0.333 333 333 333 333 314 829 616 256 247 61 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 314 829 616 256 247 61(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
0.333 333 333 333 333 314 829 616 256 247 61(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: 0.333 333 333 333 333 314 829 616 256 247 61.

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) 0.333 333 333 333 333 314 829 616 256 247 61 × 2 = 0 + 0.666 666 666 666 666 629 659 232 512 495 22;
  • 2) 0.666 666 666 666 666 629 659 232 512 495 22 × 2 = 1 + 0.333 333 333 333 333 259 318 465 024 990 44;
  • 3) 0.333 333 333 333 333 259 318 465 024 990 44 × 2 = 0 + 0.666 666 666 666 666 518 636 930 049 980 88;
  • 4) 0.666 666 666 666 666 518 636 930 049 980 88 × 2 = 1 + 0.333 333 333 333 333 037 273 860 099 961 76;
  • 5) 0.333 333 333 333 333 037 273 860 099 961 76 × 2 = 0 + 0.666 666 666 666 666 074 547 720 199 923 52;
  • 6) 0.666 666 666 666 666 074 547 720 199 923 52 × 2 = 1 + 0.333 333 333 333 332 149 095 440 399 847 04;
  • 7) 0.333 333 333 333 332 149 095 440 399 847 04 × 2 = 0 + 0.666 666 666 666 664 298 190 880 799 694 08;
  • 8) 0.666 666 666 666 664 298 190 880 799 694 08 × 2 = 1 + 0.333 333 333 333 328 596 381 761 599 388 16;
  • 9) 0.333 333 333 333 328 596 381 761 599 388 16 × 2 = 0 + 0.666 666 666 666 657 192 763 523 198 776 32;
  • 10) 0.666 666 666 666 657 192 763 523 198 776 32 × 2 = 1 + 0.333 333 333 333 314 385 527 046 397 552 64;
  • 11) 0.333 333 333 333 314 385 527 046 397 552 64 × 2 = 0 + 0.666 666 666 666 628 771 054 092 795 105 28;
  • 12) 0.666 666 666 666 628 771 054 092 795 105 28 × 2 = 1 + 0.333 333 333 333 257 542 108 185 590 210 56;
  • 13) 0.333 333 333 333 257 542 108 185 590 210 56 × 2 = 0 + 0.666 666 666 666 515 084 216 371 180 421 12;
  • 14) 0.666 666 666 666 515 084 216 371 180 421 12 × 2 = 1 + 0.333 333 333 333 030 168 432 742 360 842 24;
  • 15) 0.333 333 333 333 030 168 432 742 360 842 24 × 2 = 0 + 0.666 666 666 666 060 336 865 484 721 684 48;
  • 16) 0.666 666 666 666 060 336 865 484 721 684 48 × 2 = 1 + 0.333 333 333 332 120 673 730 969 443 368 96;
  • 17) 0.333 333 333 332 120 673 730 969 443 368 96 × 2 = 0 + 0.666 666 666 664 241 347 461 938 886 737 92;
  • 18) 0.666 666 666 664 241 347 461 938 886 737 92 × 2 = 1 + 0.333 333 333 328 482 694 923 877 773 475 84;
  • 19) 0.333 333 333 328 482 694 923 877 773 475 84 × 2 = 0 + 0.666 666 666 656 965 389 847 755 546 951 68;
  • 20) 0.666 666 666 656 965 389 847 755 546 951 68 × 2 = 1 + 0.333 333 333 313 930 779 695 511 093 903 36;
  • 21) 0.333 333 333 313 930 779 695 511 093 903 36 × 2 = 0 + 0.666 666 666 627 861 559 391 022 187 806 72;
  • 22) 0.666 666 666 627 861 559 391 022 187 806 72 × 2 = 1 + 0.333 333 333 255 723 118 782 044 375 613 44;
  • 23) 0.333 333 333 255 723 118 782 044 375 613 44 × 2 = 0 + 0.666 666 666 511 446 237 564 088 751 226 88;
  • 24) 0.666 666 666 511 446 237 564 088 751 226 88 × 2 = 1 + 0.333 333 333 022 892 475 128 177 502 453 76;
  • 25) 0.333 333 333 022 892 475 128 177 502 453 76 × 2 = 0 + 0.666 666 666 045 784 950 256 355 004 907 52;

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:


0.333 333 333 333 333 314 829 616 256 247 61(10) =

0.0101 0101 0101 0101 0101 0101 0(2)

5. Positive number before normalization:

0.333 333 333 333 333 314 829 616 256 247 61(10) =

0.0101 0101 0101 0101 0101 0101 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.333 333 333 333 333 314 829 616 256 247 61(10) =

0.0101 0101 0101 0101 0101 0101 0(2) =

0.0101 0101 0101 0101 0101 0101 0(2) × 20 =

1.0101 0101 0101 0101 0101 010(2) × 2-2


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): -2

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 127)(10) =

125(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;
  • 125 ÷ 2 = 62 + 1;
  • 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) =

125(10) =

0111 1101(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. 010 1010 1010 1010 1010 1010 =

010 1010 1010 1010 1010 1010


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 1101

Mantissa (23 bits) =
010 1010 1010 1010 1010 1010


Decimal number 0.333 333 333 333 333 314 829 616 256 247 61 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1101 - 010 1010 1010 1010 1010 1010

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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