32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 111 579 895 019 531 4 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 111 579 895 019 531 4(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: 0.000 111 579 895 019 531 4.

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) 0.000 111 579 895 019 531 4 × 2 = 0 + 0.000 223 159 790 039 062 8;
  • 2) 0.000 223 159 790 039 062 8 × 2 = 0 + 0.000 446 319 580 078 125 6;
  • 3) 0.000 446 319 580 078 125 6 × 2 = 0 + 0.000 892 639 160 156 251 2;
  • 4) 0.000 892 639 160 156 251 2 × 2 = 0 + 0.001 785 278 320 312 502 4;
  • 5) 0.001 785 278 320 312 502 4 × 2 = 0 + 0.003 570 556 640 625 004 8;
  • 6) 0.003 570 556 640 625 004 8 × 2 = 0 + 0.007 141 113 281 250 009 6;
  • 7) 0.007 141 113 281 250 009 6 × 2 = 0 + 0.014 282 226 562 500 019 2;
  • 8) 0.014 282 226 562 500 019 2 × 2 = 0 + 0.028 564 453 125 000 038 4;
  • 9) 0.028 564 453 125 000 038 4 × 2 = 0 + 0.057 128 906 250 000 076 8;
  • 10) 0.057 128 906 250 000 076 8 × 2 = 0 + 0.114 257 812 500 000 153 6;
  • 11) 0.114 257 812 500 000 153 6 × 2 = 0 + 0.228 515 625 000 000 307 2;
  • 12) 0.228 515 625 000 000 307 2 × 2 = 0 + 0.457 031 250 000 000 614 4;
  • 13) 0.457 031 250 000 000 614 4 × 2 = 0 + 0.914 062 500 000 001 228 8;
  • 14) 0.914 062 500 000 001 228 8 × 2 = 1 + 0.828 125 000 000 002 457 6;
  • 15) 0.828 125 000 000 002 457 6 × 2 = 1 + 0.656 250 000 000 004 915 2;
  • 16) 0.656 250 000 000 004 915 2 × 2 = 1 + 0.312 500 000 000 009 830 4;
  • 17) 0.312 500 000 000 009 830 4 × 2 = 0 + 0.625 000 000 000 019 660 8;
  • 18) 0.625 000 000 000 019 660 8 × 2 = 1 + 0.250 000 000 000 039 321 6;
  • 19) 0.250 000 000 000 039 321 6 × 2 = 0 + 0.500 000 000 000 078 643 2;
  • 20) 0.500 000 000 000 078 643 2 × 2 = 1 + 0.000 000 000 000 157 286 4;
  • 21) 0.000 000 000 000 157 286 4 × 2 = 0 + 0.000 000 000 000 314 572 8;
  • 22) 0.000 000 000 000 314 572 8 × 2 = 0 + 0.000 000 000 000 629 145 6;
  • 23) 0.000 000 000 000 629 145 6 × 2 = 0 + 0.000 000 000 001 258 291 2;
  • 24) 0.000 000 000 001 258 291 2 × 2 = 0 + 0.000 000 000 002 516 582 4;
  • 25) 0.000 000 000 002 516 582 4 × 2 = 0 + 0.000 000 000 005 033 164 8;
  • 26) 0.000 000 000 005 033 164 8 × 2 = 0 + 0.000 000 000 010 066 329 6;
  • 27) 0.000 000 000 010 066 329 6 × 2 = 0 + 0.000 000 000 020 132 659 2;
  • 28) 0.000 000 000 020 132 659 2 × 2 = 0 + 0.000 000 000 040 265 318 4;
  • 29) 0.000 000 000 040 265 318 4 × 2 = 0 + 0.000 000 000 080 530 636 8;
  • 30) 0.000 000 000 080 530 636 8 × 2 = 0 + 0.000 000 000 161 061 273 6;
  • 31) 0.000 000 000 161 061 273 6 × 2 = 0 + 0.000 000 000 322 122 547 2;
  • 32) 0.000 000 000 322 122 547 2 × 2 = 0 + 0.000 000 000 644 245 094 4;
  • 33) 0.000 000 000 644 245 094 4 × 2 = 0 + 0.000 000 001 288 490 188 8;
  • 34) 0.000 000 001 288 490 188 8 × 2 = 0 + 0.000 000 002 576 980 377 6;
  • 35) 0.000 000 002 576 980 377 6 × 2 = 0 + 0.000 000 005 153 960 755 2;
  • 36) 0.000 000 005 153 960 755 2 × 2 = 0 + 0.000 000 010 307 921 510 4;
  • 37) 0.000 000 010 307 921 510 4 × 2 = 0 + 0.000 000 020 615 843 020 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...)


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:


0.000 111 579 895 019 531 4(10) =

0.0000 0000 0000 0111 0101 0000 0000 0000 0000 0(2)


5. Positive number before normalization:

0.000 111 579 895 019 531 4(10) =

0.0000 0000 0000 0111 0101 0000 0000 0000 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 111 579 895 019 531 4(10) =

0.0000 0000 0000 0111 0101 0000 0000 0000 0000 0(2) =

0.0000 0000 0000 0111 0101 0000 0000 0000 0000 0(2) × 20 =

1.1101 0100 0000 0000 0000 000(2) × 2-14


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): -14

Mantissa (not normalized):
1.1101 0100 0000 0000 0000 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-14 + 2(8-1) - 1 =

(-14 + 127)(10) =

113(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;
  • 113 ÷ 2 = 56 + 1;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 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) =

113(10) =

0111 0001(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. 110 1010 0000 0000 0000 0000 =

110 1010 0000 0000 0000 0000


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 0001

Mantissa (23 bits) =
110 1010 0000 0000 0000 0000


The base ten decimal number 0.000 111 579 895 019 531 4 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 0001 - 110 1010 0000 0000 0000 0000

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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