32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.172 718 255 844 579 9 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.172 718 255 844 579 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: 0.172 718 255 844 579 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) 0.172 718 255 844 579 9 × 2 = 0 + 0.345 436 511 689 159 8;
  • 2) 0.345 436 511 689 159 8 × 2 = 0 + 0.690 873 023 378 319 6;
  • 3) 0.690 873 023 378 319 6 × 2 = 1 + 0.381 746 046 756 639 2;
  • 4) 0.381 746 046 756 639 2 × 2 = 0 + 0.763 492 093 513 278 4;
  • 5) 0.763 492 093 513 278 4 × 2 = 1 + 0.526 984 187 026 556 8;
  • 6) 0.526 984 187 026 556 8 × 2 = 1 + 0.053 968 374 053 113 6;
  • 7) 0.053 968 374 053 113 6 × 2 = 0 + 0.107 936 748 106 227 2;
  • 8) 0.107 936 748 106 227 2 × 2 = 0 + 0.215 873 496 212 454 4;
  • 9) 0.215 873 496 212 454 4 × 2 = 0 + 0.431 746 992 424 908 8;
  • 10) 0.431 746 992 424 908 8 × 2 = 0 + 0.863 493 984 849 817 6;
  • 11) 0.863 493 984 849 817 6 × 2 = 1 + 0.726 987 969 699 635 2;
  • 12) 0.726 987 969 699 635 2 × 2 = 1 + 0.453 975 939 399 270 4;
  • 13) 0.453 975 939 399 270 4 × 2 = 0 + 0.907 951 878 798 540 8;
  • 14) 0.907 951 878 798 540 8 × 2 = 1 + 0.815 903 757 597 081 6;
  • 15) 0.815 903 757 597 081 6 × 2 = 1 + 0.631 807 515 194 163 2;
  • 16) 0.631 807 515 194 163 2 × 2 = 1 + 0.263 615 030 388 326 4;
  • 17) 0.263 615 030 388 326 4 × 2 = 0 + 0.527 230 060 776 652 8;
  • 18) 0.527 230 060 776 652 8 × 2 = 1 + 0.054 460 121 553 305 6;
  • 19) 0.054 460 121 553 305 6 × 2 = 0 + 0.108 920 243 106 611 2;
  • 20) 0.108 920 243 106 611 2 × 2 = 0 + 0.217 840 486 213 222 4;
  • 21) 0.217 840 486 213 222 4 × 2 = 0 + 0.435 680 972 426 444 8;
  • 22) 0.435 680 972 426 444 8 × 2 = 0 + 0.871 361 944 852 889 6;
  • 23) 0.871 361 944 852 889 6 × 2 = 1 + 0.742 723 889 705 779 2;
  • 24) 0.742 723 889 705 779 2 × 2 = 1 + 0.485 447 779 411 558 4;
  • 25) 0.485 447 779 411 558 4 × 2 = 0 + 0.970 895 558 823 116 8;
  • 26) 0.970 895 558 823 116 8 × 2 = 1 + 0.941 791 117 646 233 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:


0.172 718 255 844 579 9(10) =

0.0010 1100 0011 0111 0100 0011 01(2)


5. Positive number before normalization:

0.172 718 255 844 579 9(10) =

0.0010 1100 0011 0111 0100 0011 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.172 718 255 844 579 9(10) =

0.0010 1100 0011 0111 0100 0011 01(2) =

0.0010 1100 0011 0111 0100 0011 01(2) × 20 =

1.0110 0001 1011 1010 0001 101(2) × 2-3


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): -3

Mantissa (not normalized):
1.0110 0001 1011 1010 0001 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(8-1) - 1 =

(-3 + 127)(10) =

124(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;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

124(10) =

0111 1100(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. 011 0000 1101 1101 0000 1101 =

011 0000 1101 1101 0000 1101


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 1100

Mantissa (23 bits) =
011 0000 1101 1101 0000 1101


The base ten decimal number 0.172 718 255 844 579 9 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 1100 - 011 0000 1101 1101 0000 1101

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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