0.000 000 110 011 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 011 9(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
0.000 000 110 011 9(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: 0.000 000 110 011 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) 0.000 000 110 011 9 × 2 = 0 + 0.000 000 220 023 8;
  • 2) 0.000 000 220 023 8 × 2 = 0 + 0.000 000 440 047 6;
  • 3) 0.000 000 440 047 6 × 2 = 0 + 0.000 000 880 095 2;
  • 4) 0.000 000 880 095 2 × 2 = 0 + 0.000 001 760 190 4;
  • 5) 0.000 001 760 190 4 × 2 = 0 + 0.000 003 520 380 8;
  • 6) 0.000 003 520 380 8 × 2 = 0 + 0.000 007 040 761 6;
  • 7) 0.000 007 040 761 6 × 2 = 0 + 0.000 014 081 523 2;
  • 8) 0.000 014 081 523 2 × 2 = 0 + 0.000 028 163 046 4;
  • 9) 0.000 028 163 046 4 × 2 = 0 + 0.000 056 326 092 8;
  • 10) 0.000 056 326 092 8 × 2 = 0 + 0.000 112 652 185 6;
  • 11) 0.000 112 652 185 6 × 2 = 0 + 0.000 225 304 371 2;
  • 12) 0.000 225 304 371 2 × 2 = 0 + 0.000 450 608 742 4;
  • 13) 0.000 450 608 742 4 × 2 = 0 + 0.000 901 217 484 8;
  • 14) 0.000 901 217 484 8 × 2 = 0 + 0.001 802 434 969 6;
  • 15) 0.001 802 434 969 6 × 2 = 0 + 0.003 604 869 939 2;
  • 16) 0.003 604 869 939 2 × 2 = 0 + 0.007 209 739 878 4;
  • 17) 0.007 209 739 878 4 × 2 = 0 + 0.014 419 479 756 8;
  • 18) 0.014 419 479 756 8 × 2 = 0 + 0.028 838 959 513 6;
  • 19) 0.028 838 959 513 6 × 2 = 0 + 0.057 677 919 027 2;
  • 20) 0.057 677 919 027 2 × 2 = 0 + 0.115 355 838 054 4;
  • 21) 0.115 355 838 054 4 × 2 = 0 + 0.230 711 676 108 8;
  • 22) 0.230 711 676 108 8 × 2 = 0 + 0.461 423 352 217 6;
  • 23) 0.461 423 352 217 6 × 2 = 0 + 0.922 846 704 435 2;
  • 24) 0.922 846 704 435 2 × 2 = 1 + 0.845 693 408 870 4;
  • 25) 0.845 693 408 870 4 × 2 = 1 + 0.691 386 817 740 8;
  • 26) 0.691 386 817 740 8 × 2 = 1 + 0.382 773 635 481 6;
  • 27) 0.382 773 635 481 6 × 2 = 0 + 0.765 547 270 963 2;
  • 28) 0.765 547 270 963 2 × 2 = 1 + 0.531 094 541 926 4;
  • 29) 0.531 094 541 926 4 × 2 = 1 + 0.062 189 083 852 8;
  • 30) 0.062 189 083 852 8 × 2 = 0 + 0.124 378 167 705 6;
  • 31) 0.124 378 167 705 6 × 2 = 0 + 0.248 756 335 411 2;
  • 32) 0.248 756 335 411 2 × 2 = 0 + 0.497 512 670 822 4;
  • 33) 0.497 512 670 822 4 × 2 = 0 + 0.995 025 341 644 8;
  • 34) 0.995 025 341 644 8 × 2 = 1 + 0.990 050 683 289 6;
  • 35) 0.990 050 683 289 6 × 2 = 1 + 0.980 101 366 579 2;
  • 36) 0.980 101 366 579 2 × 2 = 1 + 0.960 202 733 158 4;
  • 37) 0.960 202 733 158 4 × 2 = 1 + 0.920 405 466 316 8;
  • 38) 0.920 405 466 316 8 × 2 = 1 + 0.840 810 932 633 6;
  • 39) 0.840 810 932 633 6 × 2 = 1 + 0.681 621 865 267 2;
  • 40) 0.681 621 865 267 2 × 2 = 1 + 0.363 243 730 534 4;
  • 41) 0.363 243 730 534 4 × 2 = 0 + 0.726 487 461 068 8;
  • 42) 0.726 487 461 068 8 × 2 = 1 + 0.452 974 922 137 6;
  • 43) 0.452 974 922 137 6 × 2 = 0 + 0.905 949 844 275 2;
  • 44) 0.905 949 844 275 2 × 2 = 1 + 0.811 899 688 550 4;
  • 45) 0.811 899 688 550 4 × 2 = 1 + 0.623 799 377 100 8;
  • 46) 0.623 799 377 100 8 × 2 = 1 + 0.247 598 754 201 6;
  • 47) 0.247 598 754 201 6 × 2 = 0 + 0.495 197 508 403 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 - 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:


0.000 000 110 011 9(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1111 0101 110(2)

5. Positive number before normalization:

0.000 000 110 011 9(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1111 0101 110(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:


0.000 000 110 011 9(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1111 0101 110(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1111 0101 110(2) × 20 =

1.1101 1000 0111 1111 0101 110(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):
1.1101 1000 0111 1111 0101 110


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. 110 1100 0011 1111 1010 1110 =

110 1100 0011 1111 1010 1110


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) =
110 1100 0011 1111 1010 1110


Decimal number 0.000 000 110 011 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1111 1010 1110

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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