTC (GMT)
Number 0.870 954 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:08 UTC (GMT)
Number 1 010 010 001 099 999 999 999 999 999 914 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:07 UTC (GMT)
Number -129.38 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:07 UTC (GMT)
Number 824.032 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:07 UTC (GMT)
Number 134 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:07 UTC (GMT)
Number 2 147 484 390 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:07 UTC (GMT)
Number 134 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:07 UTC (GMT)
Number 101 100 038 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 01 07:07 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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